Foundation Engineering Science – Dr. Daniel Nankoo ...danny/FENGSC_A.pdf · Foundation...

Foundation Engineering Science – Dr. Daniel Nankoo Tutorial Solutions

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Solutions to Tutorial 1

1. Picture the Problem: You walk in both the

positive and negative directions along a straight line.

Strategy: The distance is the total length of travel, and the displacement is the net change in position.

Solution: (a) Add the lengths: ( ) ( )0.75 0.60 mi 0.60 mi 1.95 mi+ + =

(b) Subtract xi from xf to find the displacement.

0.75 0.00 mi 0.75 mif ix x x∆ = − = − =

Insight: The distance travelled is always positive, but the displacement can be negative.

2. Picture the Problem: Player A walks in the positive direction and player B walks in the negative direction.

Strategy: In each case the distance is the total length of travel, and the displacement is the net change in position.

Solution: (a) Note the distance travelled by player A:

5 m

The displacement of player A is positive: 5 m 0 m 5 mf ix x x∆ = − = − =

(b) Note the distance travelled by player B: 2 m

The displacement of player B is negative. Let the origin be at the initial position of player A.

5 m 7 m 2 mf ix x x∆ = − = − = −

Insight: The distance travelled is always positive, but the displacement can be negative.

3. Picture the Problem: The ball is putted in the

positive direction and then the negative direction.


Solution: (a) Add the lengths: ( )10 2.5 m 2.5 m 15 m+ + =

(b) Subtract xi from xf to find the displacement.

10 0 m 10 mf ix x x∆ = − = − =

Insight: The distance travelled is always positive, but the displacement can be negative. 4. Picture the Problem: You walk in both the

positive and negative directions along a straight line.


Solution: (a) Add the lengths: ( ) ( )0.60 0.35 mi 0.75 0.60 0.35 mi 2.65 mi+ + + + =

(b) Subtract xi from xf to find the 0.75 0.00 mi 0.75 mif ix x x∆ = − = − =


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displacement.

Insight: The distance travelled is always positive, but the displacement can be negative. 5. Picture the Problem: The runner moves along

the oval track.


Solution: 1. (a) Add the lengths: ( ) ( ) ( )15 m 100 m 15 m 130 m+ + =

2. Subtract xi from xf to find the displacement.

100 0 m 100 mf ix x x∆ = − = − =

3. (b) Add the lengths 15 100 30 100 15 m 260 m+ + + + =

4. Subtract xi from xf to find the displacement.

0 0 m 0 mf ix x x∆ = − = − =

Insight: The distance travelled is always positive, but the displacement can be negative. The displacement is always zero for a complete circuit, as in this case.

6. Picture the Problem: The pony walks around the

circular track.


Solution: (a) 1. The distance travelled is half the circumference:

( ) ( )12 2 4.5 m 14 md r rπ π π= = = =

2. The displacement is the distance from A to B: ( )2 2 4.5 m 9.0 mf ix x x r∆ = − = = =

3. (b) The distance travelled will increase when the child completes one circuit, because the pony will have taken more steps.

4. (c) The displacement will decrease when the child completes one circuit, because the displacement is maximum when the child has gone halfway around, and is zero when the child returns to the starting position.

5. (d) The distance travelled equals the circumference:

( )2 2 4.5 m 28 md rπ π= = =

6. The displacement is zero because the child has returned to her starting position.

Insight: The distance travelled is always positive, but the displacement can be negative. The displacement is always zero for a complete circuit, as in this case.

7. Picture the Problem: The kangaroo hops in the forward direction.

Strategy: The distance is the average speed multiplied by the time elapsed. The time elapsed is the distance divided by the average speed.

Solution: 1. (a) Multiply the average speed by the time elapsed:

km 1 h65 3.2 min 3.5 km

h 60 mind s t

= = × =

2. (b) Divide the distance by the average speed:

0.25 km 60 min14 s

65 km/h 1 h

dt

s= = × =

Insight: The instantaneous speed might vary from 65 km/h, but the time elapsed and the distance travelled depend only upon the average speed during the interval in question.

A B

4.5 m


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8. Picture the Problem: The radio waves propagate in a straight line.

Strategy: The time elapsed is the distance divided by the average speed. The distance to the Moon is 2.39×105 mi. We must double this distance because the signal travels there and back again.

Solution: Divide the distance by the average speed:

( )5

5

2 2.39 10 mi22.57 s

1.86 10 mi/s

dt

s

×= = =

×

Insight: The time is slightly shorter than this because the given distance is from the centre of the Earth to the centre of the Moon, but presumably any radio communications would occur between the surfaces of the Earth and Moon. When the radii of the two spheres is taken into account, the time decreases to 2.52 s.

9. Picture the Problem: The sound waves propagate in a straight line from the thunderbolt to your

ears.

Strategy: The distance is the average speed multiplied by the time elapsed. We will neglect the time it takes for the light wave to arrive at your eyes because it is vastly smaller than the time it takes the sound wave to travel.

Solution: Multiply the average speed by the time elapsed:

( )( )340 m/s 3.5 s 1200 m 1.2 kmd s t= = = =

Insight: The speed of sound, 340 m/s, works out to approximately one mile every five seconds, a useful rule of thumb for estimating the distance to an approaching thunderstorm!

10. Picture the Problem: The nerve impulses propagate at a fixed speed.

Strategy: The time elapsed is the distance divided by the average speed. The distance from your finger to your brain is on the order of one meter.

Solution: Divide the distance by the average speed:

2

1 m0.010 s

1 10 m/s

dt

s= = =

×

Insight: This nerve impulse travel time is not the limiting factor for human reaction time, which is about 0.2 s.

11. Picture the Problem: The finch travels a short distance on the back of the tortoise and a longer

distance through the air, with both displacements along the same direction.

Strategy: First find the total distance travelled by the finch and then determine the average speed by dividing by the total time elapsed.

Solution: 1. Determine the total distance travelled: ( )( ) ( )( )

1 1 2 2

0.060 m/s 1.2 min 12 m/s 1.2 min 60 s/min

870 m 0.87 km

d s t s t

d

d

= ∆ + ∆

= + ×

= =

2. Divide the distance by the time elapsed:

870 m6.0 m/s

2.4 min 60 s/min

ds

t= = =

∆ ×

Insight: Most of the distance travelled by the finch occurred by air. In fact, if we neglect the 4.3 m the finch travelled while on the tortoise’s back, we still get an average speed of 6.0 m/s over the 2.4 min time interval! The bird might as well have been at rest.

12. Picture the Problem: You travel 8.0 km on foot and then an additional 16 km by car, with both

displacements along the same direction.

Strategy: First find the total time elapsed by dividing the distance travelled by the average and divide by the total time elapsed to find the average speed. Set that average speed to the given value and solve for the car’s speed.


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Solution: 1. Use the definition of average speed to determine the total time elapsed. av

8.0 16 km1.1 h

22 km/h

dt

s

+∆ = = =

2. Find the time elapsed while in the car: 2 1 1.1 h 0.84 h 0.3 ht t t∆ = ∆ − ∆ = − =

3. Find the speed of the car: 2

22

16 km50 km/h

0.3 h

ds

t= = =

∆

Insight: This problem illustrates the limitations that significant figures occasionally impose. If you keep an extra figure in the total elapsed time (1.09 h) you’ll end up with the time elapsed for the car trip as 0.25 h, not 0.3, and the speed of the car is 64 km/h. But the rules of subtraction indicate we only know the total time to within a tenth of an hour, so we can only know the time spent in the car to within a tenth of an hour, or to within one significant digit.

13. Picture the Problem: You travel in a straight line at two different speeds during the specified time

interval.

Strategy: Determine the average speed by first calculating the total distance travelled and then dividing it by the total time elapsed.

Solution: 1. (a) Because the time intervals are the same, you spend equal times at 20 m/s and 30 m/s, and your average speed will be equal to 25.0 m/s.

2. (b) Divide the total distance by the time elapsed:

( )( ) ( ) ( )1 1 2 2av

1 2

av

20.0 m/s 10.0 min 60 s 30.0 m/s 600 s

600 600 s

25.0 m/s

s t s ts

t t

s

× +∆ + ∆= =

∆ + ∆ +

=

Insight: The average speed is a weighted average according to how much time you spend travelling at each speed.

14. Picture the Problem: You travel in a straight line at two different speeds during the specified time

interval.

Strategy: Determine the average speed by first calculating the total distance travelled and then dividing it by the total time elapsed.

Solution: 1. (a) The distance intervals are the same but the time intervals are different. You will spend more time at the lower speed than at the higher speed. Since the average speed is a time weighted average, it will be less than 25.0 m/s.

2. (b) Divide the total distance by the time elapsed:

1 2 1 2av

1 21 2

1 2

av

20.0 mi 10.0 mi 10.0 mi20.0 m/s 30.0 m/s

24.0 m/s

d d d ds

d dt t

s s

s

+ += = =

∆ + ∆ + +

=

Insight: Notice that in this case it is not necessary to convert miles to meters in both the numerator and denominator because the units cancel out and leave m/s in the numerator.


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15. Picture the Problem: The given position function indicates the particle begins travelling in the

negative direction but is accelerating in the positive direction.

Strategy: Create the x-versus-t plot using a spreadsheet, or calculate individual values by hand and sketch the curve using graph paper. Use the known x and t information to determine the average speed and velocity.

Solution: 1. (a) Use a spreadsheet or similar program to create the plot:

2. (b) Find the average velocity from t = 0 to t = 1.0 s:

( )( ) ( )( ) [ ]22

av

5 m/s 1.0 s 3 m/s 1.0 s 0.0 m

1.0 s

2.0 m/s

xv

t

− + −∆ = =

∆

= −

3. (c) The average speed is the magnitude of the average velocity:

av av 2.0 m/ss v= =

Insight: Note that the average velocity over the first second of time is equal to the slope of a straight line drawn from the origin to the curve at t = 1.0 s. At that time the position is −2.0 m.


positive direction but is accelerating in the negative direction.


Solution: 1. (a) Use a spreadsheet to create the plot shown at right:

2. (b) Find the average velocity from t = 0 to t = 1.0 s:

( )( ) ( )( ) [ ]

av

22

av

6 m/s 1.0 s 2 m/s 1.0 s 0.0 m

1.0 s

4.0 m/s

xv

t

v

∆=

∆

+ − − =

=

3. (c) The average speed is the magnitude of the average velocity:

av av 4.0 m/ss v= =

Insight: Note that the average velocity over the first second of time is equal to the slope of a straight line drawn from the origin to the curve at t = 1.0 s. At that time the position is 4.0 m.


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17. Picture the Problem: Following the motion specified

in the position-versus-time graph, the tennis player moves left, then right, then left again, if we take left to be in the negative direction.

Strategy: Determine the direction of the velocity from the slope of the graph. The speed will be greatest for the segment of the curve that has the largest slope magnitude.

Solution: 1. (a) The magnitude of the slope at B is larger than A or C so we conclude the speed is greatest at B.

2. (b) Find the slope of the graph at A: av

2.0 m1.0 m/s

2.0 s

xs

t

∆ −= = =

∆

3. (c) Find the slope of the graph at B: av

2.0 m2.0 m/s

1.0 s

xs

t

∆= = =

∆

4. (d) Find the slope of the graph at C: av

1.0 m0.50 m/s

2.0 s

xs

t

∆ −= = =

∆

Insight: The speed during segment B is larger than the speed during segments A and C, as predicted. Speeds are always positive because they do not involve direction, but velocities can be negative to indicate their direction.


positive direction but is accelerating in the negative direction.


Solution: 1. (a) Use a spreadsheet to create the plot:

2. (b) Find the average velocity from t = 0.35 to t = 0.45 s:

( )( ) ( )( ) ( )( ) ( )( )3 33 3

av

2 m/s 0.45 s 3 m/s 0.45 s 2 m/s 0.35 s 3 m/s 0.35 s

0.10 s

0.55 m/s

xv

t

− − −∆ = =∆

=

3. (c) Find the average velocity from t = 0.39 to t = 0.41 s:

( )( ) ( )( ) ( )( ) ( )( )3 33 3

av

2 m/s 0.41 s 3 m/s 0.41 s 2 m/s 0.39 s 3 m/s 0.39 s

0.41 0.39 s

0.56 m/s

xv

t

− − −∆ = =

∆ −

=


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4. (d) The instantaneous speed at t = 0.40 s will be closer to 0.56 m/s. As the time interval becomes smaller the average velocity is approaching 0.56 m/s, so we conclude the average speed over an infinitesimally small time interval will be very close to that value.

Insight: Note that the instantaneous velocity at 0.40 s is equal to the slope of a straight line drawn tangent to the curve at that point. Since it is difficult to accurately draw a tangent line, we usually resort to mathematical methods like those illustrated above to determine the instantaneous velocity.


negative direction but is accelerating in the positive direction.


Solution: 1. (a) Use a spreadsheet to create the plot:

2. (b) Find the average velocity from t = 0.150 to t = 0.250 s:

( )( ) ( )( )

( )( ) ( )( )

33

33

av

2 m/s 0.250 s 3 m/s 0.250 s

2 m/s 0.150 s 3 m/s 0.150 s1.63 m/s

0.250 0.150 s

xv

t

− + − − +∆

= = = −∆ −

3. (c) Find the average velocity from t = 0.190 to t = 0.210 s:

( )( ) ( )( )

( )( ) ( )( )

33

33

av

2 m/s 0.210 s 3 m/s 0.210 s

2 m/s 0.190 s 3 m/s 0.190 s1.64 m/s

0.210 0.190 s

xv

t

− + − − +∆

= = = −∆ −

4. (d) The instantaneous speed at t = 0.200 s will be closer to −1.64 m/s. As the time interval becomes smaller the average velocity is approaching −1.64 m/s, so we conclude the average speed over an infinitesimally small time interval will be very close to that value.

Insight: Note that the instantaneous velocity at 0.200 s is equal to the slope of a straight line drawn tangent to the curve at that point. Since it is difficult to accurately draw a tangent line, we usually resort to mathematical methods like those illustrated above to determine the instantaneous velocity.

20. Picture the Problem: The airliner accelerates uniformly along a straight runway.

Strategy: The average acceleration is the change of the velocity divided by the elapsed time.

Solution: Divide the change in velocity by the time:

2av

173 0 mi/h 0.447 m/s2.20 m/s

35.2 s mi/h

va

t

∆ −= = × =

∆

Insight: The instantaneous acceleration might vary from 2.20 m/s2, but we can calculate only average acceleration from the net change in velocity and time elapsed.


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21. Picture the Problem: The airliner slows down uniformly along a straight runway as it travels

towards the east.

Strategy: The average acceleration is the change of the velocity divided by the elapsed time. Assume that east is in the positive direction

Solution: 1. Divide the change in velocity by the time:

2av

0 115 m/s8.85 m/s

13.0 s

va

t

∆ −= = =

∆

2. We note from the previous step that the acceleration is negative. Since east is the positive direction, negative acceleration must be towards the west.

Insight: In physics we almost never talk about deceleration. Instead, we call it negative

acceleration. 22. Picture the Problem: The car travels in a straight line due north, either speeding up or slowing

down, depending upon the direction of the acceleration.

Strategy: Use the definition of acceleration to determine the final velocity over the specified time interval.

Solution: 1. (a) Evaluate v = v0 + at directly:

( )( )20 18.1 m/s 1.30 m/s 7.50 s 27.9 m/s northv v at= + = + =

2. (b) Evaluate v = v0 + at directly:

( )( )20 18.1 m/s 1.15 m/s 7.50 s 9.48 m/s northv v at= + = + − =

Insight: In physics we almost never talk about deceleration. Instead, we call it negative

acceleration. In this problem south is considered the negative direction, and in part (b) the car is slowing down or undergoing negative acceleration.

23. Picture the Problem: Following the motion specified in

the velocity-versus-time graph, the motorcycle is speeding up, then moving at constant speed, then slowing down.

Strategy: Determine the acceleration from the slope of the graph.

Solution: 1. (a) Find the slope at A:

av

2

10 m/s

5.0 s

2.0 m/s

va

t

∆= =

∆

=

2. (b) Find the slope of the graph at B:

2av

0 m/s0.0 m/s

10.0 s

va

t

∆= = =

∆

3. (c) Find the slope of the graph at C:

2av

5.0 m/s0.50 m/s

10.0 s

va

t

∆ −= = = −

∆

Insight: The acceleration during segment A is larger than the acceleration during segments B and C because the slope there has the greatest magnitude.

24. Picture the Problem: The car travels in a straight line in the positive direction while accelerating

in the negative direction (slowing down).

Strategy: Use the constant acceleration equation of motion to determine the time elapsed for the specified change in velocity.

Solution: 1. (a) The time required to come to a stop is the change in velocity divided by the acceleration. In both cases the final velocity is zero, so the change in velocity doubles when you double the initial velocity. Therefore the stopping time will increase by a factor of two when you


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double your driving speed.

2. (b) Solve for time: 0

2

0 16 m/s3.8 s

4.2 m/s

v vt

a

− −= = =

−

3. (c) Solve for time: 0

2

0 32 m/s7.6 s

4.2 m/s

v vt

a

− −= = =

−

Insight: Note that the deceleration is treated as a negative acceleration in this problem and elsewhere in the text.

25. Picture the Problem: The train travels in a straight line in the positive direction while accelerating

in the positive direction (speeding up).

Strategy: First find the acceleration and then determine the final velocity.

Solution: 1. Use the definition of acceleration: 20 4.7 0 m/s

0.94 m/s5.0 s

v va

t

− −= = =

2. Evaluate directly, using the final speed from the first segment as the initial speed of the second segment:

( ) ( )20 4.7 m/s 0.94 m/s 6.0 s

10.3 m/s

v v at

v

= + = +

=

Insight: Another way to tackle this problem is to set up similar triangles on a velocity-versus-time graph. The answer would then be calculated as (4.7 m/s) × 11 s / 5 s = 10.3 m/s. Try it!

26. Picture the Problem: The particle travels in a straight line in the positive direction while

accelerating in the positive direction (speeding up).

Strategy: Use the constant acceleration equation of motion to find the initial velocity.

Solution: Solve for 0v : ( )( )20 9.31 m/s 6.24 m/s 0.300 s 7.44 m/sv v at= − = − =

Insight: As expected the initial velocity is less than the final velocity because the particle is speeding up.

27. Picture the Problem: The car travels in a straight line toward the west while accelerating in the

easterly direction (slowing down).

Strategy: The average velocity is simply half the sum of the initial and final velocities because the acceleration is uniform.

Solution: Calculate half the sum of the velocities:

( ) ( )1 1av 02 2 12 0 m/s 6.0 m/sv v v= + = + =

Insight: The average velocity of any object that slows down and comes to a stop is just half the initial velocity.

28. Picture the Problem: The boat travels in a straight line with constant positive acceleration.

Strategy: The average velocity is simply half the sum of the initial and final velocities because the acceleration is uniform.

Solution: 1. (a) Calculate half the sum of the velocities:

( ) ( )1 1av 02 2 0 4.12 m/s 2.06 m/sv v v= + = + =

2. (b) The distance travelled is the average velocity multiplied by the time elapsed:

( )( )av 2.06 m/s 4.77 s 9.83 md v t= = =

Insight: The average velocity of any object that speeds up from rest is just half the final velocity.


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29. Picture the Problem: The cheetah runs in a straight line with constant positive acceleration.

Strategy: The average velocity is simply half the sum of the initial and final velocities because the acceleration is uniform. The distance travelled is the average velocity multiplied by the time elapsed.

Solution: 1. (a) Calculate half the sum of the velocities:

( ) ( )1 1av 02 2 0 25.0 m/s 12.5 m/sv v v= + = + =

2. Use the average velocity to find the distance:

( )( )av 12.5 m/s 6.22 s 77.8 md v t= = =

3. (b) For a constant acceleration the velocity varies linearly with time. Therefore we expect the velocity to be equal to 12.5 m/s after half the time (3.11 s) has elapsed.

4. (c) Calculate half the sum of the velocities: ( ) ( )1 1av,1 02 2 0 12.5 m/s 6.25 m/sv v v= + = + =

5. Calculate half the sum of the velocities: ( ) ( )1 1av,2 02 2 12.5 25.0 m/s 18.8 m/sv v v= + = + =

6. (d) Use the average velocity to find the distance:

( )( )1 av,1 6.25 m/s 3.11 s 19.4 md v t= = =

7. Use the average velocity to find the distance:

( )( )2 av,2 18.8 m/s 3.11 s 58.5 md v t= = =

Insight: The distance travelled is always the average velocity multiplied by the time. This stems from the definition of average velocity.

30. Picture the Problem: The air bag expands outward with constant positive acceleration.

Strategy: Assume the air bag has a thickness of 1 ft or about 0.3 m. It must expand that distance within the given time of 10 ms. Employ the relationship between acceleration, displacement, and time to find the acceleration.

Solution: Solve for a:

( )

( )2

2 2 2

2 0.3 m2 1 6000 m/s 600

9.81 m/s10 ms 0.001 s/ms

x ga g

t

∆= = = ×

×�

Insight: The very large acceleration of an expanding airbag can cause severe injury to a small child whose head is too close to the bag when it deploys. Children are safest in the back seat!

31. Picture the Problem: The bacterium accelerates from rest in the forward direction.

Strategy: Employ the definition of acceleration to find the time elapsed, and the relationship between acceleration, displacement, and velocity to find the distance travelled.

Solution: 1. (a) Solve for time: 0

2

12 0 m/s0.077 s

156 m/s

v vt

a

µ

µ

− −= = =

2. (b) Solve for displacement: ( )

( )

2 22 20

2

12 m/s 00.46 m

2 2 156 m/s

v vx

a

µµ

µ

−−∆ = = =

Insight: The accelerations are tiny but so are the bacteria! The average speed here is about 3 body lengths per second if each bacterium were 2 µm long. If this were a human that would be 6 m/s or 13 mi/h, much faster than we can swim!


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32. Picture the Problem: The rocket accelerates straight upward.

Strategy: Employ the relationship between acceleration, displacement, and time to find the acceleration. Because the rocket was at rest before blast off, the initial velocity 0v is zero, and so

is the initial position 0x . Once the acceleration is known, we can use the constant acceleration equation of motion to find the speed.

Solution: 1. (a) Use the following: 210 0 2x x v t at= + +

2. Let 0 0 0x v= = and solve for acceleration:

( )

( )2

2 2

2 77 m217 m/s upward

3.0 s

xa

t= = =

3. (b) Evaluate: ( )( )20 17 m/s 3.0 s 51 m/sv at= + = =

Insight: Notice the simple relationship between distance, acceleration, and time if the initial position and the initial velocity are zero.

33. Picture the Problem: You drive in a straight line and then slow down to a stop.

Strategy: Employ the relationship between acceleration, displacement, and velocity to find the displacement. Note that no time information is given. In this case the acceleration is negative because the car is slowing down.

Solution: 1. (a) Solve for x∆ : ( )

( )

22 2 2 2 20 0 0

2

12.0 m/s021 m

2 2 2 2 3.5 m/s

v v v vx

a a a

− −∆ = = = − = − =

−

2. (b) Since velocity is proportional to the square root of displacement, cutting the distance in half

will reduce the velocity by 2 , not 2. Therefore the speed will be greater than 6.0 m/s after travelling half the distance.

3. Solve for v:

22 2 0 00 0

12.0 m/s2 8.49 m

2 2 2 2

v vxv v a v a

a

∆= + = + − = = =

Insight: For constant acceleration, the velocity changes linearly with time but nonlinearly with distance.

34. Picture the Problem: The bicycle travels in a straight line, slowing down at a uniform rate as it

crosses the sandy patch.

Strategy: Use the time-free relationship between displacement, velocity, and acceleration to find the acceleration. The time can be determined from the average velocity and the distance across the sandy patch.

Solution: 1. (a) Solve for acceleration:

( ) ( )( )

2 22 220

6.4 m/s 8.4 m/s2.1 m/s

2 2 7.2 m

v va

x

−−= = = −

∆

where the negative sign means 2.1 m/s2 due east.

2. (b) Solve for t:

( ) ( )1 102 2

7.2 m0.97 s

8.4 6.4 m/s

xt

v v

∆= = =

+ +

3. (c) Examining 2 20 2v v a x= + ∆ in detail, we note that the acceleration is negative, and that the

final velocity is the square root of the difference between 20v and 2a x∆ . Since 2a x∆ is constant

because the sandy patch doesn’t change, it now represents a larger fraction of the smaller 20v , and

the final velocity v will be more than 2.0 m/s different than 0v . We therefore expect a final speed

of less than 3.4 m/s.


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Insight: In fact, if you try to calculate v in part (c) you end up with the square root of a negative

number, because the bicycle will come to rest in a distance ( )

( )

22 20

2

5.4 m/s06.9 m

2 2 2.1 m/s

vx

a

−−∆ = = =

−,

less than the 7.2 m length of the sandy patch. 35. Picture the Problem: David Purley travels in a straight line, slowing down at a uniform rate until

coming to rest.

Strategy: Use the time-free relationship between displacement, velocity, and acceleration to find the acceleration.

Solution: Solve for acceleration: ( )

22

2 20

22

0.278 m/s0 173 km/h

1 km/h2 2 0.66 m

1.001800 m/s 180

9.81 m/s

v va

x

ga g

− × −

= =∆

= − × =

Insight: Mr. Purley was lucky to escape death when experiencing an acceleration this large. A large acceleration implies a large force, which in this case must have been applied to his body in just the right way to produce a non-lethal injury.

36. Picture the Problem: The rocket accelerates straight upward at a constant rate.

Strategy: Since the initial and final velocities are known, the time can be determined from the average velocity and the distance travelled. The constant acceleration equation of motion can then be used to find the acceleration. Once that is known, the position of the rocket as a function of time is given, and the velocity as a function of time is also given.

Solution: 1. (a) Solve for time:

( ) ( )1 102 2

3.2 m0.25 s

0 26.0 m/s

xt

v v

∆= = =

+ +

2. (b) Solve for acceleration: 2 20 26.0 0 m/s

110 m/s 0.11 km/s0.25 s

v va

t

− −= = = =

3. (c) Evaluate directly, with

0 0 0x v= = : ( )( )

22 21 12 2 110 m/s 0.10 s 0.55 mx at= = =

4. Evaluate directly, with 0 0v = : ( )( )20 110 m/s 0.10 s 11 m/sv at= + = =

Insight: Model rockets accelerate at very large rates, but only for a very short time. Still, even inexpensive starter rockets can reach 1500 ft in altitude and can be great fun to build and launch!


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37. Picture the Problem: The distance-versus-time plot

at right shows how the bicyclist can overtake his friend by pedalling at constant acceleration.

Strategy: To find the time elapsed when the two bicyclists meet, we must set the constant velocity equation of motion of the friend equal to the constant acceleration equation of motion of the cyclist. Once the time is known, the displacement and velocity of the bicyclist can be determined.

Solution: 1. (a) Set the two equations of motion equal to each other. For the friend, set 0 0x = and for the cyclist, use

set 0 0x = and 0 0v = :

( )

friend bicyclist

2120 0 2

f b

x x

v t a t

=

= + + −

2. Solve for t: ( )( )

21friend 2

2 2friend2

2

4 4

2 3.5 m/s20 4 4 4 4

2.4 m/s

0 6.92 4

b

b

v t a t t

vt t t t

a

t t

= − +

= − + + = − + +

= − +

3. Now use the quadratic formula: ( )( )26.92 6.92 4 1 4

6.3, 0.64 s2

t+ ± −

= =

4. We choose the larger root because the time must be greater than 2.0 s, the time at which the bicyclist began pursuing his friend. The bicyclist will overtake his friend 6.3 s after his friend passes him.

5. (b) Find x: ( )( )0 3.5 m/s 6.3 s 22 mx v t= = =

6. (c) Find v. Keep in mind that 0 0v = and that the bicyclist doesn’t begin accelerating until two seconds have elapsed.:

( ) ( )( )20 2 2.4 m/s 6.3 2.0 s 10 m/sv a t= + − = − =

Insight: Even a smaller acceleration would allow the bicyclist to catch up to the friend, because the speed is always increasing for any nonzero acceleration, and so the bicyclist’s speed would eventually exceed the friend’s speed and the two would meet.

38. Picture the Problem: The velocity-versus-time plots

of the car and the truck are shown at right. The car begins with a positive position and a negative velocity, so it must be represented by the lower line. The truck begins with a negative position and a positive velocity, so it is represented by the upper line.

Strategy: The distances travelled by the car and the truck are equal to the areas under their velocity-versus-time plots. We can determine the distances travelled from the plots and use the known initial positions to find the final positions and the final separation.

Solution: 1. Find the final position of the truck:

( ) ( )( )1truck 0,truck truck 235 m 2.5 0 s 10 m/s 22.5 mx x x= + ∆ = − + − = −

2. Find the final position of the car:

( ) ( )( )1car 0,car car 215 m 3.5 0 s 15 m/s 11.25 mx x x= + ∆ = + − − = −

3. Now find the separation: ( ) ( )car truck 11.25 m 22.5 m 11.3 mx x− = − − − =


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Insight: The velocity-versus-time graph is a rich source of information. Besides velocity and time information, you can determine acceleration from the slope of the graph and distance travelled from the area under the graph. In this case, we can see the acceleration of the car (4.29 m/s2) has a greater magnitude than the acceleration of the truck (−4.00 m/s2).

39. Picture the Problem: The cart slides

down the inclined track, each time travelling a distance of 1.00 m along the track.

Strategy: The distance travelled by the cart is given by the constant-acceleration equation of motion for position as a function of time, where 0 0 0x v= = . The magnitude of the acceleration can thus be determined from the given distance travelled and the time elapsed in each case. We can then make the comparison with sina g θ= .

Solution: 1. Find the acceleration:

212 2

20 0

xx at a

t= + + ⇒ = sina g θ=

2. Now find the values for θ = 10.0°: ( )

22

2.00 m1.71 m/s

1.08 sa = = ( )2 29.81 m/s sin10.0 1.70 m/sa = =o


22

2.00 m3.37 m/s

0.770 sa = = ( )2 29.81 m/s sin 20.0 3.35 m/sa = =o


22

2.00 m4.88 m/s

0.640 sa = = ( )2 29.81 m/s sin10.0 4.91 m/sa = =o

Insight: We see very good agreement between the formula sina g θ= and the measured acceleration. The experimental accuracy gets more and more difficult to control as the angle gets bigger because the elapsed times become very small and more difficult to measure accurately. For this reason Galileo’s experimental approach (rolling balls down an incline with a small angle) gave him an opportunity to make accurate observations about free fall without fancy electronic equipment.

40. Picture the Problem: The apple falls straight downward under the influence of gravity.

Strategy: The distance of the fall is estimated to be about 3.0 m (about 10 ft). Then use the time-free equation of motion to estimate the speed of the apple.

Solution: 1. Solve for v, assuming the apple drops from rest ( 0 0v = ):

0 2v a x= + ∆

2. Let a = g and calculate v: ( )( )22 9.81 m/s 3.0 m 7.7 m/s 17 mi/hv = = =

Insight: Newton supposedly then reasoned that the same force that made the apple fall also keeps the Moon in orbit around the Earth, leading to his universal law of gravity. One lesson we might learn here is—wear a helmet when sitting under an apple tree!

41. Picture the Problem: The lava bomb travels upward, slowing down under the influence of gravity,

coming to rest momentarily before falling downward.

Strategy: Since the acceleration of gravity is known, the constant acceleration equation of motion can be used to find the speed and velocity as a function of time. Let upward be the positive direction.

Solution: 1. (a) Set a = −g: ( )( )20 28 m/s 9.81 m/s 2.0 s 8.4 m/sv v gt= − = − =

θ

1.00 m


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2. (b) Set a = −g: ( ) ( )20 28 m/s 9.81 m/s 3.0 s 1.4 m/sv v gt= − = − = −

3. We interpret the answer to (b) as a speed of 1.4 m/s but a velocity of −1.4 m/s, where the negative sign means it is travelling downward.

Insight: We can see the lava bomb must have reached its peak between 2.0 and 3.0 seconds. In

fact, it reached it at ( ) ( )20 28 m/s 9.81 m/s 2.9 st = − − = .

42. Picture the Problem: The material travels straight upward, slowing down under the influence of

gravity until it momentarily comes to rest at its maximum altitude.

Strategy: Since the maximum altitude is known, use the time-free equation of motion to find the initial velocity. Let upward be the positive direction, so that a = −1.80 m/s2.

Solution: Solve for 0v , setting

0v = : ( )( )2 2 2 5

0 2 0 2 1.80 m/s 2.00 10 m 849 m/sv v a x= − ∆ = − − × =

Insight: On Earth that speed would only hurl the material to an altitude of 37 km, as opposed to 200 km on Io. Still, that’s a very impressive initial velocity! It is equivalent to the muzzle velocity of a bullet, and is 2.5 times the speed of sound on Earth.

43. Picture the Problem: The balls fall straight down under the influence of gravity. The first ball

falls from rest but the second ball is given an initial downward velocity.

Strategy: Since the fall distance is known in each case, use the time-free equation of motion to predict the final velocity. Let downward be the positive direction for simplicity.

Solution: 1. (a) The speed increases linearly with time but nonlinearly with distance. Since the first ball has a lower initial velocity and hence a lower average velocity, it spends more time in the air. The first (dropped) ball will therefore experience a larger increase in speed.

2. (b) First ball: Solve for v, setting 0 0v = :

( )( )2 20 2 2 9.81 m/s 32.5 m 25.3 m/sv g x= + ∆ = =

3. Second ball: Solve for v: ( ) ( )( )

22 20 2 11.0 m/s 2 9.81 m/s 32.5 m 27.5 m/sv v g x= + ∆ = + =

4. Compare the v∆ values: 1 25.3 0 m/s 25.3 m/sv∆ = − = for the first ball and

2 27.5 11.0 m/s 16.5 m/sv∆ = − = for the second ball.

Insight: The second ball is certainly going faster, but its change in speed is less than the first ball. 44. Picture the Problem: The arrow rises straight upward, slowing down due to the acceleration of

gravity.

Strategy: Since the position, time, and acceleration are all known, we can use the equation of motion for position as a function of time and acceleration to find the initial velocity 0v . The same equation could be used to find the time required to rise to a height of 15.0 m above its launch point. Let the launch position 0 0x = and let upward be the positive direction.

Solution: 1. (a) Solve for 0v :

( )( )22121

220

30.0 m 9.81 m/s 2.00 s24.8 m/s

2.00 s

x atv

t

− −−= = =

2. (b) Solve with x = 15.0 m: ( ) ( )

( ) ( )

2 212

2 2

15.0 m 24.8 m/s 9.81 m/s

0 4.905 m/s 24.8 m/s 15.0 m

t t

t t

= −

= − + −


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3. Now use the quadratic formula:

( ) ( )( )22 24.8 24.8 4 4.905 15.04

2 9.81

0.702 s , 4.36 s

b b act

a

t

− ± − − −− ± −= =

−

=

Insight: The second root of the solution to part (b) corresponds to the time when the arrow, after rising to its maximum height, falls back to a position 15.0 m above the launch point.

45. Picture the Problem: The book accelerates straight downward and hits the floor of the elevator.

Strategy: The constant speed motion of the elevator does not affect the acceleration of the book. From the perspective of an observer outside the elevator, both the book and the floor have an initial downward velocity of 3.0 m/s. Therefore from your perspective the motion of the book is no different than if the elevator were at rest. Use the position as a function of time and acceleration equation to find the time, setting 0 0v = and letting downward be the positive direction. Then use velocity as a function of time to find the speed of the book when it lands.

Solution: 1. (a) Solve for t:

( )2

2 1.2 m20.49 s

9.81 m/s

xt

g= = =

2. (b) Find v: ( )( )20 0 9.81 m/s 0.49 s 4.8 m/sv v gt= + = + =

Insight: The speed in part (b) is relative to you. Relative to the ground the speed of the book is 4.8 + 3.0 = 7.8 m/s.

46. Picture the Problem: The rocket rises straight upward, accelerating over a distance of 26 m and

then slowing down and coming to rest at some altitude higher than 26 m.

Strategy: Use the given acceleration and distance and the time-free equation of motion to find the velocity of the rocket at the end of its acceleration phase, when its altitude is 26 m. Use that as the initial velocity of the free fall stage in order to find the maximum altitude. Then apply the same equation once again to find the velocity of the rocket when it returns to the ground. The given and calculated positions at various stages of the flight can then be used to find the elapsed time in each stage and the total time of flight.

Solution: 1. (a) Find the velocity at the end of the boost phase

( )( )2 2 2boost 0 2 0 2 12 m/s 26 m 25 m/sv v g x= + ∆ = + =

2. Find the height change during the boost phase and a final speed of zero:

22 2 boost

boost boost boost0 2 2

vv g x x

g= − ∆ ⇒ ∆ =

3. Now find the overall maximum height:

( )

( )

22boost

max 2

25 m/s26 m 26 m 26 32 m 58 m

2 2 9.81 m/s

vh

g= + = + = + =

4. (b) Solve again between the end of the boost phase and the point where it hits the ground:

( ) ( )( )

2 2boost

22 2boost

2

2 25 m/s 2 9.81 m/s 26 m 34 m/s

v v g x

v v g x

= − ∆

= − ∆ = − − =

5. (c) First find the duration of the boost phase. Use the known positions:

( ) ( )boost

boost 1 10 boost2 2

26 m2.1 s

0 25 m/s

xt

v v

∆= = =

+ +

6. Now find the time for the rocket to reach its maximum altitude from the end of the boost phase:

( ) ( )up

up 112boost top2

32 m2.6 s

25 0 m/s

xt

v v

∆= = =

++


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7. Now find the time for the rocket to fall back to the ground:

( ) ( )down

down 112top ground2

58 m3.4 s

0 34 m/s

xt

v v

∆= = =

++

8. Sum the times to find the time of flight:

total boost up down 2.1 2.6 3.4 s 8.1 st t t t= + + = + + =

Insight: Notice how knowledge of the initial and final velocities in each stage, and the distance travelled in each stage, allowed the calculation of the elapsed times using the relatively simple, as opposed to the quadratic equation. Learning to recognize the easiest route to the answer is an important skill to obtain.

47. Picture the Problem: Nut A is dropped

from rest. When it has fallen 2.5 m, nut B is thrown downward with an initial speed vB,0. Both nuts land at the same time after falling 10.0 m.

Strategy: First find the time it takes for nut A to fall 2.5 m using the equation of motion for position as a function of time and acceleration. Also find the time required for nut A to fall the entire 10.0 m. Subtract the first time from the second to find the time interval over which nut B must reach the ground in order to land at the same instant as nut A. Then find the initial velocity vB,0 required in order for nut B to reach the ground in that time.

Solution: 1. Find the time it takes for nut A to fall 2.5 m by solving for t and setting vA,0 = 0.

( )A,1 2

2 2.5 m20.714 s

9.81 m/s

xt

g

∆= = =

2. Find the time it takes for nut A to fall the entire 10.0 m:

( )A,total 2

2 10.0 m21.428 s

9.81 m/s

xt

g

∆= = =

3. Subtract the times to find the time over which nut B must reach the ground:

B,total A,total A,1 1.428 0.714 s 0.714 st t t= − = − =

4. Solve for vB,0:

( )( )22121

2B,total2B,0

B,total

B,0

10.0 m 9.81 m/s 0.714 s

0.714 s

10.5 m/s 11 m/s

x gtv

t

v

−∆ −= =

= ⇒

Insight: In this problem we kept an additional significant figure than is warranted in steps 1, 2, and 3 in an attempt to get a more accurate answer in step 4. However, if you choose not to do so, differences in rounding will lead to an answer of 10 m/s. The specified 2.5 m drop distance for nut A limits the answer to two significant digits, and since the answer is right between 10 and 11 m/s, it could correctly go either way.

ground

branch 2.5 m

Nut B thrown

Both land 10.0 m

vB,0 = ? Nut A


18 of 65


1. Picture the Problem: The vector direction is measured anticlockwise from the +x axis.

Strategy: In each case find the vector components.

Solution: 1. (a) Find the x and y components:

( )

( )

cos 75 m cos35.0 61 m

sin 75 m sin 35.0 43 m

x

y

r r

r r

θ

θ

= = ° =

= = ° =

2. (b) Find the x and y components: ( )

( )

cos 75 m cos 65.0 32 m

sin 75 m sin 65.0 68 m

x

y

r r

r r

θ

θ

= = ° =

= = ° =

Insight: Resolving vectors into their components is an important skill for solving physics problems.

2. Picture the Problem: The water molecule forms a triangle with the positions

of the oxygen and hydrogen nuclei as shown.

Strategy: Break the triangle up into two right triangles and use the sine function to find the distance between the hydrogen nuclei. The angle θ is half of the 104.5° bond angle, or θ = 52.25°.

Solution: 1. Use the sine function to find the distance d:

sin0.96 Å

dθ =

2. The distance between hydrogen nuclei is 2d:

( ) ( )2 2 0.96 Å sin 52.25 1.5 Åd = ° =

Insight: Identifying right triangles and manipulating the trigonometric functions are important skills to learn when solving physics problems.

3. Picture the Problem: The given vector components correspond to the vector

rr

as drawn at right.

Strategy: Use the inverse tangent function to determine the angle θ. Then use the Pythagorean Theorem to determine the magnitude of r

r.

Solution: 1. (a) Use the inverse tangent function to find the distance angle θ :

1 9.5tan 34

14θ − −

= = − °

or

34° below the +x axis

2. (b) Use the Pythagorean Theorem to determine the magnitude of r

r:

( ) ( )2 22 2 14 m 9.5 m

17 m

x yr r r

r

= + = + −

=

3. (c) If both

xr and yr are doubled, the

direction will remain the same but the magnitude will double: ( ) ( )

1

2 2

9.5 2tan 34

14 2

28 m 19 m 34 mr

θ − − × = = − °

×

= + − =

Insight: Any vector can be resolved into two components. The ability to convert a vector to and from its components is an essential skill for solving many physics problems.

x

y

rr

14 m

−9.5 m

x

y

35.0°

rr

x

y

65.0°

rr

(a) (b)


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4. Picture the Problem: The given vector components correspond to the

vector rr

as drawn at right.

Strategy: Determine the angle θ from our knowledge of analogue clocks. The given component xr together with the angle θ will allow us to calculate

the length of r and the component yr .

Solution: 1. (a) Find the angle θ : 1360 30

12θ = × ° = °

2. Find the length of r: 3.0 cm

cos 3.5 cmcos cos30

x

x

rr r rθ

θ= ⇒ = = =

o

3. (b) The components xr and

yr are only equal when θ =45°. Since in this case θ =30°, the

component yr will be

less than xr or 3.0 cm.

4. (c) Find yr : ( )sin 3.5 cm sin 30 1.7 cmyr r θ= = =o


5. Picture the Problem: The two vectors A

r (length 50 units) and B

r (length 120

units) are drawn at right.

Strategy: Resolve Br

into its x and y components to answer the questions.

Solution: 1. (a) Find Bx: ( )120 units cos70 41 units

xB = ° =

2. Since the vector Ar

points entirely in the x direction, we can see that Ax = 50

units and that vector Ar

has the greater x component.

3. (b) Find By: ( )120 units sin 70 113 unitsx

B = ° =

4. The vector Ar

has no y component, so it is clear that vector Br

has the greater y component.


6. Picture the Problem: The four possible locations of

the treasure are labelled A, B, C, and D in the figure at right. The position vector for location A is also drawn. North is up and east is to the right.

Strategy: Use the vector components to find the magnitude and direction of each vector.

Solution: 1. Find the magnitude of A

r: ( ) ( )

2 225.0 m 12.0 5.00 m 30.2 mA = + + =

2. Find the direction from north of A

r: 1 25.0 m

tan 55.8 west of north12.0 5.00 mAθ −

= = ° +

3. Find the magnitude of B

r: ( ) ( )

2 225.0 m 12.0 5.00 m 26.0 mB = + − =

4. Find the direction from north of B

r: 1 25.0 m

tan 74.4 west of north12.0 5.00 mBθ −

= = ° −

5. Find the magnitude of C

r: ( ) ( )

2 225.0 m 12.0 5.00 m 30.2 mC = + + =

x

y

Br

Ar

70°

12.0 m

25.0 m 25.0 m

5.00 m

A

B

C

D

palm tree

Ar

θA

x

y

rr θ

II

III

I XII

3.0 cm


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6. Find the direction from north of C

r: 1 25.0 m

tan 55.8 east of north12.0 5.00 mCθ −

= = ° +

7. Find the magnitude of D

r: ( ) ( )

2 225.0 m 12.0 5.00 m 26.0 mD = + − =

8. Find the direction from north of D

r: 1 25.0 m

tan 74.4 east of north12.0 5.00 mDθ −

= = ° −

Insight: If you ever find a treasure map like this one, you’ll be glad you mastered vectors in physics!

7. Picture the Problem: The whale dives

along a straight line tilted 20.0° below horizontal for 150 m as shown in the figure.

Strategy: Resolve the whale’s displacement vector into horizontal and vertical components in order to find its depth ry and its horizontal travel distance rx.

Solution: 1. (a) The depth is given by ry: ( ) ( )sin 150 m sin 20.0 51 myr r θ= = ° =

2. (b) The horizontal travel distance is given by rx:

( ) ( )cos 150 m cos 20.0 140 m 0.14 kmxr r θ= = ° = =

Insight: Note that both answers are limited to two significant figures, because although “20.0°” has three, “150 m” has only two significant figures.

8. Picture the Problem: The two vectors A

r (length 50.0 m) and B

r

(length 70.0 m) are drawn at right.

Strategy: Add vectors A

r and B

r using the vector component method.

Solution: 1. (a) A sketch of the vectors and their sum is shown at right.

2. (b) Add the x components:

( ) ( ) ( ) ( )50.0 m cos 20.0 70.0 m cos 50.0 92.0 mx x x

C A B= + = − ° + ° =

3. Add the y components: ( ) ( ) ( ) ( )50.0 m sin 20.0 70.0 m sin 50.0 36.5 my y y

C A B= + = − ° + ° =

4. Find the magnitude of C:

( ) ( )2 22 2 92.0 m 36.5 m 99.0 m

x yC C C= + = + =

5. Find the direction of C: 1 1 36.5 m

tan tan 21.792.0 m

y

C

x

C

Cθ − −

= = = °

Insight: Resolving vectors into components takes a little bit of extra effort, but you can get much more accurate answers using this approach than by using a ruler and protractor to add the vectors graphically.

Cr

x

y Br

Ar

20.0°

50.0°


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9. Picture the Problem: The vectors involved in the

problem are depicted at right. The control tower (CT) is at the origin and north is up in the diagram.

Strategy: Subtract vector Br

from Ar

using the vector component method.

Solution: 1. (a) A sketch of the vectors and their difference is shown at right.

2. (b) Subtract the x components:

( ) ( ) ( ) ( )220 km cos 180 32 140 km cos 90 65 310 kmx x x

D A B= − = − ° − − ° = −

3. Subtract the y components:

( ) ( ) ( ) ( )220 km sin 180 32 140 km sin 90 65 57 kmy y y

D A B= − = − ° − − ° =

4. Find the magnitude of D:

( ) ( )2 22 2 5310 km 57 km 320 km 3.2 10 m

x yD D D= + = + = = ×

5. Find the direction of D:

1 1 57 kmtan tan 10 180 170 or 10 north of west

310 kmy

D

x

D

Dθ − −

= = = − ° + ° = ° ° −

Insight: Resolving vectors into components takes a little bit of extra effort, but you can get much more accurate answers using this approach than by adding the vectors graphically. Notice, however, that when your calculator returns −10° as the angle in step 5, you must have a picture of the vectors in your head (or on paper) to correctly determine the direction.

10. Picture the Problem: The vectors involved in the problem are

depicted at right.

Strategy: Deduce the x and y components of Br

from the

information given about Ar

and Cr

. Use the known components

to estimate the length and direction of Br

as well as calculate them precisely.

Solution: 1. (a) A sketch of the vectors is shown at right.

2. (b) The vector Br

must have an x component of −75 m so that when it is added to A

r the x components will cancel out. It must

also have a y component of 95 m because that is the length of Cr

and Ar

has no y component to contribute. Therefore Br

must be

longer than either Ar

or Cr

and it must have an angle of greater than 90°. I estimate that its length is about 120 m and that it points at about 130°.

3. (c) Using the known components of B

r

we can find its magnitude:

2 2( 75 m) (95 m) 121 mB = − + =

4. Find the direction of B

r: 1 95 m

tan 52 180 128–75 mBθ −

= = − ° + ° = °

Insight: Here the length and direction of Br

are determined by its x and y components, which are

determined from Ar

and Cr

. Learning to manipulate vector components will be a useful skill when tackling many physics problems.


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problem are depicted at right.

Strategy: Use the vector component method of addition and subtraction to determine the components of each combination of A

r and B

r.

Once the components are known, the length and direction of each combination can be determined fairly easily.

Solution: 1. (a) Determine the components of +A B

r r:

( ) ( )ˆ ˆ ˆ ˆ–5 10 10 – 5+ = + =A B y x x yr r

2. Find the magnitude of +A Br r

: ( ) ( )2 2

10 5 11 units+ = + − =A Br r

3. Determine the direction of +A Br r

, measured anticlockwise from the positive x axis.

1 5tan 27 or 333

10θ −

+

− = = − ° °

A Br r

4. (b) Determine the components of −A B

r r:

( ) ( )ˆ ˆ ˆ ˆ–5 10 10 – 5− = − = −A B y x x yr r

5. Find the magnitude of −A Br r

: ( ) ( )2 2

10 5 11 units− = − + − =A Br r

6. Determine the direction of −A Br r


1 5tan 27 180 207

10θ −

−

− = = ° + ° = °

− A Br r

7. (c) Determine the components of −B Arr

: ( ) ( )ˆ ˆ ˆ ˆ10 –5 10 5− = − − = +B A x y x y

rr

8. Find the magnitude of −B Arr

: ( ) ( )2 2

10 5 11 units− = + =B Arr

9. Determine the direction of −B Arr


1 5tan 27

10θ −

−

= = °

B A

rr

Insight: This problem is simplified by the fact that Ar

and Br

have only one component each, but a similar approach will work even with more complicated vectors. Notice that you must have a picture of the vectors in your head (or on paper) in order to correctly interpret the directions in steps 3, 6, and 9.

12. Picture the Problem: The vectors involved in

the problem are depicted at right.

Strategy: Add the vectors using the component method in order to find the components of the vector sum. Use the components to find the magnitude and the direction of the vector sum.

Solution: 1. (a) Make estimates from the drawing:

20 m 1.5θ+ + ≈ ≈ °A B Cr rr

+A Br r

θ+A B

r rθ

−A Br r

θ−B Arr

x

y

Br

Ar

O 10

5

−B Arr

−A Br r

Cr

x

y

Br

Ar

+ +A B Cr rr

θ

45°

30°


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2. (b) Add the vector components:

( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( )

ˆ0 20.0 m cos 45 7.0 m cos 30

ˆ10.0 m 20.0 m sin 45 7.0 m sin 30

ˆ ˆ20.2 m 0.64 m

+ + = + ° + − ° + − + ° + − °

+ + = +

A B C x

y

A B C x y

r rr

r rr

3. Use the components to find the magnitude:

( ) ( )2 220.2 m 0.64 m 20.2 m+ + = + =A B Cr rr

4. Use the components to find the angle:

1 0.64 mtan 1.8

20.2 mθ −

= = °

Insight: Resolving vectors into components takes a little bit of extra effort, but you can get much more accurate answers using this approach than by adding the vectors graphically. Notice, however, that when your calculator returns the angle of 1.8° in step 4, you must have a picture of the vectors in your head (or on paper) to correctly determine the direction.

13. Picture the Problem: The vector involved in the problem is depicted at

right.

Strategy: Determine the x and y components of and then express them in terms of the unit vectors.

Solution: 1. Find the x and y components of ∆r

r:

( ) ( )( ) ( )

54 m cos 42 40 m

54 m sin 42 36 m

x

y

r

r

= − =

= − = −

o

o

2. Now express ∆rr

in terms of the unit vectors:

( ) ( )ˆ ˆ40 m 36 m∆ = + −r x yr

Insight: In general, an arbitrary two-dimensional vector Ar

can always be written as the sum of a vector component in the x direction and a vector component in the y direction.

14. Picture the Problem: The vector A

r has a length of 6.1 m and points in the negative x direction.

Strategy: In order to multiply a vector by a scalar, you need only multiply each component of the vector by the same scalar.

Solution: 1. (a) Multiply each component of A

r by −3.7:

( )

( )( ) ( )

ˆ6.1 m

ˆ ˆ3.7 3.7 6.1 m 23 m so 23 mxA

= −

− = − − = =

A x

A x x

r

r

2. (b) Since Ar

has only one component, its magnitude is simply 23 m.

Insight: Multiplying both components of a vector by a scalar will change the length of the vector but not its direction.


depicted at right.

Strategy: Determine the lengths and directions of the various vectors by using their x and y components.

Solution: 1. (a) Find the direction of A

r

from its components:

1 2.0 mtan –22

5.0 mθ − −

= = °

Ar

2. Find the magnitude of A

r:

( ) ( )2 2

5.0 m 2.0 m 5.4 mA = + − =

3. (b) Find the direction of B

r from its

components:

1 5.0 mtan 68 180 110

–2.0 mθ −

= = − ° + ° = °

Br

+A Br r

θ+A B

r r x

y

Br

Ar

2.0 m 5.0 m O

2.0 m

∆rr

x

y

54 m 42°


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4. Find the magnitude of B

r:

( ) ( )2 2

2.0 m 5.0 m 5.4 mB = − + =

5. (c) Find the components of +A B

r r:

( ) ( ) ( ) ( )ˆ ˆ ˆ ˆ5.0 2.0 m 2.0 5.0 m 3.0 m 3.0 m+ = − + − + = +A B x y x yr r

6. Find the direction of +A B

r r from its

components:

1 3.0 mtan 45

3.0 mθ −

+

= = °

A Br r

7. Find the magnitude of +A B

r r:

( ) ( )2 23.0 m 3.0 m 4.2 m+ = + =A Br r

Insight: In the world of vectors 5.4+5.4 m can be anything between 0 m and 10.8 m, depending upon the directions that the vectors point. In this case their sum is 4.2 m.


depicted at right.

Strategy: Determine the lengths and directions of the various vectors by using their x and y components.

Solution: 1. (a) Find the direction

of Ar


1 12 mtan –26

25 mθ − −

= = °

Ar

2. Find the

magnitude of Ar

: ( ) ( )2 225 m –12 m 28 mA = + =

3. (b) Find the

direction of Br


1 15 mtan 82

2.0 mθ −

= = °

Br

4. Find the

magnitude of Br

: ( ) ( )2 22.0 m 15 m 15 mB = + =

5. (c) Find the components of

+A Br r

:

( ) ( ) ( ) ( )ˆ ˆ ˆ ˆ25 2.0 m 12 15 m 27 m 3.0 m+ = + + − + = +A B x y x yr r

6. Find the direction

of +A Br r


1 3.0 mtan 6.3

27 mθ −

+

= = °

A Br r

7. Find the magnitude of

+A Br r

:

( ) ( )2 227 m 3.0 m 27 m+ = + =A Br r

Insight: In the world of vectors 28 + 15 m can be anything between 13 m and 43 m, depending upon the directions that the vectors point. In this case their sum is 27 m.

+A Br r

x

y

Br

Ar

O 25 m

12 m

θ+A B

r r


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17.

Picture the Problem: The vectors involved in the problem are depicted at right.

Strategy: Use the information given in the figure to determine the components of each vector.

Solution

: 1. ( ) ( ) ( ) ( )

( ) ( )

ˆ ˆ1.5 m cos 40 1.5 m sin 40

ˆ ˆ1.1 m 0.96 m

= ° + °

= +

A x y

A x y

r

r

2. ( ) ( ) ( ) ( )

( ) ( )

2.0 m cos 19 2.0 m sin 19

ˆ ˆ1.9 m 0.65 m

= − ° − − °

= + −

B

B x y

r

r

3. ( ) ( ) ( ) ( )

( ) ( )

ˆ ˆ1.0 m cos 180 25 1.0 m sin 180 25

ˆ ˆ0.91 m 0.42 m

= − ° + − °

= − +

C x y

C x y

r

r

4. ( )ˆ ˆ0 1.5 m= +D x yr



problem are depicted at right.

Strategy: Use the information given in the figure

to determine the components of vectors , , and A B Cr rr

. Then add the components.

Solution: 1. Add the x component of each vector:

( ) ( )

( ) ( )

( ) ( )

( )

1.5 m cos 40 1.1 m

2.0 m cos 19 1.9 m

1.0 m cos 180 25 0.91 m

2.1 m

x

x

x

x

A

B

C

= ° =

= − ° =

= − ° = −

+ + =A B Cr rr

2. Add the y component of each vector:

( ) ( )

( ) ( )

( ) ( )

( )

1.5 m sin 40 0.96 m

2.0 m sin 19 0.65 m

1.0 m sin 180 25 0.42 m

0.74 m

y

y

y

y

A

B

C

= ° =

= − ° = −

= − ° =

+ + =A B Cr rr

3. Express the sum in unit vector notation:

( ) ( )ˆ ˆ2.1 m 0.74 m+ + = +A B C x yr rr

Insight: In this problem the vector component method of addition is much quicker than the graphical method.

19. Picture the Problem: The displacement vectors are depicted at

right. North is in the y direction and east is in the x direction.

Strategy: Sum the components of the vectors in order to determine +A B

r r. Multiply that vector by −1 in order to reverse

its direction. Then find the magnitude and direction of the reversed vector.

Solution: 1. (a) Add the two displacement vectors:

( ) ( )ˆ ˆ72 m 120 m+ = − +A B x yr r

2. Multiply by −1 in order to ( ) ( ) ( )ˆ ˆ72 m 120 m− + = + −A B x yr r

( )− +A Br r

x

y

Br

Ar

−72 m O

120 m θ


26 of 65

reverse the direction of the net displacement and bring the cat back home:

3. Find the magnitude of the desired displacement:

( ) ( ) ( )2 2

72 m 120 m 140 m− + = + =A Br r

4. Find the direction of the desired displacement:

1 120 mtan 59 59 south of east

72 mθ − −

= = − ° = °

5. (b) Vector addition is independent of the order in which the addition is accomplished. The initial displacement is the same, so there is no change in the displacement for the homeward part of the trip.

Insight: In this problem we could claim the cat’s initial displacement is a single vector with the given components. The answers wouldn’t change, but it would simplify the solution a little bit.

20. Picture the Problem: You travel due west for 130 s at 27 m/s then due south at 14 m/s for 62 s.

Strategy: Find the components of the displacement vector. Once the components are known the magnitude and direction can be easily found. Let north be the positive y direction and east be the positive x direction.

Solution: 1. Find the westward displacement:

( )( )27 m/s 130 s 3500 mx xr v t= = − = −

2. Find the southward displacement:

( ) ( )14 m/s 62 s 870 my yr v t= = − = −

3. Find the direction of the displacement:

1 1 870 mtan tan 14 180 194

3500 m

or 14 south of west

y

x

r

rθ − − −

= = = ° + ° = ° −

°

4. Find the magnitude of the displacement:

( ) ( )2 2

3500 m 870 m 3600 m 3.6 kmr = − + − = =

Insight: The 14° refers to the angle below the negative x axis (west) because the argument of the inverse tangent function is y xr r , or south divided by west.

21. Picture the Problem: The jogger runs at 3.25 m/s in a direction 30.0° above the positive x axis.

Strategy: Find the components of the velocity vector.

Solution: 1. (a) Find the x component of vr

: ( ) ( )3.25 m/s cos 30.0 2.81 m/sxv = ° =

2. Find the y component of vr

: ( ) ( )3.25 m/s sin 30.0 1.63 m/syv = ° =

3. (b) If the jogger’s speed is halved, the direction will remain unchanged but the x and y components will be halved.

Insight: In this case the angle of 30.0° corresponds to the standard angle θ. 22. Picture the Problem: The ball rises straight upward, momentarily comes to rest, and then falls

straight downward.

Strategy: After it leaves your hand the only acceleration of the ball is due to gravity, so we expect the answer to be 9.81 m/s2. To calculate the acceleration we need only consider the initial and final velocities and the time elapsed. Because of the symmetry of the situation, the final velocity downward will have the same magnitude as the initial velocity upward. Apply the relevant equation, taking upward to be the positive direction.


27 of 65

Solution: Apply equation:

( ) ( ) 2av

ˆ ˆ4.5 m/s 4.5 m/sˆ9.8 m/s

0.92 sf i

t

− − −= = = −

∆

v v y ya y

r rr

Insight: A uniform acceleration will produce a symmetric trajectory, with the time to rise to the peak of flight equalling the time to fall back down, and with equal initial and final speeds.

23. Picture the Problem: The skateboarder rolls down the ramp that is inclined 20.0° above the

horizontal.

Strategy: To calculate the acceleration we need only consider the initial and final velocities and the time elapsed. Apply the relevant equation, taking the direction down the ramp to be the positive direction.

Solution: 1. Apply equation:

( ) ( ) 2av

10.0 m/s 0 m/s3.33 m/s

3.00 sf iv v

at

− −= = =

∆

2. Compare with g sin θ : ( ) ( )2 2sin 9.81 m/s sin 20.0 3.36 m/sg θ = ° =

Insight: The two are equal to within rounding errors. Or perhaps there was a small amount of friction between the skateboard wheels and the ramp.

24. Picture the Problem: The initial and final displacement vectors

are depicted at right.

Strategy: Use the given formulae to determine the components of the initial and final positions. Then use those components to find the displacement vector. Divide the displacement vector by the elapsed time to find the velocity vector, and then determine its magnitude and direction.

Solution: 1. (a) Find the initial position vector:

( ) [ ] [ ]{ }

( )

8i

8

ˆ ˆ3.84 10 m cos 0 sin 0

ˆ3.84 10 m

= × +

= ×

r x y

x

r

2. Find the arguments of the sine and cosine functions for t = 7.38 days. Let 6 radians/s2.46 10ω −= × :

( )( )62.46 10 radians/s 7.38 d 86400 s/d

1.57 radians

tω −= × ×

=

3. Find the final position vector: ( ) [ ] [ ]{ }

( ) [ ] [ ]{ }

( ) ( ) ( )

8f

8

5 8 8f

ˆ ˆ3.84 10 m cos sin

ˆ ˆ3.84 10 m cos 1.57 radians sin 1.57 radians

ˆ ˆ ˆ3.06 10 m 3.84 10 m 3.84 10 m

t tω ω= × +

= × +

= × + × ≅ ×

r x y

x y

r x y y

r

r

4. Find the displacement vector: ( ) ( )8 8f i ˆ ˆ3.84 10 m 3.84 10 m∆ = − = × − ×r r r y x

r r r

5. Find the vector avv

r:

( ) ( )

( ) ( )

8 8

av

av

ˆ ˆ3.84 10 m 3.84 10 m

7.38 d 86400 s/dˆ ˆ602 m/s 602 m/s

t

− × + ×∆= =

∆ ×= − +

x yrv

v x y

rr

r

6. Find the magnitude of avv

r: ( ) ( )

2 2

av 602 m/s 602 m/s 852 m/sv = − + =

7. Find the direction of avv

r:

av, 1 1

av,

602 m/stan tan 45 180 135

602 m/sy

x

v

vθ − −

= = = − ° + ° = ° −

8. (b) The instantaneous speed of the Moon is greater than the average velocity because the distance travelled is greater than the displacement in this case.

Insight: If the Moon had completed an entire orbit, instead of just one-quarter of an orbit, its displacement and its average velocity would have been zero. Its speed remains constant, however, at about 947 m/s using the data given in this problem. The given data correspond to a coordinate system where the x direction always points towards the centre of the Sun even as the Earth orbits the Sun.


28 of 65

25.

Picture the Problem: The initial and final velocity vectors are depicted at right.

Strategy: Use the given formulae to determine the components of the initial and final velocities. Then use those components to find the change in velocity vector. Divide the change in velocity vector by the elapsed time to find the acceleration vector, and then determine its magnitude and direction.

Solution: 1. (a) Find the initial velocity vector:

( ) [ ] [ ]{ }( )

i ˆ ˆ945 m/s sin 0 cos 0

ˆ945 m/s

= − +

=

v x y

y

r

2. Find the arguments of the sine and cosine functions for t = 0.100 days. Let

6 radians/s2.46 10ω −= ×:

( )( )62.46 10 radians/s 0.100 d 86400 s/d

0.0213 radians

tω −= × ×

=

3. Find the final position vector:

( ) [ ] [ ]{ }

( ) [ ] [ ]{ }( ) ( )

f

f

ˆ ˆ945 m/s sin cos

ˆ ˆ945 m/s sin 0.0213 radians cos 0.0213 radians

ˆ ˆ20.1 m/s 945 m/s

t tω ω= − +

= − +

= − +

v x y

x y

v x y

r

r

4. Find the acceleration vector:

( ) ( ) ( )( )2f i

ˆ ˆ ˆ20.1 m/s 945 m/s 945 m/s0.00233 m/s

0.100 d 86400 s/dt

− − − − = = = −

∆ ×

x y yv va x

r rr

5. (b) Find the arguments of the sine and cosine functions for t = 0.0100 days:

( )( )62.46 10 radians/s 0.0100 d 86400 s/d

0.00213 radians

tω −= × ×

=

6. Find the final position vector:

( ) [ ] [ ]{ }

( ) [ ] [ ]{ }( ) ( )

f

f

ˆ ˆ945 m/s sin cos

ˆ ˆ945 m/s sin 0.00213 radians cos 0.00213 radians

ˆ ˆ2.01 m/s 945 m/s

t tω ω= − +

= − +

= − +

v x y

x y

v x y

r

r

7. Find the acceleration vector:

( ) ( ) ( )( )2f i

ˆ ˆ ˆ2.01 m/s 945 m/s 945 m/s0.00233 m/s

0.0100 d 86400 s/dt

− − − − = = = −∆ ×

x y yv va x

r rr

Insight: The two answers ended up being the same because both time intervals are fairly small. If instead we had examined an interval of 1.00 days there would have been a ˆ−y component of ∆v

r and a slightly different acceleration. The given data in this problem correspond to a coordinate system where the x direction always points towards the centre of the Sun even as the Earth orbits the Sun.

26. Picture the Problem: The vectors involved in this problem are

depicted at right.

Strategy: Let pg =vr

plane’s velocity with respect to the

ground, ap =vr

attendant’s velocity with respect to the plane,

and add the vectors to find ag =vr

attendant’s velocity with

respect to the ground.

apvr

pgvr

agvr

∆vr

945 m/s

fvr

ivr

y

O x


29 of 65

Solution: 1. Apply equation:

( ) ( ) ( )ag ap pg

ag

ˆ ˆ ˆ1.22 m/s 16.5 m/s 15.3 m/s

15.3 m/sv

= + = − + =

=

v v v x x xr r r

Insight: If the attendant were walking towards the front of the plane, her speed relative to the ground would be 17.7 m/s, slightly faster than the airplane’s speed.


depicted at right.

Strategy: Let yw =vr

your velocity with respect to the walkway,

wg =vr

walkway’s velocity with respect to the ground, and add the

vectors to find yg =vr

your velocity with respect to the ground.

Then find the time it takes you to travel the 85-m distance.

Solution: 1. Find your velocity with respect to the walkway:

( )yw

85 mˆ ˆ ˆ1.25 m/s

68 s

x

t

∆ = = =

∆ v x x xr

2. Apply equation to find your velocity with respect to the ground:

( ) ( ) ( )yg yw wg ˆ ˆ ˆ1.25 m/s 2.2 m/s 3.45 m/s= + = + =v v v x x xr r r

3. Now find the time of travel: yg

85 m25 s

3.45 m/s

xt

v

∆= = =

Insight: The moving walkway slashed your time of travel from 68 s to 25 s, a factor of 2.7! Note that we bent the significant figures rules a little bit by not rounding ywv

r to 1.3 m/s. This helped us

avoid rounding error.

28. Picture the Problem: The vectors involved in this problem are depicted at right.

Strategy: Let pg =v

rvelocity of the plane relative to

the ground, pa =vr

velocity of the plane relative to the

air, and ag =vr

velocity of the air relative to the

ground. The drawing at right depicts the vectors added, pg pa ag= +v v v

r r r. Determine the angle of the

triangle from the inverse sine function.

Solution: 1. (a) Use the inverse sine function to find θ:

ag1 1

pa

65 km/hsin sin 11 west of north

340 km/h

v

vθ − −

= = = °

2. (b) The drawing above depicts the vectors.

3. (c) If the plane reduces its speed but the wind velocity remains the same, the angle found in part (a) should be increased in order for the plane to continue flying due north.

Insight: If the plane’s speed were to be reduced to 240 km/h, the required angle would become 16°.

wgvr

ygvr

ywvr


30 of 65


depicted at right.

Strategy: Let pf =vr

the passenger’s velocity relative to the ferry,

pw =vr

the passenger’s velocity relative to the water, and fw =vr

the

ferry’s velocity relative to the water. Apply the relevant equation and solve for fwv

r. Once the components of fwv

r are known, its

magnitude and direction θ can be determined.

Solution: 1. Solve equation for

fwvr

: pw pf fw

fw pw pf

= +

= −

v v v

v v v

r r r

r r r

2. Determine the components of fwv

r:

( ) ( ) ( )

( ) ( )fw

fw

ˆ ˆ ˆ4.50 m/s sin 30 4.50 m/s cos30 1.50 m/s

ˆ ˆ2.25 m/s 2.40 m/s

= − ° + ° −

= − +

v x y y

v x y

r

r

3. Find the direction of fwv

r: fw, 1 1

fw,

2.25 m/stan tan 43 west of north

2.40 m/sx

y

v

vθ − −

= = = °

4. Find the magnitude of fwv

r: ( ) ( )

2 22 2fw fw, fw, 2.25 m/s 2.40 m/s 3.29 m/s

x yv v v= + = − + =

Insight: If the person were to walk even faster with respect to the ferry, then fwvr

would have to be shorter and point more in the westerly direction.

30. Picture the Problem: The situation is similar to that depicted

in the figure at right, except the boat is supposed to be a jet ski.

Strategy: Place the x-axis perpendicular to the flow of the river, such that the river is flowing in the negative y-direction. Let bw =v

r jet ski’s velocity relative to the water, bg =v

r jet

ski’s velocity relative to the ground, and wg =vr

water’s

velocity relative to the ground. Find the vector bwvr

, and then determine its magnitude.

Solution: 1. Solve for

bwvr

: bg bw wg bw bg wg = + ⇒ = −v v v v v v

r r r r r r

2. Find the components of bgv

r:

( ) ( )

( ) ( )bg ˆ ˆ9.5 m/s cos 20.0 9.5 m/s sin 20.0

ˆ ˆ8.9 m/s 3.2 m/s

= ° + °

= +

v x y

x y

r

3. Subtract to find

bwvr

: ( ) ( ) ( )

( ) ( )bw bg wg ˆ ˆ ˆ8.9 m/s 3.2 m/s 2.8 m/s

ˆ ˆ8.9 m/s 6.0 m/s

= − = + − −

= +

v v v x y y

x y

r r r

4. Find the magnitude of bwv

r:

( ) ( )2 2

bw 8.9 m/s 6.0 m/s 11 m/sv = + =

Insight: Note that the 35° angle is extraneous information for this problem. If we work backwards to find the angle from the components of bwv

r we get ( )1tan 6.0 8.9 34θ −= = ° , not exactly 35° due

to rounding errors.

pwvr

fwvr

pfvr

θ

30°

E

N


31 of 65

31. Picture the Problem: The situation is depicted in the figure at

right, except the boat is supposed to be a jet ski.

Strategy: Place the x-axis perpendicular to the flow of the river, such that the river is flowing in the negative y-direction. Let bw =v

r jet ski’s velocity relative to the

water, bg =vr

jet ski’s velocity relative to the ground, and

wg =vr

water’s velocity relative to the ground. Set the y

component of bwvr

equal to the magnitude of wgvr

so that

they cancel, leaving only an x component of bgvr

. Then

determine the angle θ.

Solution: 1. (a) Set bw, wg, 0y yv v+ = and

solve for θ: ( )

bw, bw wg,

wg, 1 1

bw

sin

2.8 m/ssin sin 13

12 m/s

y y

y

v v v

v

v

θ

θ − −

= = −

− − − = = = °

2. (b) Increasing the jet ski’s speed relative to the water will increase bwv and therefore decrease the angle θ.

Insight: Aeroplanes must also make heading adjustments like the jet ski’s in order to fly in a certain direction when there is a steady wind present.

32. Picture the Problem: The vectors for Jet Ski A

and B are depicted at right. Note that bwvr

has the same magnitude for each jet ski but if you inspect the diagram for Jet Ski B you can see that

bw bg,Bv v> .

Strategy: Use the x components of the velocities of each jet ski relative to the ground to determine the time required for each jet ski to cross the river.

Solution: 1. (a) The time required for each jet ski to cross the river equals the width of the river divided by the x component of the jet ski’s velocity relative to the ground. From the diagrams you can see that the x component of

bg,Avr

equals bwvr

, but that the x component

of bg,Bvr

is shorter than bwvr

. Therefore Jet Ski A has a higher velocity in the x direction relative

to the ground, and will cross the river first.

2. (b) Find the ratio of the times:

river

bg, A, bg, B, bw

river bg, A, bw

bg, B,

cos350.82x xA

B x

x

x

v v vt

xt v v

v

∆

°∆= = = =

∆∆

Insight: The ratio is less than one, so A Bt t∆ < ∆ and Jet Ski A reaches the opposite shore first.

Remember this the next time you race jet skis across a flowing river!

bg,Avr

wgvr

bwvr

bg,Bvr

wgvr bwv

r

Jet Ski A

Jet Ski B 35°


32 of 65


1. Picture the Problem: The vector involved in this problem is

depicted at right.

Strategy: Separate vr

into x- and y-components. Let north be along the x-axis, west along the y axis. Find the components of the velocity in each direction and use them to find the distances travelled.

Solution: 1. (a) Find xv

and yv :

( )

( )

3.2 m/s sin 42 2.1 m/s

3.2 m/s cos 42 2.4 m/s

x

y

v

v

= ° =

= ° =

2. Find the westward distance travelled:

( ) ( )2.4 m/s 25 min 60 s/min

3600 m 3.6 km

yy v t= = ×

= =

3. (b) Find the northward distance travelled:

( )( )2.1 m/s 25 min 60 s/min

3200 m 3.2 km

xx v t= = ×

= =

Insight: The northward and westward motions can be considered separately. In this case they are each described by constant velocity motion.


depicted at right.

Strategy: Let north be along the y-axis and east along the x axis. Find the components of the velocity in each direction, and use them to find the times elapsed.

Solution: 1. (a) Find the x component of v

r:

( )1.60 m/s cos15.01.55 m/s

xv = °

=

2. Find the time elapsed to travel east 20.0 m:

20.0 m12.9 s

1.55 m/sx

xt

v= = =

3. (b) Find the y component of vr

: ( )1.60 m/s sin15.0 0.414 m/syv = ° =

4. Find the time elapsed to travel north 30.0 m:

30.0 m72.4 s

0.414 m/sy

yt

v= = =

Insight: The northward and eastward motions can be considered separately. In both cases the actual distance travelled is greater than 20.0 or 30.0 m, respectively. For instance, in the second case you must actually travel 30.0 m sin15.0 116 m° = to change your displacement by 30.0 m north.

3. Picture the Problem: The car moves up the 5.5° incline with constant acceleration, changing both

its horizontal and vertical displacement simultaneously.

Strategy: Find the magnitude of the displacement along the incline, and then independently find the horizontal and vertical components of the displacement.

Solution: 1. (a) Find the magnitude of the displacement along the incline:

( )( )22 21 1

0 2 20 0 0 2.0 m/s 12 s 140 mr v t at∆ = + + = + + =

2. Find the horizontal component of r∆ :

( )cos 140 m cos5.5 140 mx d θ= = ° =

vr

W

N 42°

xv

yv

vr

N

E 15°

xv

yv


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3. (b) Find the vertical component r∆ : ( )sin 140 m sin 5.5 14 my d θ= = ° =

Insight: The horizontal and vertical motions can be considered separately. In this case they are each described by constant acceleration motion, but the vertical acceleration is less than the horizontal. The two accelerations would be equal if the angle of the incline were 45°.

4. Picture the Problem: The motion of the electron is depicted at right.

Strategy: Use the given information to independently write the equations of motion in the x and y directions. There will be a pair of equations for the position of the particle only the acceleration will not be the same as g

r.

Use the equations of motion to find the requested time and position information. Let the x direction correspond to horizontal, and the y direction to vertical.

Solution: 1. (a) The horizontal motion is characterized by constant velocity.

–909

6.20 0 cm2.95 10 s

2.10 10 cm/sx

x xt

v

− −= = = ×

×

2. (b) Use the time to find the vertical deflection , again using except substituting for .a g

r r

( )

210 0 2

17 2 9 2120 cm 0 cm 5.30 10 cm/s (2.95 10 s)

2.31 cm

y yy y v t a t

−

= + +

= + + × ×

=

Insight: This problem is very much akin to projectile motion, with uniform acceleration in one direction and constant velocity the perpendicular direction. The only difference is the acceleration is 5.30×1017 cm/s2 upward instead of 9.81 m/s2 downward.

5. Picture the Problem: The paths of the two canoeists are

shown at right.

Strategy: Canoeist 1’s 45° path determines an isosceles right triangle whose legs measure 1.0 km. So canoeist 2’s path determines a right triangle whose legs measure 1.0 km north and 0.5 km west. Use the right triangle to find the angle θ. Find the distance travelled by each canoeist and set the times of travel equal to each other to determine the appropriate speed of canoeist 2:

Solution: 1. (a) Find the angle θ from the right triangle of canoeist 2:

1 0.5 kmtan 27

1.0 kmθ −

= = °

2. (b) Set the travel times equal to each other:

1 21 2

1 2

r r

t tv v

∆ ∆= ⇒ =

3. Use the resulting ratio to find the appropriate speed of canoeist 2:

( ) ( )

( ) ( )

2 2

22 1 2 2

1

0.5km 1.0 km m 1.35 1.1 m/s

s1.0 km 1.0 km

dv v

d

+ = = =

+

Insight: There are other ways to solve this problem. For instance, because the motions are independent, we could set the time it takes canoeist 1 to travel 1.0 km horizontally equal to the time it takes canoeist 2 to travel 0.5 km horizontally.


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6. Picture the Problem: The arrow falls below the

target centre as it flies from the bow to the target.

Strategy: Treat the vertical and horizontal motions separately. First find the time required for the arrow to drop straight down 52 cm from rest. Then use that time together with the horizontal target distance to find the horizontal speed of the arrow. That must also equal the initial speed because the arrow was launched horizontally.

Solution: 1. Find the time to drop 52 cm: ( )

2

2 0.52 m20.326 s

9.81 m/s

yt

g

∆∆ = = =

2. Find the speed of the arrow from the horizontal distance and time elapsed:

0

15 m46 m/s

0.326 sx

xv

t

∆= = =

∆

Insight: We had to bend the significant figures rules a bit to obtain an accurate answer. Another way to solve this problem and avoid the rounding error is to solve for velocity:

( )20 2v g x h y= − . In this case we would set h = 0.52 m , y = 0.0 m, and x = 15 m.

7. Picture the Problem: The diver falls down along a parabolic arc, maintaining her horizontal

velocity but gaining vertical speed as she falls.

Strategy: Find the vertical speed of the diver after falling 3.00 m. The horizontal velocity remains constant throughout the dive. Then find the magnitude of the velocity from the horizontal and vertical components.

Solution: 1. Use to find yv : ( )( )2 2 2 2 20 2 0 2 9.81 m/s 0 m – 3.00 m 58.9 m /sy yv v g y= − ∆ = − =

2. Use the components and x yv v to

find the speed:

( )22 2 2 21.75 m/s 58.9 m /s 7.87 m s

x yv v v= + = + =

Insight: Projectile problems are often solved by first considering the vertical motion, which determines the time of flight and the vertical speed, and then considering the horizontal motion.

8. Picture the Problem: The rock falls down along a parabolic arc, maintaining its horizontal

velocity but gaining vertical speed as it falls.

Strategy: There is no initial component of velocity in the y-direction. Therefore we can solve for the acceleration of gravity on the planet Zircon.

Solution: 1. Solve for g: ( )2

0

2

2v h yg

x

−=

2. Find the acceleration of gravity on Zircon:

( ) ( )

( )

2

22

2 6.95 m/s 1.40 0 m1.77 m/s

8.75 mg

−= =

Insight: Another way to solve this problem is to first find the time required to travel the horizontal distance of 8.75 m, knowing the horizontal speed is 6.95 m/s. Use that time (1.26 s) together with the vertical fall distance of 1.40 m to find the acceleration of gravity.

0.52 m 0v

r


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9. Picture the Problem: The trajectory of the clam is indicated in

the figure at right.

Strategy: Analyze the motion of the clam as it is launched horizontally and falls to the rocks below. Let upward be the positive y direction.

Solution: 1. (a) The horizontal velocity remains constant at 2.70 m/s.

2. (b) The vertical speed increases due to the acceleration of gravity:

( )( )29.81 m/s 2.10 s

–20.6 m s

yv gt= − = −

=

3. (c) If the speed of the crow were to increase, the speed of the clam in the x-direction would increase, but the speed in the y-direction would stay the same.

Insight: The speed of the crow determines xv and gravity determines yv . As in most of the other

problems in this section, the answers are only true as long as we neglect the effects of air friction. 10. Picture the Problem: The trajectory of the climber

is indicated in the figure at right.

Strategy: The 45− ° direction of motion indicates that, just prior to landing, the climber is falling with a speed equal to his horizontal speed. Use this fact (since the initial velocity is horizontal) to find the height difference of the crevasse and the landing point of the climber.

Solution: 1. (a) Use the fact that 0yv v=

to find the time of flight:

2

7.8 m/s

9.81 m/s

0.80 s

yvt

g= − = −

=

2. Find the vertical drop during the flight, which is also the height difference between the two sides of the crevasse:

( )( )22 21 1

2 2 9.81 m/s 0.80 s 3.1 mh y gt− = = =

3. (b) Find the horizontal distance travelled:

( )( )7.8 m/s 0.80 s 6.2 mx

x v t= = =

4. The climber lands 6.2 – 2.8 m = 3.4 m beyond the far edge of the w = 2.8 m wide crevasse.

Insight: The climber impacts the other side of the crevasse at about 25 mi/h (verify this for yourself!). It would be much safer to cross the crevasse with a ladder!


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11. Picture the Problem: The sparrow falls down along a parabolic arc, maintaining its horizontal velocity but gaining vertical speed as it falls.

Strategy: Find the time it takes the sparrow to travel a horizontal distance of 0.500 m given that its horizontal velocity remains unchanged at 1.80 m/s. Then find the distance the sparrow falls during that time interval.

Solution: 1. (a) Find the time to travel 0.500 m horizontally. m

s

0.500 m0.2778 s

1.80 x

xt

v= = =

2. Find the vertical drop distance: ( )( )22 21 1

2 2 9.81 m/s 0.2778 s 0.378 mh y gt− = = =

3. (b) If the sparrow’s initial speed increases, the time interval required for it to travel 0.500 m horizontally decreases. The distance of fall decreases for a shorter time interval.

Insight: The speed of the sparrow determines xv and gravity determines

yv . Flying faster will

increase xv but not

yv .

12. Picture the Problem: The pumpkin’s trajectory is depicted in

the figure at right.

Strategy: Because the pumpkin’s initial velocity is horizontal, we can find the required initial speed of the pumpkin.

Solution: Solve for

0v : ( )

( )( )

( )

222

0

9.81 m/s 3.5 m

2 2 9.0 0 m

2.6 m/s

g xv

h y= =

− −

=

Insight: Another way to solve this problem is to first find the time of fall using the height of the tower (1.36 s). Then calculate the horizontal speed required to hit the target during the time of fall.

13.

Picture the Problem: The stuffed animal’s trajectory is depicted in the figure at right.

Strategy: Determine the average speed of the riders on the Ferris wheel by dividing the circumference of the wheel by the time to complete a revolution. This becomes the initial speed of the stuffed animal that is launched horizontally. The initial speed and height of the stuffed animal determines the location it lands.

Solution: 1. (a) Find the circumference of the Ferris wheel:

( )2 2 5.00 m 31.4 mC rπ π= = =

2. Find the average speed of a rider:

31.416 m0.982 m/s

32.0 s

Cv

t= = =

∆

3. (c) Solve equation 4-8 for x given that

( )2 5.00 m 1.75 m 11.75 mh = + = :

( ) ( ) ( )22

02

2 2 0.982 m/s 11.75 0 m1.52 m

9.81 m/s

v h yx

g

− −= = =

Insight: Another way to solve this problem is to first find the time of fall (1.55 s) using the h = 11.75 m. Then calculate the horizontal distance travelled given a horizontal velocity of 0.982 m/s.


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14. Picture the Problem: The swimmer’s trajectory is depicted

in the figure at right.

Strategy: Because the swimmer’s initial velocity was horizontal we can find the initial height h.

Solution: 1. (a) Solve for h:

( )( )

( )

222

2 20

9.81 m/s 1.68 m0

2 2 3.62 m/s

1.06 m

gxh y

v= + = +

=

2. (b) It takes the same time to reach the water. Gravity and the vertical distance, not the horizontal speed, determine the time of flight.

Insight: In order to increase the time of flight the swimmer should launch herself at least partly in the upward direction.

15. Picture the Problem: The basketball’s trajectory is

depicted at right.

Strategy: Find the x and y positions of the basketball as a function of time. Use the right triangle formed by the floor and the basketball’s release and landing points to write a ratio that allows us to calculate the time of flight and therefore the initial height.

Solution: 1. Find the y position as a function of time:

212

212

0y h gt

h gt

= − =

=

2. Find the x position as a function of time:

0x v t=

3. Use the tangent function for the right triangle:

212

0 0

tan2

gth gt

x v t vθ = = =

4. Now solve for the flight time t: ( )0

2

2 4.20 m/s tan 30.02 tan0.495 s

9.81 m/s

vt

g

θ °= = =

5. Find the initial height: ( )( )22 21 1

2 2 9.81 m/s 0.495 s 1.20 mh gt= = =

Insight: If the basketball player throws the ball from the same height but with a higher initial speed, the 30.0° angle will decrease. For instance, 0 8.40 m/sv = produces an angle of 16.1°. Dropping the ball from rest makes the angle 90.0°.

16. Picture the Problem: The ball falls down along a parabolic arc, maintaining its horizontal velocity

but gaining vertical speed as it falls.

Strategy: The ball is accelerated only by gravity. The initial height of the ball determines the vertical component of its final velocity, which together with the final speed can be used to find the horizontal component of the velocity. The initial speed equals the horizontal component because it remains the same throughout the flight.

Solution: 1. (a) The acceleration is due only to gravity:

29.81 m/s downward=ar

2. (b) Find the vertical component of the final velocity: ( )( )

2

2

2

2 9.81 m/s – 0.75 m 3.84 m/s

y

y

v g y

v

= − ∆

= − =


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3. Find the horizontal component of the ball’s velocity, which is also its initial speed:

( ) ( )2 22 2 4.0 m/s 3.84 m/s 1.1 m/s

x yv v v= − = − =

4. (c) Repeat for a final speed of 5.0 m/s: ( ) ( )

2 22 2 5.0 m/s 3.84 m/s 3.2 m/sx y

v v v= − = − =

Insight: The vertical motion of an object dropped from rest or launched horizontally is determined solely by the initial height and the acceleration of gravity. That is why the vertical component of the final velocity is the same in part (c) as it is in part (b).

17. Picture the Problem: The cork travels along a parabolic arc, maintaining its horizontal velocity

but changing its vertical speed due to the constant downward acceleration of gravity.

Strategy: Find the horizontal component of the initial velocity by dividing the horizontal distance travelled by the time of flight. Then use the cosine function to find the initial speed of the cork.

Solution: 1. Find the horizontal speed of the cork:

0

1.30 m1.04 m/s

1.25 sx x

xv v

t= = = =

2. Use the cosine function to find the initial speed:

00

1.04 m/s1.27 m/s

cos cos35.0xv

vθ

= = =°

Insight: Because gravity acts only in the vertical direction, the horizontal component of the cork’s velocity remains unchanged throughout the flight.

18. Picture the Problem: The ball travels along a parabolic arc, maintaining its horizontal velocity but

changing its vertical speed due to the constant downward acceleration of gravity.

Strategy: The given angle of the kick allows us to calculate the vertical component of the initial velocity by using the sine function. The time it takes the acceleration of gravity to slow down the vertical speed, bring it to zero, and speed it up again to its initial value is the same as the time the ball is in the air.

Solution: 1. Find the y component of the initial velocity:

( )0 0 sin 9.50 m/s sin 25.0 4.01 m/sy

v v θ= = ° =

2. Let 0y yv v= − and find the time of flight:

( )0

2

4.01 4.01 m/s0.818 s

9.81 m/sy yv v

tg

− − −= = =

− −

Insight: Another way to solve this problem is to find the time it takes gravity to bring 0 yv to zero.

That time corresponds to the peak of flight, and since the flight is symmetric we need only double that time to find the time of flight.

19. Picture the Problem: The trajectory of the ball is

shown at right.

Strategy: Find the time the ball is in the air. Then use the known initial speed and angle to find the horizontal speed of the ball, which together with the time of flight can be used to find the horizontal distance travelled. In this case y = 0 corresponds to the point where the ball leaves the forward’s hands, and the floor corresponds to 0.80 m.y = −

Solution: 1. Find the time: ( )( ) ( ) ( )

( ) ( )

210 2

2 212

2 2

sin

0.80 m 4.3 m/s sin 15 9.81 m/s

0 0.80 m 1.1 m/s 4.905 m/s

y v t gt

t t

t t

θ= −

− = − ° −

= + − + −


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2. Apply the quadratic formula: ( ) ( )( )

( )

22 1.1 1.1 4 4.905 0.804

2 2 4.905

0.53, 0.31 s

b b act

a

t

± − − −− ± −= =

−

= −

3. Find the horizontal distance travelled:

( ) ( ) ( )( )0 0 cos 4.3 m/s cos –15 0.31 s 1.3 mxx v t v tθ= = = ° =

Insight: Another way to solve this problem is to use 2 20 2y yv v g y= − ∆ to find yv , then use yv , 0 yv

and the acceleration of gravity to find the time of flight and hence the horizontal distance travelled. Such an approach requires an extra step but avoids the quadratic formula.

20. Picture the Problem: The trajectories of the snowballs are

depicted at right.

Strategy: Find the vertical component of the final velocity of each snowball, as well as the horizontal component of each velocity. Use the known components to determine the landing speed.

Solution: 1. (a) The landing speed of snowball A is the same as that of snowball B, because the landing speed is independent of launch angle.

2. (b) Find the vertical component of the final velocity for snowball A: ( ) ( ) ( )( )

( )

2 20

2 2 2

sin 2

13 m/s sin 90 2 9.81 m/s 7.0 m

18 m/s The snowball is traveling downward

yv v g yθ= ± − ∆

= ± − ° − −

= −

3. Find the horizontal component of the velocity for snowball A:

( )0 0 cos 90 0 m/sx xv v v= = − ° = .

4. Find the landing speed of snowball A:

( )22 2 218 m/s 0 18 m/s

y xv v v= + = − + =

5. Find the vertical component of the final velocity for snowball B: ( ) ( ) ( )( )

( )

2 20

2 2 2

sin 2

13 m/s sin 25 2 9.81 m/s 7.0 m

13 m/s The snowball is traveling downward

yv v g yθ= ± − ∆

= ± ° − −

= −

6. Find the horizontal component of the velocity for snowball B:

( ) ( )0 13 m/s cos 25 12 m/sx x

v v= = ° =

7. Find the landing speed of snowball B:

( ) ( )2 22 2 13 m/s 12 m/s 18 m/s

y xv v v= + = − + =

Insight: Conservation of mechanical energy provides an even clearer explanation of why the landing speeds of the snowballs are the same regardless of launch angle.

21. Picture the Problem: The golf ball travels along a parabolic arc, landing at the same level from

which it was launched.

Strategy: The maximum range of a projectile launched from level ground occurs when the launch angle is 45°. Predict the range of the golf ball when launched at 45°. The minimum speed of the ball will occur when the ball reaches the peak of its flight. At that point the vertical component of the velocity is zero and the speed equals the horizontal component of the velocity, which remains unchanged throughout the flight.


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Solution: 1. (a) Find the range of the ball when it is launched at 45°:

( )22

02

30.0 m/ssin 2 sin 90 91.7 m

9.81 m/s

vR

gθ

= = ° =

2. (b) Find the x component of the ball’s velocity, which corresponds to the minimum speed during the flight:

( )0 0 cos 30.0 m/s cos 45 21.2 m/sxv v θ= = ° =

Insight: The maximum range will occur at θ =45° only in the absence of air resistance. In the presence of the atmosphere you must launch the ball at a lower angle than that in order to maximize the range of the ball.


which it was launched.

Strategy: The highest tree the ball in the previous problem could clear would be a tree that is just shorter than the maximum altitude achieved by the ball. Use the maximum height formula to find the maximum height of a ball launched at 45° above the horizontal with a speed of 30.0 m/s.

Solution: 1. Find the maximum height of the ball:

( ) ( )

( )

22

0max 2

30.0 m/s sin 45sin22.9 m

2 2 9.81 m/s

vy

g

θ ° = = =

Insight: To hit a hole-in-one is amazing enough, but to clear a 23m tree along the way is almost unbelievable!

23. Picture the Problem: The path of the ball is depicted at

right.

Strategy: Use the horizontal component of the ball’s velocity together with the horizontal distance d to find the time elapsed between the hit and its collision with the wall. Then use the time to determine the vertical position h of the ball when it collides with the wall.

Solution: 1. (a) Find the time: ( )0

3.8 m

cos 14 m/s cos34

0.33 s

dt

v θ= =

°

=

2. (b) Find h: ( )

( ) ( ) ( )( )

210 2

2212

sin

14 m/s sin 34 0.33 s 9.81 m/s 0.33 s 2.0 m

h v t gtθ= −

= ° − =

Insight: In many cases the vertical motion determines the time of flight, but in this case it is the horizontal distance between the point where the ball is struck and the wall that limits the time of flight.

24. Picture the Problem: The trajectory of the girl

is depicted at right.

Strategy: Use the given time of flight, initial speed, and launch angle to determine the initial height of the girl at the release point.

Solution: Find the initial

height of the girl at the release point. If we let the release height correspond to y = 0, then the landing height is:

( )

( )( )( ) ( )( )

210 2

2212

sin

2.25 m/s sin 35.0 0.616 s 9.81 m/s 0.616 s

1.07 m

y v t gt

y

θ= −

= ° −

= −

In other words, she was 1.07 m above the water when she let go of the rope.

Insight: The girl’s speed upon impact with the water is 5.10 m/s.

d

h

θ

y

x

0vr


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25. Picture the Problem: The projectile travels along a parabolic arc, maintaining its horizontal

velocity but changing its vertical speed due to the constant downward acceleration of gravity.

Strategy: At the projectile’s highest point, 0.yv = Therefore the 0 4v corresponds to the

horizontal component of the velocity 0xv , which remains constant throughout the flight. Use the

cosine function together with 0v and 0xv to determine the launch angle θ.

Solution: Use the known 0v and 0xv to find θ:

0 0

1 1 10 0

0 0

cos

4 1cos cos cos 75.5

4

x

x

v v

v v

v v

θ

θ − − −

=

= = = = °

Insight: If the launch angle were to increase, the minimum speed would decrease until it becomes zero when the launch angle is 90° (straight upward). When the launch angle is 0° (horizontal) then the minimum speed is 0 0xv v= .

26. Picture the Problem: The dolphin travels along a parabolic arc as it glides through the air and lands in the water.

Strategy: Because the dolphin is travelling horizontally as it passes through the hoop, we conclude that 0yv = at that point and that the dolphin must be at the peak of its flight. Use the appropriate

formula to find how high the centre of the hoop is above the surface of the water.

Solution: Find the maximum height of the dolphin above the water:

( ) ( )

( )

22

0max 2

12.0 m/s sin 40.0sin3.03 m

2 2 9.81 m/s

vy

g

θ ° = = =

Insight: In order to pass through a higher hoop the dolphin must either increase the launch angle or jump with a higher initial speed.

27. Picture the Problem: The ball travels along a parabolic arc, landing at the same level from which it was launched.

Strategy: Because the ball lands at the same level from which it was launched, we can find the launch angle θ. The time of flight can then be found from the horizontal component of the velocity and the range.

Solution: 1. (a) Solve for θ:

( )( )

( )

2

1 12 20

9.81 m/s 4.6 m1 1sin sin 32

2 2 7.1 m/s

gR

vθ − −

= = = °

2. (b) Find the time of flight:

( )0

4.6 m0.76 s

cos 7.1 m/s cos32

xt

v θ= = =

°

Insight: Verify for yourself that the maximum height above the level of the throw and catch is 0.72 m, and that for a launch speed of 7.1 m/s the maximum range of the ball would be 5.1 m.

28. Picture the Problem: The ball travels along a parabolic arc, landing at the same level from which it was struck.

Strategy: Because the ball lands at the same level from which it was struck, we can use the relevant formulas to solve this problem. We can get a single formula for the range as a function of time of flight and launch angle, and solve it for launch angle. Once the launch angle is known, the magnitude of the initial velocity can be found.


42 of 65

Solution: 1. (a) Combine:

2 2 2200 2

2 2 2

2

2sin cos Eq. 4-12 and Eq. 4-11

4sin

2sin cos

2 tan4sin

v g tR v

g

g t gtR

g

θ θθ

θ θθθ

= =

= =

2. Solve the equation for θ:

( )( )

( )

2221 1

9.81 m/s 4.30 stan tan 45.5

2 2 92.2 m

g t

Rθ − −

= = = °

3. (b) Solve for 0v :

( )( )2

0

9.81 m/s 4.30 s30.1 m/s

2sin 2sin 44.5

g tv

θ= = =

°

Insight: Another way to solve this problem is to figure out the x component of the velocity (21.4 m/s) from the range and the time of flight. The y component of the initial velocity (21.1 m/s) can be determined from the fact that gravity reduces it to zero in half the time of flight (2.15 s). The two components can be used to find the direction and speed.

29. Picture the Problem: The lava that reaches maximum altitude is hurled straight upward, comes to

rest momentarily, and falls straight downward again.

Strategy: Determine the launch speed required to achieve the maximum altitude.

Solution: 1. (a) Find 0v :

( )( )

2 20

2 50

2 0 at the peak of flight

2 2 1.80 m/s 2.00 10 m 849 m/s

y y

y

v v g y

v g y

= − ∆ =

= ∆ = × =

2. (b) If the launch speed remains the same, the maximum height of the ejected lava on Earth would be less than it is on Io because the acceleration of gravity on Earth is much greater than 1.80 m/s2.

Insight: On Earth a launch speed of 849 m/s would be 2.5 times the speed of sound—roughly equal to the muzzle velocity of a rifle bullet! Tidal forces from Jupiter produce these violent volcanic eruptions on Io.

30. Picture the Problem: The football travels along a parabolic arc, maintaining its horizontal velocity

but changing its vertical speed due to the constant downward acceleration of gravity.

Strategy: Find the components of the football’s velocity as a function of time. The components can then be used to find the magnitude and direction of the velocity at each instant. The horizontal component of the velocity remains the same throughout the flight.

Solution: 1. (a) Find the horizontal component of the velocity:

( )0 0 cos 10.2 m/s cos 25.0 9.24 m/sx x

v v v θ= = = ° =

2. Find yv at 0.250

s:

( ) ( )( )20 sin 10.2 m/s sin 25.0 9.81 m/s 0.250 s 1.86 m/syv v gtθ= − = ° − =

3. Find the magnitude of the velocity:

( ) ( )2 22 2 9.24 m/s 1.86 m/s 9.43 m/s

x yv v v= + = + =

4. Find the direction of the velocity:

1 1.86 m/stan 11.4

9.24 m/sθ −

= = °

or 11.4° above horizontal

5. (b) Find yv at

0.500 s:

( ) ( )( )20 sin 10.2 m/s sin 25.0 9.81 m/s 0.500 s 0.589 m/s

yv v gtθ= − = ° − = −

6. Find the magnitude of the velocity:

( ) ( )2 22 2 9.24 m/s 0.589 m/s 9.26 m/s

x yv v v= + = + − =


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7. Find the direction of the velocity:

1 0.589 m/stan 3.68

9.24 m/sθ − −

= = − °

or 3.68° below horizontal

8. (c) The ball is at its greatest height before 0.500 s because at 0.500 s the vertical component of its velocity is already negative, or downward.

Insight: You can verify for yourself that the ball reaches its maximum altitude of 0.947 m at 0.440 s after the kick.

31. Picture the Problem: The soccer ball travels along a parabolic arc, maintaining its horizontal


Strategy: Find the initial vertical velocity of the ball from the time interval between the kick and the peak of the flight. Then use the known magnitude of the initial velocity together with its vertical component in order to find the direction of the kick.

Solution: 1. When the ball is at the peak of flight its vertical speed is zero:

00 sinyv v gtθ= = −

2. Solve the equation for θ:

( )( )2

1 1

0

9.81 m/s 0.750 ssin sin 46.2

10.2 m/s

gt

vθ − −

= = = °

Insight: You can verify for yourself that with the new angle of the kick the maximum height is 2.76 m and the downfield range is 10.6 m.


which it was struck.

Strategy: Because the ball lands at the same level from which it was struck, we can use the range formula to solve this problem. We can find the ball’s downfield range, and utilize the fact that the range is proportional to ( )sin 2θ in order to determine the second angle that will produce the same

range.

Solution: 1. (a) Find the rang: ( )

( )

220

2

42.0 m/ssin 2 sin 70.0 169 m

9.81 m/s

vR

gθ= = ° =

2. (b) Because the range depends upon ( )sin 2θ , there are two launch angles that

will produce the same range: θ and 90°-θ.

2 190 90 35.0 55.0θ θ= ° − = ° − ° = °

Insight: You can verify for yourself that ( ) ( )sin 70 sin 110° = ° and that is why the two ranges are

the same for θ =35.0° and θ =55.0°. 33. Picture the Problem: The hay bale travels along a parabolic arc, maintaining its horizontal


Strategy: The initial velocity of the bale is given as ( ) ( )0 ˆ ˆ1.12 m/s 8.85 m/s= +v x yr

. Use the fact

that the horizontal component of the bale’s velocity never changes throughout the flight in order to find the vertical component of the velocity when the total speed is 5.00 m/s. Then find the time elapsed between the initial throw and the instant the bale has that vertical speed. For part (b) set the vertical speed equal to the (constant) horizontal speed in magnitude but negative in direction (pointing downward). Find the time elapsed between initial throw and the instant the bale has that new vertical speed.

Solution: 1. (a) Determine the y component of the velocity when the total speed is 5.00 m/s: ( ) ( )

2 2 20

2 22 20 5.00 m/s 1.12 m/s 4.873 m/s

x y

y x

v v v

v v v

= +

= ± − = ± − = ±


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2. Use the positive value of yv since

the bale is rising when the speed first equals 5.00 m/s. Find the time elapsed to this point from equations 4-6:

0

2

4.87 m/s 8.85 m/s0.405 s

9.81 m/sy y

v vt

g

− −= = =

− −

3. (b) If the bale’s velocity points 45.0° below the horizontal then we know the vertical velocity:

0 1.12 m/sy xv v= − = −

4. Find the time elapsed: 0

2

1.12 m/s 8.85 m/s1.02 s

9.81 m/sy yv v

tg

− − −= = =

− −

5. (c) If 0vr

is pointed straight upward then the initial vertical velocity component will be larger, so

it will rise higher and its time in the air will increase.

Insight: The bale will have a speed of 5.00 m/s again after 1.40 s has elapsed. If it were thrown straight upward with the same initial speed (8.92 m/s) it would rise to a height of 4.06 m, as opposed to 3.99 m as in the original case.


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1. Picture the Problem: The image shows a wave with

the given wave dimensions.

Strategy: Set the wavelength equal to the horizontal crest-to-crest distance, or double the horizontal crest-to-trough distance. Set the amplitude equal to the vertical crest-to-midline distance, or half the vertical crest-to-trough distance.

Solution: 1. (a) Double the horizontal crest-to-trough distance:

( )2 26 cm 52 cmλ = =

2. (b) Halve the vertical crest-to-trough distance:

( )12 11 cm 5.5 cmA = =

Insight: Note the difference in wavelength and amplitude. The wavelength is the entire distance from crest to crest, but amplitude is only from the equilibrium point to the crest.

2. Picture the Problem: A surfer measures the frequency and length of the waves that pass her.

From this information we wish to calculate the wave speed.

Strategy: Write the wave speed as the product of the wavelength and frequency.

Solution: Multiply wavelength by frequency:

( )( )1 min

34 m 14 /min 7.9 m/s60 sec

v fλ

= = =

Insight: The wave speed can increase by either an increase in wavelength or an increase in frequency.

3. Picture the Problem: The image shows

water waves passing to a shallow region where the speed decreases. We need to calculate the wavelength in the shallow area.

Strategy: Calculate the frequency in the deep water. Then use the constant frequency and the speed in the shallow water to calculate the new wavelength.

Solution: 1. Calculate the frequency:

1

1

2.0 m/s1.333 Hz

1.5 m

vf

λ= = =

2. Calculate the new wavelength: 2

2

1.6 m/s1.2 m

1.333 Hz

v

fλ = = =

Insight: Note that decreasing the speed, with constant frequency, will decrease the wavelength. 4. Picture the Problem: The speed and wavelength of a tsunami are given and we wish to calculate

the frequency.

Strategy: Solve for the frequency.

Solution: Calculate the frequency: ( ) 4750 km/h 1 h

6.7 10 Hz310 km 3600 s

vf

λ−

= = = ×

Insight: Although the tsunami has a very high speed, the long wavelength gives the tsunami a low frequency.


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5. Picture the Problem: A wave of known amplitude, frequency, and wavelength travels along a string. We wish to calculate the distance travelled horizontally by the wave in 0.5 s and the distance travelled by a point on the string in the same time period.

Strategy: Multiply the time by the wave speed, where the wave speed is given, to calculate the horizontal distance traveled by the wave. A point on the string travels up and down a distance four times the amplitude during each period. Calculate the fraction of a period by dividing the time by the time of a full period. Set the period equal to the inverse of the period and multiply by four times the amplitude to calculate the distance traveled by a point on the string.

Solution: 1. (a) Calculate the horizontal distance:

( ) ( )( )( )2w 27 10 m 4.5 Hz 0.50 s 0.61 md vt f tλ −= = = × =

2. (b) Calculate the vertical distance:

( ) ( )( )( )2k 4 4 4 12 10 m 4.5 Hz 0.50 s 1.1 m

td A Aft

T

− = = = × =

3. (c) The distance travelled by a wave peak is independent of the amplitude, so the answer in part (a) is unchanged. The distance travelled by the knot varies directly with the amplitude, so the answer in part (b) is halved.

Insight: A point on the string travels four times the wave amplitude in the same time that the crest travels one wavelength.

6. Picture the Problem: Using the equation for the speed of deep water waves given in the problem we

want to calculate the speed and frequency of the waves.

Strategy: Insert the given data into the equation 2v gλ π= to solve for the speed of the waves.

Then calculate the wave frequency.

Solution: 1. (a) Insert the frequency into the deep water velocity equation:

( ) ( )( ) ( )2/ 2 9.81 m/s 4.5 m / 2 2.65 m sv gλ π π= = =

2. (b) Solve for the frequency: 2.651 m/s

0.59 Hz4.5 m

vf

λ= = =

Insight: Since the velocity is proportional to the square-root of the wavelength, the frequency is inversely proportional to the square-root of the wavelength. Increasing the wavelength by a factor of four will double the wave speed and cut the frequency in half.

7. Picture the Problem: The speed of shallow water waves is proportional to the square-root of the

water depth. We wish to calculate the speed and frequency of some shallow water waves.

Strategy: Use the speed equation v gd= given in the problem, where d is the water depth, to calculate the wave speed. Then calculate the wave frequency.

Solution: 1. (a) Calculate the wave speed:

( )( )29.81 m/s 0.026 m 0.51 m sv gd= = =

2. (b) Calculate the wave frequency: 0.505 m/s

67 Hz0.0075 m

vf

λ= = =

Insight: As the wave approaches shallower water, with constant frequency, its wavelength decreases. In this problem, if the depth drops to 1.3 cm, the wavelength will decrease to 0.59 cm.

8. Picture the Problem: The string tension is changed until the wave speed doubles.

Strategy: The speed of a wave on a string is given. Solve the equation for the tension in the string. Then use a ratio to find the factor by which the tension increases.

Solution: 1. Solve for the tension: 2 v F F vµ µ= ⇒ =

2. Divide the tension at higher velocity by the initial tension:

2 222 2 2

2

32 m/s4

16 m/s

F v v

F vv

µ

µ

= = = =

The tension increases by a factor of 4.

Insight: The tension increases by a factor equal to the square of the fractional increase in velocity.


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9. Picture the Problem: The image shows two

people talking on tin can telephone. The cans are connected by a 9.5-meter-long string weighing 32 grams. We wish to calculate the time it takes for a message to travel across the string.

Strategy: Set the time equal to the distance divided by the velocity, where the velocity is given. The linear mass density is the total mass divided by the length.

Solution: 1. Set the time equal to the distance divided by velocity:

dt d

v F

µ= =

2. Substitute m dµ = and insert numerical values:

( )( )0.032 kg 9.5 m/0.19 s

8.6 N

m d mdt d

F F= = = =

Insight: The message travels the same distance in the air in 0.028 seconds, about 7 times faster. 10. Picture the Problem: The image shows two

people talking on tin can telephone. The cans are connected by a 9.5-meter-long string weighing 32 grams. We wish to determine how the tension in the string affects the time for the message to travel across the string.

Strategy: In problem 9, we found that the travel time across the string is given by

/t md F= . Use this equation to calculate the time for the different tensions.

Solution: 1. (a) Since the time is inversely related to the tension, increasing the tension will result in less time.

2. (b) Set the tension equal to 9.0 N: ( )( ) ( )0.032 kg 9.5 m / 9.0 N 0.18 st = =

3. (c) Set the tension equal to 10.0 N: ( ) ( ) ( )0.032 kg 9.5 m / 10.0 N 0.17 st = =

Insight: As predicted, increasing the tension decreases the time for the message to travel the string.

11. Picture the Problem: Sound takes 0.94 seconds to travel across a wire of known length and

density. We want to calculate the tension in the wire.

Strategy: Solve for the tension in the wire. The velocity is given by the length of the wire divided by the time for the sound to travel across it. The linear mass density is the mass divided by the length.

Solution: 1. (a) Solve for the tension:

22

2

Fv

m L mLF v

L t t

µ

µ

=

= = =


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2. Insert the given mass, length and time:

( )( )

( )2

0.085 kg 7.3 m0.70 N

0.94 sF = =

3. (b) The mass is proportional to the tension (if L and t remain constant). So increased mass means increased tension.

4. (c) Solve with a mass of 0.095 kg.

( )( )

( )2

0.095 kg 7.3 m0.78 N

0.94 sF = =

Insight: A heavier string requires greater tension for a wave to travel across it in the same time. 12. Picture the Problem: Waves travel down two strings, made of the same material and having the

same length, but having different diameters and tensions. We wish to calculate the ratio of the wave speeds on these two strings.

Strategy: Calculate the ratio of the velocities. Set the linear mass densities equal to the density of steel times the cross-sectional area of the wires.

Solution: 1. (a) Write the ratio of the velocities: A A A A

B B B B

/

/B

A

v F F

v F F

µ µ

µ µ= =

2. Write the linear mass density in terms of density and area:

A A A

B B B

B B

A A

v F A F A

v F A F A

ρ

ρ= =

3. Write the area in terms of the diameter: ( )

( )

2

A A A2

B B B

2

2

B B

AA

dv F F d

v F F dd

π

π= =

4. Insert the given tensions and diameters: A

B

410 N 1.0 mm2 1.4

820 N 0.50 mm

v

v

= = =

Insight: The ratio of the velocities is proportional to the square root of the tensions and inversely proportional to the diameters.

13.

Picture the Problem: The dolphin sends a signal to the ocean floor and hears its echo.

Strategy: We want to calculate the time the elapses before the dolphin hears the echo and the wavelength of the sound in the ocean. The wave must travel to the ocean floor and back before it is heard. So the distance travelled is twice the distance to the floor. Divide this distance by the speed of sound in water to calculate the time. Calculate the wavelength from equation 14-1.

Solution: 1. (a) Divide the distance by the speed of sound in water:

( )2 75 m20.098 s

1530 m/s

dt

v= = =

2. (b) Solve for the wavelength: 31530 m/s28 10 m 28 mm

55 kHz

v

fλ −= = = × =

Insight: In air the wavelength would be 6.2 mm. The wavelength is longer in the water because the wave travels faster in water, while the frequency is the same.

14. Picture the Problem: We need to calculate the wavelength of sound in air from its frequency.

Strategy: Solve for the wavelength, using 343 m/s for the speed of sound in air.

Solution: 1. (a) Solve for the wavelength:

343 m/s0.807 m

425 Hz

v

fλ = = =


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2. (b) Examine the relationship between wavelength and frequency:

Wavelength is inversely related to frequency so, if the frequency increases the wavelength decreases.

3. (c) Calculate the wavelength at 450 Hz:

343 m/s0.722 m

475 Hzλ = =

Insight: As predicted, an increase in frequency corresponds to a decrease in wavelength. 15. Picture the Problem: The figure represents you dropping a rock

down a well and listening for the splash. From the time lapse between dropping the rock and hearing the splash we want to calculate the depth of the well.

Strategy: The time to hear the splash, t = 1.2 s, is the sum of the time for the rock to fall to the water, t1, and the time for the sound of the splash to reach you, t2. Solve the free-fall equation (equation 2-13) for the time to fall and displacement at constant velocity to calculate the time for the sound to return. Set the sum of these times equal to the time to hear the splash and solve for the distance.

Solution: 1. (a) Solve for the falling time: 1

2dt

g=

2. Solve for the time for the sound to travel up the well:

2s

dt

v=

3. Sum the two times to equal the total time:

1 2

2

s

d dt t t

g v= + = +

4. Rewrite as a quadratic equation in terms of the

variable d :

( )

( )

2

2

2

1 20

1 20 1.2s

343 m/s 9.81 m/s

s

d d tv g

d d

= + −

= + −

5. Solve for d using the quadratic formula and square the result:

3.2537 m 10.587 m 11 md d= ⇒ = =

6. (b) The time to hear the sound would be less then 3.0 seconds because, although the sound travel time would double, the fall time would less than double.

Insight: The time to hear the sound for a 21-meter-deep well is 2.1 s, which is indeed less than 3.0 s.

16. Picture the Problem: The figure shows a person throwing a rock

down an 8.80-m deep well. The sound of the splash reaches the person’s ear 1.20 seconds after the rock is thrown. We want to calculate the speed of the rock.

Strategy: Solve for the initial velocity of the rock, where the fall time is equal to the total time minus the sound travel time. The sound travel time is the depth of the well divided by the speed of sound.

Solution: 1. Calculate st : s

8.80 m0.0257 s

343m/s

dt

v= = =

2. Subtract st from the total time:

f 1.2 s 0.0257 s 1.1743 st = − =


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3. Solve for 0v :

( )( )

20 0 f f

00 f

f

2

1

21

2

8.8 m 19.81 m/s 1.1743 s 1.7 m s

1.1743 s 2

y y v t at

y yv at

t

= + +

−= −

−= − − = −

The initial velocity of the rock is 1.7 m/s downward

Insight: Even though the speed of sound is much larger than the speed of the rock, the time for the sound to travel up the well is significant. If the sound travel time was not included, the initial velocity of the rock would incorrectly be calculated as 1.4 m/s, which is 15% off of the actual velocity.

17. Picture the Problem: We are given the intensity level in a truck and need to calculate the sound

intensity.

Strategy: Convert the sound intensity level to sound intensity.

Solution: Solve for the sound intensity:

( )( )

0

/100

12 2 92 /10 2

10 log

10

10 W/m 10 1.6 mW m

I

I

I Iβ

β

−

=

=

= =

Insight: Note that doubling the sound intensity from 1.6 mW/m2 to 3.2 mW/m2 increases the intensity level from 92 dB to 95 dB.

18. Picture the Problem: We want to determine how tripling the distance to the sound source affects

the sound intensity and the intensity level.

Strategy: Find the ratio of the final sound intensity to initial intensity. Insert the new intensity to calculate the change in intensity level.

Solution: 1. (a) Calculate the ratio of intensities:

2 222 2 1 1

21 2 11

I 4 1

I 3 94

P r r r

r rP r

π

π

= = = =

2. (b) Insert the new intensity 1

2 19I I= :

1192

20 0

10 log 10 log II

I Iβ

= =

3. (c) Solve for the change in intensity levels:

12 1

0

2 1

1 110 log 10 log 10 log

9 9

110 log 9.54

9

I

Iβ β

β β

= + = +

− = = −

Insight: If the initial intensity was 1.0 µW/m2, the initial intensity level would be 60.0 dB. The intensity three times the distance away from the source would be 0.11 µW/m2 and the intensity level would be 50.5 dB.

19. Picture the Problem: We are given a sound intensity and asked to calculate the sound intensity

which would create a intensity level 2.5 dB higher than that produced by the first sound intensity.

Strategy: Calculate the intensity level for Sound 1. Add 2.5 dB to the intensity level and then solve for the intensity of Sound 2.

Solution: 1. Calculate the intensity level for sound 1:

21

1 12 20

38.0 W/m10log 10log 135.80 dB

10 W/m

I

Iβ

−

= = =

2. Add 2.5 dB to the intensity level: 2 1 2.5 dB 135.80 dB 2.5 dB 138.30 dBβ β= + = + =


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3. Invert to calculate the intensity:

( )2

22

0

/10 12 2 138.30 /10 22 0

10log

10 10 W/m 10 67.6 W/m

I

I

I Iβ

β

−

=

= = =

Insight: An increase of 2.5 dB increases the intensity by a factor of 2.5 /1010 1.78.= 20. Picture the Problem: The intensity level at a distance of 2.0 meters is given. We want to find the

intensity levels at 12 m and 21 meters. We also want to find the distance for which the intensity level is 0, the furthest point at which the siren can be heard.

Strategy: Create an equation for intensity level in relation to distance. Use this relationship to calculate the intensity levels at 12 m and 21 m. To calculate the furthest distance at which the siren can be heard, set the intensity level to zero, and solve the relation for distance.

Solution: 1. Create equation:

2

2 1 1

0 2 0

2

1 1

0 2

10 log 10 log

10 log 10 log

I r I

I r I

I r

I r

β = =

= +

2. (a) Insert β at r1 = 2.0 m and set r2=12 m:

22.0 m

120 10 log 104 dB12 m

β

= + =

3. (b) Repeat for r2=21 m:

22.0 m

120 10 log 99.6 dB21 m

β

= + =

4. (c) Set the intensity level equal to zero:

2

1 1

0 2

2

10 log 10 log

2.0 m0 120 + 10 log

I r

I r

r

β

= +

=

5. Solve for r: 2

212 6

6

2.0 m10 log 120

2.0 m 2.0 m10 2.0 10 m

10

r

rr

−

−

− =

= ⇒ = = ×

Insight: This is a theoretical limit that could be realized in an ideal case. In a more realistic scenario, ambient noise, as well as energy losses when the sound waves are reflected or absorbed by surfaces, would prevent us from hearing the sound 2000 km away. Sometimes the real-world factors we ignore make a huge difference!

21. Picture the Problem: We are given the sound intensity of one hog caller and are asked to

calculate how many hog callers are needed to increase the intensity level by 10 dB.

Strategy: Multiply the intensity by N callers, setting the intensity level to 120 dB and solve for N.

Solution: 1. Write the intensity level for N callers:

( )0 0

10log 10log 10logNI I

NI I

β

= = +

2. Insert the intensity levels and solve for N:

( )

( )10 /10

120 10log 110

10 10log

10 10 callers

N

N

N

= +

=

= =

Insight: Increasing intensity level by 10 dB increases the intensity by a factor of 10. Therefore 10 callers, each with intensity level 110 dB, would produce a net intensity level of 120 db. 100 callers (10 × 10 callers) would be needed to produce an intensity level of 130 dB (120 dB + 10 dB).


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22. Picture the Problem: Twenty identical violins play simultaneously.

Strategy: Write equation for the intensity level of 20 violins and, using the rules of logarithms solve for the intensity level of one violin.

Solution: 1. (a) Write equation for 20 violins:

( )

( )

200 0

1

2010log 10log 10log 20

10log 20

I I

I Iβ

β

= = +

= +

2. Solve for the intensity level of one violin:

( )1 20 10log 20 82.5 13 69.5 dBβ β= − = − =

3. (b) Doubling the number of violins will increase the intensity level by ( )10log 2 3.0 dB= ,

therefore the intensity of 40 violins will be 82.5 dB + 3.0 dB = 85.5 dB, which is less than 165 dB.

Insight: The intensity is proportional to the number of instruments. However, the intensity level is related to the logarithm of the intensity, so it is not proportional to the number of instruments.

23. Picture the Problem: We are given the size of the eardrum and the intensity of sound hitting the

eardrum. From this information we want to calculate the energy per second (or power) received by the eardrum.

Strategy: Solve for the power in terms of the intensity and area. The area of the eardrum is the area of a circle with radius given in the problem. The intensities for the threshold of hearing (10-12

W/m2) and pain (1 W/m2) are given.

Solution: 1. Calculate the area of the eardrum:

( )22 3 5 24.0 10 m 5.0 10 m .A rπ π − −= = × = ×

2. (a) Solve for the power at the threshold of hearing: ( )12 5 2 17

2

W10 5.0 10 m 5.0 10 W

mP

− − − = × = ×

3. (b) Solve for the power at the threshold of pain: ( )5 2 –5

2

W1 5.0 10 m 5.0 10 W

mP

− = × = ×

Insight: The ear can receive acoustical power over a range of 12 orders of magnitude! 24.

Picture the Problem: The image shows Brittany 12.5 m north of a sound source and Phillip a distance RP

east of the sound source. The sound intensity at Brittany is twice that at Phillip. From this information we wish to calculate the distance RBP from Phillip to Brittany.

Strategy: Calculate the distance from Phillip to the sound source. Then use the Pythagorean Theorem to calculate the distance between Brittany and Phillip.

Solution: 1. Solve for the distance to Phillip:

( )

2

BP B

P

BP B

P

12.5 m 2 17.68 m

RI I

R

IR R

I

=

= = =

2. Find the hypotenuse of the right triangle: ( ) ( )

2 2

BP 12.5 m 17.68 m 21.7 mR = + =

Insight: Since the intensity at Brittany is greater than the intensity at Phillip, Brittany must be closer to the source than Phillip.


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25.

Picture the Problem: The train, a moving source, sounds its horn. We wish to calculate the frequency heard by a person standing near the tracks.

Strategy: Solve for the observed frequency, using the negative sign since the train is moving toward the observer.

Solution: Insert the given data into equation: ( ) ( )

( )

2

1 1136 Hz

1 – 1 31.8 m/s 343 m/s

1.50 10 Hz

f fu v

′ = =

−

= ×

Insight: If the train were moving away from the observer, he would hear a frequency of 124 Hz. 26. Picture the Problem: A stationary person

sounds a 136-Hz horn as a train approaches him at 31.8 m/s. We want to know at what frequency a passenger on the train hears the horn.

Strategy: This problem has a stationary source and an approaching observer, the passenger. Use equation with a plus sign for the observed frequency.

Solution: Insert the given data into equation:

( )

( ) ( )

1

1 31.8 m/s 343 m/s 136 Hz 149 Hz

f u v f′ = +

= + =

Insight: The frequency is slightly lower than the frequency found in problem 35, where the source was moving.

27. Picture the Problem: Two jets, one moving close to the speed of sound and the other stationary,

emit the same frequency of sound. We wish to calculate the frequencies heard by the pilots.

Strategy: Assume that the moving jet is headed toward the stationary jet. Use the equation for the frequency heard by the pilot in the moving jet, because he is a moving observer of a stationary source. Use the equation for the frequency heard by the pilot in the stationary jet, because she is a stationary observer of a moving source.

Solution: 1. (a) The pilot of the jet on the ground hears a greater Doppler shift, because Doppler effects are greater with a moving source.

2. (b) Solve for the frequency heard by the moving pilot:

( ) ( )( ) ( )( )1 1 0.825 495 Hz 1.825 495 Hz 903 Hzf u v f v v′ = + = + = =

3. (c) Solve for the frequency heard by the stationary pilot:

( )1 1 495 Hz

495 Hz 2.83 kHz1 1 0.825 0.175

f fu v v v

′ = = = =

− −

Insight: Note that as the speed of the jet approaches the speed of sound, the frequency heard by the moving pilot approaches 2f, while the frequency heard by the stationary pilot approaches ∞.


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28.

Picture the Problem: The figure shows two cyclists approaching each other at the same speed.

Strategy: We want to calculate the frequency at which Cyclist B hears Cyclist A’s horn. Both the source and the observer are moving. Since they are approaching each other, the plus sign is used in the numerator and the minus sign in the denominator.

Solution: 1. (a) Solve for the observed frequency:

( )( )

( )1 8.50 m/s 343 m/s1

315 Hz 0.33 kHz1 1 8.50 m/s 343 m/s

o

s

u vf f

u v

+ +′ = = =

− −

2. (b) Cyclist A speeding up will have the greatest effect on the Doppler shift. For equal changes in speed, moving-source effects are greater than moving-observer effects.

Insight: Increasing the Cyclist B’s speed by 1.5 m/s results in an observed frequency of 332 Hz, while increasing Cyclist A’s speed by 1.5 m/s results in an observed frequency of 333 Hz.

29. Picture the Problem: The image shows two trains travelling in the same direction at different speeds. The front train sounds a horn. We want to calculate the speed of the second train from the frequency heard by a passenger on the second train.

Strategy: Since both the source train and the observer train are moving, to solve for the speed of the second train. The observer train is moving toward the source train, so use the plus sign in the numerator. The source train is moving away from the observer train, so use the plus sign in the denominator also.

Solution: 1. Write equation with plus signs in numerator and denominator:

o

s

1

1

u vf f

u v

+′ =

+

2. Solve for the observer speed:

( )( )

( )( )s o

o s

1 1

1 1

f f u v u v

u v f f u v

′ + = +

= ′ + −

3. Insert the given data: ( ) ( )( )o 343 m/s 133 Hz 124 Hz 1 35.8 m/s 343 m/s 1

63.3 m/s

u = + −

=

Insight: Since the observed frequency is greater than the source frequency, the two trains have to be getting closer together. Therefore, the speed of the second train has to be greater than the speed of the first train. After the second train passes the first, the observed frequency will drop to 91.5 Hz.

30. Picture the Problem: Two cars travel away from each

other at the same speed, as shown in the figure. We want to calculate the speed of the cars based on the difference in emitted and observed horn frequencies.

Strategy: Since both the observer and source are moving, find the relationship between the source and observed frequencies. Since both vehicles are moving away from

each other, use the minus sign in the numerator and the plus sign in the dominator. Set the source and observer speeds equal to each other and solve for the speed.


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Solution: 1. Write equation 14-11 with o su u u= = :

( ) ( )

1

1

u vf f

u v

f v u f v u

−′ =

+

′ + = −

2. Solve for the speed u: ( )

205 Hz 192 Hz343 m/s 11.2 m s

205 Hz 192 Hz

f fu v

f f

′ − − = = = ′+ +

Insight: The relative speed of the two cars is 22.4 m/s. If the source car were stationary and the observer car was moving away at 22.4 m/s, the observed frequency would be slightly less than 192 Hz. If the observer were stationary while the source moved away at 22.4 m/s, the observed frequency would be slightly greater than 192 Hz.

31. Picture the Problem: As a train passes a crossing signal the observers on the train hear the pitch

drop to 2/3 of the pitch they heard as the train approached the signal.

Strategy: Assume the frequency emitted by the crossing signal remains constant. Use equation when the source is stationary. Use the plus sign in the equation to calculate f, the observed frequency as the train approaches the signal, and use the minus sign to calculate 2/3f, the frequency as the train moves away from the signal. Combine the two equations to solve for the speed.

Solution: 1. Write the observed frequency as the train approaches the signal:

( ) 01f u v f= +

2. Write the observed frequency as the train leaves the signal:

( )203 1f u v f= −

3. Divide the leaving equation by the approaching equation:

( )( )

23 0

0

1 2 1

1 3 1

u vf f u v

f u v f u v

− −= ⇒ =

+ +

4. Solve the expression from step 3 for u: ( ) ( )

( )

2 3

1 1343 m/s 68.6 m s

5 5

v u v u

u v

+ = −

= = =

Insight: The train is travelling about 150 mph.


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1. Picture the Problem: This is a units conversion problem

Strategy: Use the appropriate unit conversions to measure the distance of one light year in kilometres, the speed of light in light years per year, and the speed of light in feet per nanosecond.

Solution: 1. (a) Express 1 ly in km:

( )8 12m 86,400 s 1 km1.00 ly 3.00 10 365 d 9.46 10 km

s d 1000 m

= × = ×

2. (b) Write the speed of light in lightyears per year:

1.00 ly yc =

3. (c) Convert the speed of light to units of feet per nanosecond:

89

m 1 s 1 ft3.00 10 0.984 ft/ns

s 0.3048 m10 nsc

= × =

Insight: In solving problems involving the light year it is helpful to remember that the unit ly is a unity of distance, where ( )1 y ,d c t c= = the speed of light times one year.

2. Picture the Problem: The distance to Alpha Centauri is given in lightyears.

Strategy: Convert the distance from light years to meters.

Solution: Convert to meters:

15169.46 10 m

4.3 ly 4.1 10 m1 ly

d ×

= = ×

Insight: The light that we see from Alpha Centauri, left the star 4.3 years before we see it. 3. Picture the Problem: Radio waves travel from Mars to Earth at the speed of light.

Strategy: Multiply the speed of light by the time to calculate the distance to Mars.

Solution: Calculate the distance: ( )( )8 1160 s

3.00 10 m/s 12 min 2.2 10 m1 min

d c t

= = × = ×

Insight: The distance between Earth and Mars varies between 107.8 10 m× and 113.8 10 m× , so the time for a signal to reach Earth from Mars can vary between 4.3 minutes and 21 minutes.

4. Picture the Problem: The wavelength of light from a star that is travelling away from Earth is

Doppler-shifted to a longer wavelength.

Strategy: Replace the frequency with the wavelength given in c = fλ. Solve the resulting expression for the ratio of the shifted wavelength to the unshifted wavelength.

Solution: 1. Write the Doppler shift in terms of wavelength:

1 1u c c u

f fc cλ λ

′ = − ⇒ = − ′

2. Solve for the ratio of wavelength: ( ) ( )7 8

1 11.14

1 1 3.65 10 m/s 3.00 10 m/su c

λ

λ

′= = =

− − × ×

Insight: Note that the wavelength is increased when the star moves away from Earth and decreased when the star moves toward Earth.


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5. Picture the Problem: The wavelength of light from a star that is moving toward Earth is Doppler-shifted to a shorter wavelength.

Strategy: Replace the frequency with the wavelength given in c = fλ. Solve the resulting expression for the shifted wavelength divided by the unshifted wavelength.

Solution: 1. (a) Because the star is moving toward Earth, the frequencies of the electromagnetic waves are increased. However, the measured wavelengths are less than what they would be if the star were at rest relative to us because the wavelengths are inversely proportional to the frequencies

2. (b) Write the Doppler shift in terms of wavelength:

1 1u c c u

f fc cλ λ

′ = + ⇒ = + ′

3. Solve for the ratio of wavelength:

( ) ( )7 8

1 10.892

1 1 3.65 10 m/s 3.00 10 m/su c

λ

λ

′= = =

+ + × ×

Insight: The wavelength is shorter than it is when the star is at rest relative to Earth, as predicted. 6. Picture the Problem: The frequency of light from a distant galaxy is reduced by 15% because of

the Doppler effect.

Strategy: Solve for the speed of the galaxy when the frequency equals 85% of the emitted frequency.

Solution: 1. (a) According to the Doppler effect, the frequency of electromagnetic radiation as measured by an observer is less than what it was when emitted if the source is receding. Therefore, the galaxy is moving away from Earth.

2. (b) Set the shift frequency to 85% initial frequency:

( )1 0.15 0.85 1u

f f f fc

′ = − = = −

3. Solve for the speed:

( )

0.85 1

1 0.85 0.15

u

c

u c c

= −

= − =

Insight: Note that the percent change in frequency is equal to fraction of the speed of light that the galaxy is travelling.

7. Picture the Problem: Light travels

from one lantern a distance d and then back to the first lantern. The distance d is to be large enough that the reaction time of 0.2 s will produce an error less than 15% in the speed of light.

Strategy: Set the speed of light equal to twice the distance divided by the travel time. Add the reaction time to the

total time and set the result equal to 85% of the speed of light. Solve the resulting equation for the minimum distance.

Solution: 1. Calculate the travel time without the reaction time:

2 2

dc t d c

t= ⇒ =

2. Add the reaction time and set the result equal to 85% of the speed of light:

( )2

1 0.15 0.85d

c ct t

= − =+ ∆


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3. Eliminate the variable t and solve for the distance:

( )

( )( )

( )

8

8

1 1

20.85

2

3.00 10 m/s 0.2 s2 10 m

2 0.85 1 2 0.85 1

dc

d c t

c td

− −

=+ ∆

×∆= = = ×

− −

Insight: This distance is about half way to the Moon!

8. Picture the Problem: The figure shows Michelson’s eight-sided mirror for determining the speed of light.

Strategy: Set the speed of light equal to twice the distance to the mirror divided by the time to rotate the mirror one eighth of a revolution, were the time is the angular distance divided by the angular speed.

Solution: 1. Write an expression for c:

2 2 2

/

d d dc

t

ω

θ ω θ= = =

∆ ∆ ∆

2. Insert numerical values:

( )( )3

8

18

2 35.5 10 m 528 rev/s3.00 10 m/s

revc

×= = ×

Insight: If the mirrors rotated at only 415 rev/s the distance to the fixed mirror would need to increase to 45.2 km.

9. Picture the Problem: Radio signals travel from Earth to a distant spacecraft.

Strategy: Divide the distance by the speed of light to calculate the time for the signal to reach the craft.

Solution: Calculate the time:

124

8

4.5 10 m1.5 10 s

3.00 10 m/s

dt

c

×∆ = = = ×

×

Insight: This time delay is 4 hours and 10 minutes. When NASA sends a signal to the craft it takes 8 hours and 20 minutes for NASA to receive a confirmation from the satellite.

10. Picture the Problem: Sound travels from the cricket bat to the daughter in the stands 115 m away

at the speed of sound. The sound also travels as a radio wave (at the speed of light) to the father a distance 132 km away.

Strategy: Calculate the time at which each person hears the sound by dividing the distance by the speed.

Solution: 1. Calculate the time for the daughter:

daughter

115 m0.335 s

343 m/s

dt

v∆ = = =

2. Calculate the time for the father:

34

father 8

132 10 m4.40 10 s

3.00 10 m/s

dt

c

−×∆ = = = ×

×

3. The father hears the six first.

Insight: In the time the sound took the reach the daughter in the stands, the radio signal could have travelled around the world two and a half times!


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11. Picture the Problem: As a motorist approaches a yellow signal light, the motorist sees the light

Doppler shifted to green.

Strategy: Calculate the two frequencies from the wavelengths, and then insert the frequencies into c = fλ to solve for the speed.

Solution: 1. (a) Calculate the frequencies:

( ) ( )( ) ( )

8 9 14

8 9 14

3.00 10 m/s 590 10 m 5.0847 10 Hz

3.00 10 m/s 550 10 m 5.4545 10 Hz

f c

f c

λ

λ

−

−

= = × × = ×

′ ′= = × × = ×

2. Solve for the speed: ( )14

8 714

1

5.45 10 Hz1 3.00 10 m/s 1 2.2 10 m/s

5.08 10 Hz

f f u c

fu c

f

′ = +

′ ×= − = × − = ×

×

3. (b) The wavelength decreases if the motorist travels toward the traffic light.

Insight: If the motorist were travelling away from the light, it would have a wavelength of 640 nm and appear red.

12. Picture the Problem: Light of frequency 5.000×1014 Hz is emitted from a galaxy that is receding

from Earth at a speed of 3325 km/s.

Strategy: Solve for the observed frequency, using the negative sign because the galaxy is receding.

Solution: Calculate the observed frequency:

314 14

8

3325 10 m/s1 5.000 10 Hz 1 4.945 10 Hz

3.00 10 m/s u

f fc

× ′ = − = × − = ×

×

Insight: The observed frequency is lower than the emitted frequency because the galaxy is receding.

13. Picture the Problem: The wavelength of light that is sent from the Enterprise to the Constitution

is shortened due to the Doppler effect as the two starships approach each other.

Strategy: Replace the frequency with the wavelength given. Solve the resulting equation for the emitted wavelength:

Solution: 1. Write the Doppler shift in terms of wavelengths:

( ) ( )1 1c c

f f u c u cλ λ

′ = + ⇒ = +′

2. Solve for the emitted wavelength:

3

8

722.5 10 m/s1 670.3 nm 1 671.9 nm

3.00 10 m/s

u

cλ λ

× = ′ + = + =

×

Insight: If the ships were receding at the same speed, the Constitution would observe a wavelength of 673.5 nm.

14. Picture the Problem: As a ball travels toward a radar gun, it receives a Doppler-shifted frequency

and reflects that frequency back to the gun. The frequency is again shifted as the ball’s a moving source.

Strategy: Solve for the change in frequency received by the ball. Double this change to calculate the change in frequency received back by the radar gun.

Solution: 1. Calculate the change in frequency received by the ball:

( )( )9

8

1

10.525 10 Hz 90.0 mph 0.447 m/s1410 Hz

mph3.00 10 m/s

u f uf f f

c c

f uf f

c

′ = + = +

× ′ − = = =

×

2. Double change in frequency: ( ) ( )2 2 1410 Hz 2.82 kHzf f f f′′ ′− = − = =

Insight: This represents only a 2.7×10−5 % change in frequency.


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15. Picture the Problem: As a car travels away from a radar gun, it receives a Doppler-shifted frequency and reflects that frequency back toward the gun. The frequency is again shifted because the car is a moving source.

Strategy: Solve for the change in frequency received by the car. Double this change to calculate the change in frequency received back by the radar gun.

Solution: 1. Calculate the change in frequency received by the car:

( )( )9

8

1

8.00 10 Hz 44.5 m/s1190 Hz

3.00 10 m/s

u f uf f f

c c

f uf f

c

′ = − = −

×′ − = − = − = −

×

2. Double the change in frequency: ( ) ( )2 2 1187 Hz 2.37 kHzf f f f′′ ′− = − = − = −

Insight: This represents only a 3.0×10-5 % change in frequency.

16. Picture the Problem: The image

shows a spinning spiral galaxy that is receding from Earth. The spinning of the galaxy produces a Doppler shift of the light emitted by each arm.

Strategy: Solve for the observed frequency, where the speed is (a) u V v= − and (b) u V v= + . In both cases, use the negative sign in because the galaxy is receding from Earth.

Solution: 1. (a) Calculate the Doppler shift for the approaching arm:

5 514

8

14

3.600 10 m/s 6.400 10 m/s1 8.230 10 Hz 1

3.00 10 m/s

8.238 10 Hz

uf f

c

× − × ′ = − = × −

×

= ×

2. (b) Calculate the Doppler shift for the receding arm:

5 514

8

14

3.600 10 m/s 6.400 10 m/s8.230 10 Hz 1

3.00 10 m/s

8.203 10 Hz

f × + ×

′ = × − ×

= ×

Insight: The motion of stars in the galaxy will vary between the two limits V v− and .V v+ This results in the 8.230×1014 Hz frequency being observed as a range of frequencies between 8.203×1014 Hz and 8.238×1014 Hz.

17. Picture the Problem: As a car travels away from a radar gun, it receives a Doppler shifted

frequency and reflects that frequency back toward the gun. The frequency is again shifted because the car is a moving source.

Strategy: Divide the total Doppler shift in half to calculate the change in frequency between the radar gun and the frequency received by the speeding car. Insert this change in frequency and solve for the speed of the car.

Solution: 1. (a) The car is moving away from the radar gun because observed frequencies decrease for receding objects.

2. (b) Divide the change in frequency in half:

( ) ( )1 12 2 4.04 kHz 2.02 kHzf f f f′ ′′− = − = − = −

3. Solve for the speed:

89

1

2020 Hz3.00 10 m/s 25.1 m/s

24.150 10 Hz

u f uf f f

c c

f fu c

f

′ = − = −

′ − − = − = − × =

×

Insight: When the reflected wave interferes with the transmitted wave it creates a beat frequency that determines the speed of the car. The speed of the car is equivalent to 56.2 mi/h.


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18. Picture the Problem: Dental X-rays have a wavelength of 0.30 nm.


Solution: Calculate the frequency:

818

9

3.00 10 m/s1.0 10 Hz

0.30 10 mc

fλ −

×= = = ×

×

Insight: This frequency falls in the X-ray part of the spectrum. 19. Picture the Problem: Blue light has a wavelength of 460 nm.


Solution: Calculate the frequency:

814

9

3.00 10 m/s6.52 10 Hz

460 10 m

cf

λ −

×= = = ×

×

Insight: This frequency falls in the visible light region of the spectrum. 20. Picture the Problem: The frequency of a cell phone transmission is 1.25×108 Hz.

Strategy: Solve for the wavelength.

Solution: Calculate the wavelength:

8

8

3.00 10 m/s2.40 m

1.25 10 Hzc

fλ

×= = =

×

Insight: We can see that this wavelength corresponds to a radio/TV wave.

21. Picture the Problem: The radiation emitted by humans has a wavelength of about 9.0 µm.

Strategy: Solve to calculate the frequency. Then compare the frequencies to the ranges given in section 25-3 of the text.

Solution: 1. (a) Calculate the frequency:

813

6

3.00 10 m/s3.3 10 Hz

9.0 10 m

cf

λ −

×= = = ×

×

2. (b) This frequency falls in the infrared range 12 14(10 Hz to 4.3 10 Hz).×

Insight: This frequency can be detected with the infrared goggles that are often used by hunters and the military.

22. Picture the Problem: The ultraviolet range of the spectrum is subdivided into three regions.

Strategy: Calculate the frequencies of each range of ultraviolet radiation.

Solution: 1. (a) Calculate the minimum frequency of UV-A:

814

UVA-min 9

3.00 10 m/s7.50 10 Hz

400 10 mf

×= = ×

×

2. Calculate the frequency division between UV-A and UV-B:

814

UVA-UVB 9

3.00 10 m/s9.38 10 Hz

320 10 mf

×= = ×

×

3. Calculate the frequency division between UV-B and UV-C:

815

UVB-UVC 9

3.00 10 m/s1.07 10 Hz

280 10 mf

×= = ×

×

4. Calculate the maximum frequency of UV-C:

815

UVC-max 9

3.00 10 m/s3.00 10 Hz

100 10 mf

×= = ×

×

5. UV-B falls in the range: 14 15UV-B9.38 10 Hz 1.07 10 Hzf× < < ×

6. (b) 7.9×1014 Hz falls into the UV-A range.

Insight: Frequencies smaller than 7.50×1014 Hz fall in the visible spectrum and frequencies greater than 3.00×1015 Hz are X-rays.


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23. Picture the Problem: The very low frequency radio waves have a frequency of 10.0 kHz.

Strategy: Solve for the wavelength.

Solution: Calculate the wavelength:

8

3

3.00 10 m/s30.0 km

10.0 10 Hzc

fλ

×= = =

×

Insight: The intensity of the wave as it enters water decreases exponentially on a scale inversely proportional to the wavelength. The longer the wavelength, the deeper the signal will penetrate the water. However, longer waves cannot carry information as quickly as shorter waves, so that communication with submarines tends to be very slow.

24. Picture the Problem: As a wave enters a medium in which the speed of light decreases, the frequency remains constant, but the wavelength changes.

Strategy: Write the ratio of the new wavelength in terms of the initial wavelength.

Solution: 1. (a) The wavelength of an electromagnetic wave is directly proportional to its speed. Therefore, if the wave speed decreases, the wavelength decreases.

2. (b) Write the ratio:

34 3

4

c f

c f

λ

λ

′= =

Insight: The speed of light in water is 34 of the speed of light in a vacuum, and its index of

refraction is 43 .

25. Picture the Problem: Blue light has a wavelength of 470 nm and red light a wavelength of 680 nm.

Strategy: Calculate the frequencies of both colours.

Solution: 1. (a) Frequency varies inversely with wavelength, so that a shorter wavelength corresponds to a higher frequency. Since the wavelength of violet light is shorter than that for red light, violet light has the higher frequency.

2. (b) Calculate the frequency of blue light:

814

blue 9blue

3.00 10 m/s6.4 10 Hz

470 10 m

cf

λ −

×= = = ×

×

3. Calculate the frequency of red light:

814

red 9red

3.00 10 m/s4.4 10 Hz

680 10 mc

fλ −

×= = = ×

×

Insight: Violet light (λ = 420 nm) has a frequency 7.1×1014 Hz, even greater than the frequency of blue light.

26. Picture the Problem: Ultra low frequency electromagnetic waves have a wavelength of 29 million kilometres.

Strategy: Find the period by taking the inverse of the frequency.

Solution: Calculate the period:

9

8

1 29 10 m97 s

3.00 10 m/sT

f c

λ ×= = = =

×

Insight: A sound wave with this same period would have a wavelength of 33 km.


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27. Picture the Problem: To obtain optimal reception of a 66.0-MHz TV signal, the length of the antenna should be equal to half a wavelength.

Strategy: Set the length of the antenna equal to half a wavelength, where the wavelength is given in terms of the frequency.

Solution: Calculate the length of the antenna: ( )

8

7

1 3.00 10 m/s2.27 m

2 2 2 6.60 10 Hz

cL

fλ

×= = = =

×

Insight: To listen to an FM radio station at 107 MHz the antenna should be 1.40 m long.

28. Picture the Problem: A quarter-wavelength antenna is designed to broadcast an 880-kHz AM radio station.

Strategy: Set the length of the antenna equal to one quarter of a wavelength, where the wavelength is given in terms of the frequency.

Solution: Calculate the length of the antenna: ( )

8

3

1 3.00 10 m/s85.2 m

4 4 4 880 10 Hz

cL

fλ

×= = = =

×

Insight: This is almost the length of a football field and is equivalent to the height of a 28-story building.

29. Picture the Problem: A quarter-wavelength radio antenna is 112 m high.

Strategy: Set the wavelength equal to four times the antenna length and calculate the frequency.

Solution: Calculate the broadcast frequency: ( )

83.00 10 m/s670 kHz

4 4 112 mc c

fLλ

×= = = =

Insight: For a higher frequency AM station (such as 1400 kHz) a quarter-wavelength antenna would not need to be as tall (53.6 m).

30. Picture the Problem: Compare the difference in wavelengths for radio stations that differ in

frequency by 2 kHz.

Strategy: Subtract the two wavelengths to find the difference in wavelength, and write the wavelength in terms of the frequency.

Solution: 1. (a) Calculate the wavelength difference for the 50 kHz and 52 kHz frequencies:

( )81 2 3 3

1 2

1 13.00 10 m/s 0.2 km

50 10 Hz 52 10 Hzc c

f fλ λ

− = − = × − =

× ×

2. (b) Repeat for the 500 kHz and 502 kHz frequencies:

( )81 2 3 3

1 13.00 10 m/s 2 m

500 10 Hz 502 10 Hzλ λ

− = × − =

× ×

Insight: Another approach to solving this problem is to use the relation, .f fλ λ∆ = ∆ For

instance, you could find that ( )( )

( )

8 3

23

3.00 10 m/s 2 10 Hz0.2 km

51 10 Hz

f c f

f f fλ λ

× ×∆ ∆∆ = = = =

×as we

calculated in part (a) above.


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31. Picture the Problem: Compare the difference in frequencies for radio stations that differ in

wavelengths by 0.5 m.

Strategy: Subtract the two frequencies to find the difference in frequency, and write the frequency in terms of the wavelength.

Solution: 1. (a) Calculate the frequency difference for the 300.0-m and 300.5-m wavelengths:

( )81 2

1 2

1 13.00 10 m/s 2 kHz

300.0 m 300.5 m

c cf f

λ λ

− = − = × − =

2. (b) Repeat for the 30.0-m and 30.5-m wavelengths: ( )8

1 2

1 13.00 10 m/s 0.2 MHz

30.0 m 30.5 mf f

− = × − =

Insight: Another approach to solving this problem is to use the relation, .f fλ λ∆ = ∆ For

instance, you could find that ( )( )

( )

8

2

3.00 10 m/s 0.5 m0.2 MHz

30.25 m

cf f

λ λ

λ λ λ

×∆ ∆∆ = = = = as we

calculated in part (b) above.


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Solutions to Tutorial 6 The solutions to Tutorial 6 can be found in Lecture 6: Introduction to Semiconductors.

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