Instructor's Solution Manuals to Principles of Physics: A Calculus-Based Text

1 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
1 Introduction and Vectors
1.2 Dimensional Analysis
1.9 Components of a Vector and Unit Vectors
1.10 Modeling, Alternative Representations, and Problem-Solving Strategy
* An asterisk indicates an item new to this edition.
ANSWERS TO OBJECTIVE QUESTIONS
*OQ1.1 The answer is yes for (a), (c), and (e). You cannot add or subtract a number of apples and a number of jokes. The answer is no for (b) and (d). Consider the gauge of a sausage, 4 kg/2 m, or the volume of a cube, (2 m)3. Thus we have (a) yes (b) no (c) yes (d) no (e) yes.
*OQ1.2 41 € ≈ 41 € (1 L/1.3 €)(1 qt/1 L)(1 gal/4 qt) ≈ (10/1.3) gal ≈ 8 gallons, answer (c).
*OQ1.3 In the base unit we have (a) 0.032 kg (b) 0.015 kg (c) 0.270 kg (d) 0.041 kg (e) 0.27 kg. Then the ranking is c = e > d > a > b.
*OQ1.4 Answer (c). The vector has no y component given. It is therefore 0.
*OQ1.5 The population is about 6 billion = 6 × 109. Assuming about 100 lb per person = about 50 kg per person (1 kg has the weight of about 2.2 lb),
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
the total mass is about (6 × 109)(50 kg) = 3 × 1011 kg, answer (d).
*OQ1.6 The number of decimal places in a sum of numbers should be the same as the smallest number of decimal places in the numbers summed.
21.4 s 15 s 17.17 s 4.003 s 57.573 s = 58 s, answer (d).
*OQ1.7 The meterstick measurement, (a), and (b) can all be 4.31 cm. The meterstick measurement and (c) can both be 4.24 cm. Only (d) does not overlap. Thus (a), (b), and (c) all agree with the meterstick measurement.
*OQ1.8 Mass is measured in kg; acceleration is measured in m/s2. Force = mass × acceleration, so the units of force are answer (a) kg⋅m/s2.
*OQ1.9 Answer (d). Take the difference of the x coordinates of the ends of the vector, head minus tail: –4 – 2 = –6 cm.
*OQ1.10 Answer (a). Take the difference of the y coordinates of the ends of the vector, head minus tail: 1 − (−2) = 3 cm.
*OQ1.11 The answers are (a) yes (b) no (c) no (d) no (e) no (f) yes (g) no. Only force and velocity are vectors. None of the other quantities requires a direction to be described.
*OQ1.12 Answers (a), (b), and (c). The magnitude can range from the sum of the individual magnitudes, 8 + 6 =14, to the difference of the individual magnitudes, 8 − 6 = 2. Because magnitude is the “length” of a vector, it is always positive.
*OQ1.13 Answer (a). The vector −2 D1 will be twice as long as
D1 and in the
opposite direction, namely northeast. Adding D2 , which is about
equally long and southwest, we get a sum that is still longer and due east.
*OQ1.14 Answer (c). A vector in the second quadrant has a negative x component and a positive y component.
*OQ1.15 Answer (e). The magnitude is 102 + 102 m/s.
*OQ1.16 Answer (c). The signs of the components of a vector are the same as the signs of the points in the quadrant into which it points. If a vector arrow is drawn to scale, the coordinates of the point of the arrow equal the components of the vector. All x and y values in the third quadrant are negative.
Chapter 1 3
ANSWERS TO CONCEPTUAL QUESTIONS
*CQ1.1 A unit of time should be based on a reproducible standard so it can be used everywhere. The more accuracy required of the standard, the less the standard should change with time. The current, very accurate standard is the period of vibration of light emitted by a cesium atom. Depending on the accuracy required, other standards could be: the period of light emitted by a different atom, the period of the swing of a pendulum at a certain place on Earth, the period of vibration of a sound wave produced by a string of a specific length, density and tension, and the time interval from full Moon to full Moon.
*CQ1.2 (a) 0.3 millimeters (b) 50 microseconds (c) 7.2 kilograms.
*CQ1.3 Density varies with temperature and pressure. It would be necessary to measure both mass and volume very accurately in order to use the density of water as a standard.
*CQ1.4 Vectors A and
B are perpendicular to each other.
*CQ1.5 (a) The book’s displacement is zero, as it ends up at the point from which it started. (b) The distance traveled is 6.0 meters.
*CQ1.6 No, the magnitude of a vector is always positive. A minus sign in a vector only indicates direction, not magnitude.
*CQ1.7 The inverse tangent function gives the correct angle, relative to the +x axis, for vectors in the first or fourth quadrant, and it gives an incorrect answer for vectors in the second or third quadrant. If the x and y components are both positive, their ratio y/x is positive and the vector lies in the first quadrant; if the x component is positive and the y component negative, their ratio y/x is negative and the vector lies in the fourth quadrant. If the x and y components are both negative, their ratio y/x is positive but the vector lies in the third quadrant; if the x component is negative and the y component positive, their ratio y/x is negative but the vector lies in the second quadrant.
*CQ1.8 Addition of a vector to a scalar is not defined. Try adding the speed and velocity, 8.0 m/s + (15.0 m/s i) : Should you consider the sum to be a vector or a scaler? What meaning would it have?
SOLUTIONS TO END-OF-CHAPTER PROBLEMS
4 3 πr3 and the mass is
m = ρV = ρ 4
3 πr3. We divide this equation for the larger sphere by the
same equation for the smaller:
m ms
Then r = rs 53 = 4.50cm 1.71( ) = 7.69cm .
P1.2 (a) Modeling the Earth as a sphere, we find its volume as
4 3 πr3 =
= 1.08 × 1021 m3
= 5.98 × 1024 kg 1.08 × 1021 m3 = 5.52 × 103 kg/m3
(b) This value is intermediate between the tabulated densities of aluminum and iron. Typical rocks have densities around 2000 to 3000 kg/m3. The average density of the Earth is significantly higher, so higher-density material must be down below the surface.
P1.3 Let V represent the volume of the model, the same in ρ =
m V
, for both.
mgold
V .
Next,
ρgold
ρiron
= mgold
7.87 × 103 kg/m3
= 22.9 kg
P1.4 The volume of a spherical shell can be calculated from
V = Vo −Vi =
V , so
Chapter 1 5
m = ρV = ρ 4
3 − r1 3( )
3
Section 1.2 Dimensional Analysis P1.5 (a) This is incorrect since the units of [ax] are m2/s2, while the units
of [v] are m/s.
(b) This is correct since the units of [y] are m, cos(kx) is
dimensionless if [k] is in m–1, and the constant multiplying cos (kx) is in units of m.
P1.6 Circumference has dimensions L, area has dimensions L2, and volume has dimensions L3. Expression (a) has dimensions L(L2)1/2 = L2, expression (b) has dimensions L, and expression (c) has dimensions L(L2) = L3.
The matches are: (a) and (f), (b) and (d), and (c) and (e).
P1.7 The term x has dimensions of L, a has dimensions of LT−2 , and t has dimensions of T. Therefore, the equation x = kamtn has dimensions of
L = LT−2( )m T( )n or L1T0 = LmTn−2m
The powers of L and T must be the same on each side of the equation. Therefore,
L1 = Lm and m = 1
Likewise, equating terms in T, we see that n – 2m must equal 0. Thus,
n = 2 . The value of k, a dimensionless constant,
cannot be obtained by dimensional analysis .
Section 1.3 Conversion of Units P1.8 It is often useful to remember that the 1 600-m race at track and field
events is approximately 1 mile in length. To be precise, there are 1 609 meters in a mile. Thus, 1 acre is equal in area to
1 acre( ) 1 mi2
P1.9 V = At so t =
V A
P1.10 Apply the following conversion factors:
1 in = 2.54 cm, 1 d = 86 400 s, 100 cm = 1 m, and 109 nm = 1 m
1 32
= 9.19 nm/s
This means the proteins are assembled at a rate of many layers of atoms each second!
P1.11 The weight flow rate is
1200
= 667 lb/s .
P1.12 We obtain the number of atoms in the Sun by dividing its mass by the mass of a single hydrogen atom:
Natoms =
mSun
matom
= 1.19 × 1057 atoms
P1.13 The masses given are for a 1.00 m3 volume. Density is defined as mass per unit volume, so ρAl = 2.70 × 103 kg/m3 and ρFe = 7.86 × 103 kg/m3. For the spheres to balance, mFe = mA1 or ρFeVFe = ρA1VA1 :
ρFe 4 3
4πρAlrAl 3
4 3 πρALrAl
3 = rFe(1.43)
The resulting expression shows that the radius of the aluminum sphere is directly proportional to the radius of the balancing iron sphere. The sphere of lower density has larger radius.
Chapter 1 7
P1.15 (a) rate =
231 in3
1 gal

s
(c) To find the time to fill a 1.00-m3 tank, find the rate time/volume:
2.70 × 10−4 m3

1.06 × 10−10 m
dnucleus,scale = 6.79 × 10−3 ft( ) 304.8 mm/1 ft( ) = 2.07 mm
(b)
Vatom
Vnucleus
= 4πratom

Section 1.4 Order-of-Magnitude Calculations P1.17 Model the room as a rectangular solid with dimensions 4 m by 4 m by
3 m, and each ping-pong ball as a sphere of diameter 0.038 m. The volume of the room is 4 × 4 × 3 = 48 m3, while the volume of one ball is
4π 3
48 2.87 × 10−5 106 ping-pong balls in the
room.
As an aside, the actual number is smaller than this because there will be a lot of space in the room that cannot be covered by balls. In fact, even in the best arrangement, the so-called “best packing fraction” is
1 6 π 2 = 0.74 , so that at least 26% of the space will be empty.
Therefore, the above estimate reduces to 1.67 × 106 × 0.740 ~ 106.
*P1.18 (a) We estimate the mass of the water in the bathtub. Assume the tub measures 1.3 m by 0.5 m by 0.3 m. One-half of its volume is then
V = (0.5)(1.3)(0.5)(0.3) = 0.10 m3.
The mass of this volume of water is
mwater = ρwaterV = 1 000 kg/m3( ) 0.10 m3( ) = 100 kg 102 kg
(b) Pennies are now mostly zinc, but consider copper pennies filling 50% of the volume of the tub. The mass of copper required is
mcopper = ρcopperV = 8 920 kg/m3( ) 0.10 m3( ) = 892 kg ~ 103 kg
P1.19 Assume: Total population = 107; one out of every 100 people has a piano; one tuner can serve about 1000 pianos (about 4 per day for 250 weekdays, assuming each piano is tuned once per year). Therefore,
# tuners ~

107 people( ) = 100 tuners
P1.20 A reasonable guess for the diameter of a tire might be 2.5 ft, with a circumference of about 8 ft. Thus, the tire would make
50 000 mi( ) 5 280 ft/mi( ) 1 rev/8 ft( ) = 3 × 107 rev ~ 107 rev
Chapter 1 9
ANS. FIG. P1.25
Section 1.5 Significant Figures P1.21 We work to nine significant digits:
1 yr = 1 yr 365. 242 199 d
1 yr
= 315 569 26.0 s
P1.22 (a) 756 + 37.2 + 0.83 + 2 = 796.03→ 796 , since the number with the fewest decimal places is 2.
(b) 0.003 2( ) 2 s.f.( ) × 356.3( ) 4 s.f.( ) = 1.140 16 = 2 s.f.( ) 1.1
(c) 5.620 4s.f.( ) × π > 4s.f.( ) = 17.656 = 4s.f.( ) 17.66
P1.23 (a) 3 (b) 4 (c) 3 (d) 2
P1.24 r = 6.50 ± 0.20( )cm = 6.50 ± 0.20( ) × 10−2 m
m = 1.85 + 0.02( )kg
In other words, the percentages of uncertainty are cumulative. Therefore,
δρ ρ
= 0.02 1.85
+ 3 0.20( )
6.50 = 0.103,
ρ = 1.85
4 3( )π 6.5 × 10−2 m( )3 = 1.61× 103 kg/m3
then δρ = 0.103ρ = 0.166 × 103 kg/m3
and ρ ± δρ = 1.61 ± 0.17( ) × 103 kg/m3 = 1.6 ± 0.2( ) × 103 kg/m3.
P1.25 The volume of concrete needed is the sum of the four sides of sidewalk, or
V = 2V1 + 2V2 = 2 V1 +V2( )
The figure on the right gives the dimensions needed to determine the volume of each portion of sidewalk:
V1 = 17.0 m + 1.0 m + 1.0 m( ) 1.0 m( ) 0.09 m( ) = 1.70 m3
V2 = 10.0 m( ) 1.0 m( ) 0.090 m( ) = 0.900 m3
V = 2 1.70 m3 + 0.900 m3( ) = 5.2 m3
The uncertainty in the volume is the sum of the uncertainties in each dimension:
δ 1
= 0.006 + 0.010 + 0.011 = 0.027 = 3%
P1.26 Using substitution is to solve simultaneous equations. We substitute p = 3q into each of the other two equations to eliminate p:
3qr = qs 1 2
, assuming q ≠ 0.
We substitute the upper relation into the lower equation to eliminate s:
3r2 + 3r( )2 = t2 → 12r2 = t2 →
t2
t r = ± 12 = ±3.46
P1.27 We draw the radius to the initial point and the radius to the final point. The angle θ between these two radii has its sides perpendicular, right side to right side and left side to left side, to the 35° angle between the original and final tangential directions of travel. A most useful theorem from geometry then identifies these angles as equal: θ = 35°. The whole circumference of a 360° circle of the same radius is 2πR. By proportion, then
2πR 360°
= 840 m
ANS. FIG. P1.27
Chapter 1 11
R =
840 m 35°
= 840 m 0.611
= 1.38 × 103 m
We could equally well say that the measure of the angle in radians is
θ = 35° = 35°
2π radians 360°
Solving yields R = 1.38 km.
P1.28 For those who are not familiar with solving equations numerically, we provide a detailed solution. It goes beyond proving that the suggested answer works.
The equation 2x4 – 3x3 + 5x – 70 = 0 is quartic, so we do not attempt to solve it with algebra. To find how many real solutions the equation has and to estimate them, we graph the expression:
x –3 –2 –1 0 1 2 3 4
y = 2x4 – 3x3 + 5x – 70 158 –24 –70 –70 –66 –52 26 270
We see that the equation y = 0 has two roots, one around x = –2.2 and the other near x = +2.7. To home in on the first of these solutions we compute in sequence:
When x = –2.2, y = –2.20. The root must be between x = –2.2 and x = –3. When x = –2.3, y = 11.0. The root is between x = –2.2 and x = –2.3. When x = –2.23, y = 1.58. The root is between x = –2.20 and x = –2.23. When x = –2.22, y = 0.301. The root is between x = –2.20 and –2.22. When x = –2.215, y = –0.331. The root is between x = –2.215 and –2.22. We could next try x = –2.218, but we already know to three-digit precision that the root is x = –2.22.
P1.29 We require
sin θ cos θ
= tan θ = −3
For tan–1(–3) = arctan(–3), your calculator may return –71.6°, but this angle is not between 0° and 360° as the problem requires. The tangent function is negative in the second quadrant (between 90° and 180°) and in the fourth quadrant (from 270° to 360°). The solutions to the equation are then
360° − 71.6° = 288° and 180° − 71.6 = 108°
ANS. FIG. P1.28
ANS. FIG. P1.29
Section 1.6 Coordinate Systems P1.30 (a) The distance between the points is given by:
d = x2 − x1( )2 + y2 − y1( )2
= 2.00 − −3.00[ ]( )2 + −4.00 − 3.00( )2
d = 25.0 + 49.0 = 8.60 m
(b) To find the polar coordinates of each point, we measure the radial distance to that point and the angle it makes with the +x axis:
r1 = 2.00( )2 + −4.00( )2 = 20.0 = 4.47 m
θ1 = tan−1 − 4.00
θ2 = 135° measured from the +x axis.
P1.31
x = r cosθ = 5.50 m( )cos240° = 5.50 m( ) −0.5( ) = −2.75 m
y = r sinθ = 5.50 m( )sin 240° = 5.50 m( ) −0.866( ) = −4.76 m
P1.32 We have r = x2 + y2 and θ = tan−1 y
x
(−x)2 + y2 = x2 + y2 = r
and its angle is
= 180° −θ
(b) (−2x)2 + (−2y)2 = 2r . This point is in the third quadrant if (x, y) is in the first quadrant or in the fourth quadrant if (x, y) is in the second quadrant. It is at an angle of 180° +θ .
(c) (3x)2 + (−3y)2 = 3r . This point is in the fourth quadrant if (x, y) is in the first quadrant or in the third quadrant if (x, y) is in the second quadrant. It is at an angle of −θ or 360 −θ .
P1.33 The x distance out to the fly is 2.00 m and the y distance up to the fly is
Chapter 1 13
1.00 m.
(a) We can use the Pythagorean theorem to find the distance from the origin to the fly.
distance = x2 + y2 = (2.00 m)2 + (1.00 m)2 = 5.00 m2 = 2.24 m
(b) θ = tan−1 1
2
= 26.6°; r = 2.24 m, 26.6°
Section 1.7 Vectors and Scalars Section 1.8 Some Properties of Vectors P1.34 To find these vector expressions
graphically, we draw each set of vectors. Measurements of the results are taken using a ruler and protractor. (Scale: 1 unit = 0.5 m)
(a) A + B = 5.2 m at 60o
(b) A − B = 3.0 m at 330o
(c) B − A = 3.0 m at 150o
(d) A − 2B = 5.2 m at 300o
P1.35
From the figure, we note that the length of the skater's path along the
.
ANS. FIG. P1.36
P1.36 Ans. Fig. P1.36 shows the graphical addition of the vector from the base camp to lake A to the vector connecting lakes A and B, with a scale of 1 unit = 20 km. The distance from lake B to base camp is then the negative of this resultant vector, or
− R = 310 km at 57° S of W .
P1.37 The scale drawing for the graphical solution should be similar to the figure to the right. The magnitude and direction of the final displacement from the starting point are obtained by measuring d and θ on the drawing and applying the scale factor used in making the drawing. The results should be d = 420 ft and θ = –3° .
Section 1.9 Components of a Vector and Unit Vectors P1.38 (a) See figure to the right.
(b)
= 5.00i + 4.00 j D =
= −1.00i + 8.00 j
5
−1.00

ANS. FIG. P1.37
ANS. FIG. P1.38
Chapter 1 15
P1.39 (a) Taking components along i and j , we get two equations:
6.00a – 8.00b +26.0 = 0
and –8.00a + 3.00b + 19.0 = 0
Substituting a = 1.33 b – 4.33 into the second equation, we find
−8 1.33 b − 4.33( ) + 3 b + 19 = 0→ 7.67b = 53.67 → b = 7.00
and so a = 1.33(7) – 4.33 = 5.00.
Thus a = 5.00, b = 7.00 . Therefore, 5.00
A + 7.00
(b)

P1.40 The superhero follows a straight-line path at 30.0° below the horizontal. If his displacement is 100 m, then the coordinates of the superhero are:
x = 100 m( )cos −30.0°( ) = 86.6 m
y = 100 m( )sin −30.0°( ) = −50.0 m
P1.41 Ax = –25.0
We observe that
= tan−1(1.60) = 58.0°
The diagram shows that the angle from the +x axis can be found by subtracting from 180°:
θ = 180° − 58° = 122°
ANS. FIG. P1.44
ANS. FIG. P1.45
P1.42
B = Bx i + By j + Bzk = 4.00i + 6.00 j + 3.00k B = 4.002 + 6.002 + 3.002 = 7.81
α = cos−1 4.00 7.81

β = cos−1 6.00 7.81

γ = cos−1 3.00 7.81

(b)
(c) C = −3
P1.44 (a)

θ = tan−1 27.1 49.5

Therefore,
B = [115 − (−75)i + [80.3 − 130]j = 190i − 49.7 j( ) cm B = 1902 + 49.72 = 196 cm
θ = tan−1 − 49.7 190

Chapter 1 17
P1.46
B = 13.2i − 6.60 j
3 C = B − A = 21.9i − 21.6 j

P1.47 (a)
A + B( ) = 3i − 2 j( ) + − i − 4 j( ) = 2i − 6 j
(b)
A − B( ) = 3i − 2 j( ) − − i − 4 j( ) = 4i + 2 j
(c) A + B = 22 + 62 = 6.32
(d) A − B = 42 + 22 = 4.47
(e)

A = 3.00 m, θA = 30.0°
Ax = A cos θA = 3.00 cos 30.0° = 2.60 m,
Ay = A sin θA = 3.00 sin 30.0° = 1.50 m
So A = Ax i + Ay j = (2.60i + 1.50 j) m
B = 3.00 m, θB = 90.0°
Bx = 0, By = 3.00 m →
B = 3.00 j m
A + B = 2.60i + 1.50 j( ) + 3.00 j = 2.60i + 4.50 j( )m
P1.49 Let the positive x direction be eastward, the positive y direction be vertically upward, and the positive z direction be southward. The total displacement is then
d = 4.80i + 4.80 j( ) cm + 3.70 j − 3.70k( ) cm
= 4.80i + 8.50 j − 3.70k( ) cm
(a) The magnitude is d = 4.80( )2 + 8.50( )2 + −3.70( )2 cm = 10.4 cm .
ANS. FIG. P1.51
ANS. FIG. P1.52
cosθ =
D = 62 + 22 = 6.32 m at θ = 342°

E = 22 + 122 = 12.2 m at θ = 99.5°

P1.51 d1 = 100i
d2 = −300i d3 = −150cos 30.0°( ) i − 150sin 30.0°( ) j = −130i − 75.0 j d4 = −200cos 60.0°( ) i + 200sin 60.0°( ) j = −100i + 173 j R = d1 +
d2 +
d3 +
R = −130( )2 + −202( )2 = 240 m
φ = tan−1 202 130

P1.52 (a)

(b)

F = −7.72i + 15.1j( ) cm Note that we
did not need to explicitly identify the angle with the positive x axis, but by doing so, we don’t have to keep track of minus signs for the components.
(c)

Chapter 1 19
ANS. FIG. P1.54
Section 1.10 Modeling, Alternative Representations, and Problem-Solving Strategy
P1.53 From the figure, we may see that the spacing between diagonal planes is half the distance between diagonally adjacent atoms on a flat plane. This diagonal distance may be obtained from the Pythagorean
theorem, Ldiag = L2 + L2 . Thus, since the atoms are separated by a distance L = 0.200 nm, the diagonal planes are separated by
1 2
L2 + L2 = 0.141 nm .
P1.54 We note that − i = west and − j = south. The given mathematical representation of the trip can be written as 6.30 b west + 4.00 b at 40° south of west +3.00 b at 50° south of east +5.00 b south.
(a) The figure on the right shows a map of the successive displacements that the bus undergoes.
(b) The total odometer distance is the sum of the magnitudes of the four displacements:
6.30 b + 4.00 b + 3.00 b + 5.00 b = 18.3 b
(c)
R = (−6.30 − 3.06 + 1.93) b i + (−2.57 − 2.30 − 5.00) b j
= −7.44 b i − 9.87 b j
= (7.44 b)2 + (9.87 b)2 at tan−1 9.87 7.44

south of west

P1.55 Figure P1.55 suggests a right triangle where, relative to angle θ, its adjacent side has length d and its opposite side is equal to the width of the river, y; thus,
tanθ =
y = (100 m) tan 35.0° = 70.0 m
The width of the river is 70.0 m .
P1.56 The volume of the galaxy is
πr2t = π 1021 m( )2 1019 m( ) ~ 1061 m3
If the distance between stars is 4 × 1016, then there is one star in a volume on the order of
4 × 1016 m( )3
1061 m3
1050 m3/star ~ 1011 stars .
P1.57 It is desired to find the distance x such that
x 100 m
= 1 000 m
x
(i.e., such that x is the same multiple of 100 m as the multiple that 1 000 m is of x). Thus, it is seen that
x2 = (100 m)(1 000 m) = 1.00 × 105 m2
and therefore
P1.58 One month is
Applying units to the equation,
V = (1.50 Mft3/mo)t + (0.008 00 Mft3/mo2 )t2
Since 1 Mft3 = 106 ft3,
V = (1.50 × 106 ft3/mo)t + (0.008 00 × 106 ft3/mo2 )t2
Converting months to seconds,
t + 0.008 00 × 106 ft3/mo2
(2.592 × 106 s/mo)2 t2
V = 0.579t + (1.19 × 10−9 )t2
for V in cubic feet and t in seconds.
P1.59 Since
ANS. FIG. P1.59
Chapter 1 21
ANS. FIG. P1.61
giving Ax + Bx = 0→ Ax = −Bx .
Because the vectors have the same magnitude and x components of equal magnitude but of opposite sign, the vectors are reflections of each other in the y axis, as shown in the diagram. Therefore, the two vectors have the same y components:
Ay = By = (1/2)(6.00) = 3.00
B and the y axis, it is seen
that
cosθ =
Ay
A =
By
B =
B is then
φ = 2θ = 106° .
P1.60 The table below shows α in degrees, α in radians, tan(α), and sin(α) for angles from 15.0° to 31.1°:
α′(deg) α(rad) tan(α) sin(α) difference between
α and tan α
15.0 0.262 0.268 0.259 2.30%
20.0 0.349 0.364 0.342 4.09%
30.0 0.524 0.577 0.500 9.32%
33.0 0.576 0.649 0.545 11.3%
31.0 0.541 0.601 0.515 9.95%
31.1 0.543 0.603 0.516 10.02%
We see that α in radians, tan(α), and sin(α) start out together from zero and diverge only slightly in value for small angles. Thus 31.0° is the
largest angle for which
*P1.61 Let θ represent the angle between the directions of
A and
B . Since
A and
B have
R = A + B
are 180° −θ ,
. The magnitude of R
ANS. FIG. P1.62
. This can be seen from applying the law of
cosines to the isosceles triangle and using the fact that B = A.
Again, A , −
B , and
angle θ. Applying the law of cosines and the identity
1− cosθ( ) = 2sin2 θ
D = 2Asin
θ 2
Thus, 2Acos
θ 2
P1.62 Let θ represent the angle between the directions of
A and
B . Since
A and
B have
R = A + B
are 180° −θ ,
then R = 2Acos
. This can be seen by
applying the law of cosines to the isosceles triangle and using the fact that B = A.
Again, A , −
B , and
angle θ. Applying the law of cosines and the identity
1− cosθ( ) = 2sin2 θ
D = 2Asin
θ 2
cos
n
.
The larger R is to be compared to D, the smaller the angle between A
and B becomes.
Chapter 1 23
P1.63 The actual number of seconds in a year is
86 400 s day( ) 365.25 day yr( ) = 31 557 600 s yr
The percent error in the approximation is
π × 107 s yr( ) − 31 557 600 s yr( ) 31 557 600 s yr
× 100% = 0.449%
P1.64 The position vector from the ground under the controller of the first airplane is
r1 = (19.2 km)(cos25°)i + (19.2 km)(sin 25°)j + (0.8 km)k
= 17.4i + 8.11j + 0.8k( ) km
The second is at
r2 = (17.6 km)(cos20°)i + (17.6 km)(sin 20°)j + (1.1 km)k
= 16.5i + 6.02 j + 1.1k( ) km
Now the displacement from the first plane to the second is
r2 − r1 = −0.863i − 2.09 j + 0.3k( ) km
with magnitude
0.863( )2 + 2.09( )2 + 0.3( )2 km = 2.29 km
P1.65 Observe in Fig. 1.65 that the radius of the horizontal cross section of the bottle is a relative maximum or minimum at the two radii cited in the problem; thus, we recognize that as the liquid level rises, the time rate of change of the diameter of the cross section will be zero at these positions.
The volume of a particular thin cross section of the shampoo of thickness h and area A is V = Ah, where A = πr2 = πD2/4. Differentiate the volume with respect to time:
dV dt
+ 2πhr dr dt
Because the radii given are a maximum and a minimum value, dr/dt = 0, so
dV dt
dV dt
where v = dh/dt is the speed with which the level of the fluid rises.
(a) For D = 6.30 cm,
v =
(b) For D = 1.35 cm,
v =
P1.66 (a) 1 cubic meter of water has a mass
m = ρV = 1.00 × 10−3 kg cm3( ) 1.00 m3( ) 102 cm m( )3
= 1 000 kg
(b) As a rough calculation, we treat each item as if it were 100% water.
For the cell:
6 π D3( )
= 1 000 kg m3( ) 1 6 π( ) 1.0 × 10−6 m( )3
= 5.24 × 10−16 kg
= 1.00 × 10−3 kg cm3( ) 4 3 π( )(4.0 cm)3
= 0.268 kg

F = 120 cos (60.0°)i + 120 sin (60.0°)j

Chapter 1 25
F = 39.32 + 1812 N = 185 N
θ = tan−1 181 39.3

ANS. FIG. P1.64(a) ANS. FIG. P1.64(b)
(c) From the triangles,
and tan 14° =
→ y = (x − 1.00 km)tan 14° .
(d) Equating the two expressions for y, we solve to find y = 1.44 km.
P1.69 (a) You start at point A: r1 = rA = 30.0i − 20.0 j( ) m.
The displacement to B is
rB − rA = 60.0i + 80.0 j − 30.0i + 20.0 j = 30.0i + 100 j
You cover half of this, 15.0i + 50.0 j( ) to move to
r2 = 30.0i − 20.0 j + 15.0i + 50.0 j = 45.0i + 30.0 j
Now the displacement from your current position to C is
rC − r2 = −10.0i − 10.0 j − 45.0i − 30.0 j = −55.0i − 40.0 j
You cover one-third, moving to
r3 = r2 + Δr23 = 45.0i + 30.0 j +
1 3
The displacement from where you are to D is
rD − r3 = 40.0i − 30.0 j − 26.7 i − 16.7 j = 13.3i − 46.7 j
You traverse one-quarter of it, moving to
r4 = r3 +
1 4

The displacement from your new location to E is
rE − r4 = −70.0i + 60.0 j − 30.0i − 5.00 j = −100i + 55.0 j
of which you cover one-fifth the distance, −20.0i + 11.0 j, moving to
r4 + Δr45 = 30.0i + 5.00 j − 20.0i + 11.0 j = 10.0i + 16.0 j
The treasure is at
10.0 m, 16.0 m( ) .
(b) Following the directions brings you to the average position of the trees. The steps we took numerically in part (a) bring you to
rA +
3 = rA + rB +

*P1.70 The geometry of the problem suggests we use the law of cosines to relate known sides and angles of a triangle to the unknown sides and angles. Recall that the sides a, b, and c with opposite angles A, B, and C have the following relationships:
Chapter 1 27
ANS. FIG. P1.70
ANS. FIG. P1.71
b 2 = c2 + a2 − 2cacosB
c 2 = a2 + b2 − 2abcosC
For the cows in the meadow, the triangle has sides a = 25.0 m and b = 15.0 m, and angle C = 20.0°, where object A = cow A, object B = cow B, and object C = you.
(a) Find side c:
c2 = a2 + b2 − 2abcosC
c = 12.1 m
cos A = a2 − b2 − c2
2bc =
2(15.0 m)(12.1 m)
→ A = 134.8° = 135°
2(25.0 m)(12.1 m)
→ B = 25.2°
(d) For the situation, object A = star A, object B = star B, and object C = our Sun (or Earth); so, the triangle has sides a = 25.0 ly, b = 15.0 ly, and angle C = 20.0°. The numbers are the same, except for units, as in part (b); thus,
angle A = 135.
R1 = ai + bj.

ANSWERS TO EVEN-NUMBERED PROBLEMS
P1.2 (a) 5.52 × 103 kg/m3 ; (b) value is intermediate to tabulated densities of aluminum and iron
P1.4
3( ) 3
P1.6 (a) and (f); (b) and (d); (c) and (e)
P1.8 4.05 × 103 m2
P1.16 (a) 2.07 mm; (b) 8.62 × 1013 times as large
P1.18 (a) ~ 102 kg; (b) ~ 103 kg
P1.20 107 rev
P1.24 1.61× 103 kg/m3 , 1.61 ± 0.17( ) × 103 kg/m3
P1.26 ±3.46
P1.28 See complete description in P1.28
P1.30 (a) 8.60 m; (b) 4.47 m, –63.4°; 4.24 m, 135°
P1.32 (a) r, 180° – θ; (b) 180° + θ ; (c) –θ or 360 – θ
P1.34 See graphs in P1.34, and (a) 5.2 m at 60°; (b) 3.0 m at 330°; (c) 3.0 m at 150°; (d) 5.2 at 300°
P1.36 310 km at 57° S of W
P1.38 (a) See Ans. Fig. P1.38; (b) 5.00i + 4.00 j,−1.00i + 8.00 j; (c) 6.40 at 38.7°, 8.06 at 97.2°
P1.40 86.6 m, –50.0 m
P1.42 59.2° with the x axis, 39.8° with the y axis, 67.4° with the z axis
P1.44 (a) 45.5i + 27.1j; (b) 56.4, 28.7°
P1.46 Cx = 7.30 cm; Cy = −7.20 cm
P1.48 2.60i + 4.50 j( )m
Chapter 1 29
P1.50 (a) 6.32 m at 342°; (b) 12.2 m at 99.5°
P1.52 (a) (15.1i + 7.72 j) cm ; (b) −7.72i + 15.1j( ) cm ; (c)
−7.72i − 15.1j( ) cm
P1.54 (a) See Ans. Fig. P1.54; (b) 18.3 b; 12.4 b at 233° counterclockwise from east
P1.56 1011 stars
P1.60 31.0°
n
P1.64 2.29 km
P1.66 (a) 1 000 kg; (b) cell: 5.24 × 10−16 kg , kidney: 0.268 kg, fly: 1.26 × 10−5
kg
P1.68 (a-b) see triangles; (c) y = x tan 12° and y = (x − 1.00 km) tan 14°; (d) y = 1.44 km
P1.70 (a) 12.1 m; (b) 135°; (c) 25.2°; (d) 135°
2 Motion in One Dimension
CHAPTER OUTLINE
2.4 Acceleration
2.7 Freely Falling Objects
*OQ2.1 Count spaces (intervals), not dots. Count 5, not 6. The first drop falls at time zero and the last drop at 5 × 5 s = 25 s. The average speed is 600 m/25 s = 24 m/s, answer (b).
*OQ2.2 The derivation of the equations of kinematics for an object moving in one dimension was based on the assumption that the object had a constant acceleration. Thus, (b) is the correct answer. An object would have constant velocity if its acceleration were zero, so (a) applies to cases of zero acceleration only. The speed (magnitude of the velocity) will increase in time only in cases when the velocity is in the same direction as the constant acceleration, so (c) is not a correct response. An object projected straight upward into the air has a constant downward acceleration, yet its position (altitude) does not always increase in time (it eventually starts to fall back downward) nor is its velocity always directed downward (the direction of the constant acceleration). Thus, neither (d) nor (e) can be correct.
Chapter 2 31
*OQ2.3 The bowling pin has a constant downward acceleration while in flight. The velocity of the pin is directed upward on the ascending part of its flight and is directed downward on the descending part of its flight. Thus, only (d) is a true statement.
*OQ2.4 Once the arrow has left the bow, it has a constant downward acceleration equal to the free-fall acceleration, g. Taking upward as the positive direction, the elapsed time required for the velocity to change from an initial value of 15.0 m/s upward (v0 = +15.0 m/s) to a value of 8.00 m/s downward (vf = −8.00 m/s) is given by
Δt =
−9.80 m/s2 = 2.35 s
Thus, the correct choice is (d).
*OQ2.5 (c) The object has an initial positive (northward) velocity and a negative (southward) acceleration; so, a graph of velocity versus time slopes down steadily from an original positive velocity. Eventually, the graph cuts through zero and goes through increasing-magnitude- negative values.
*OQ2.6 We take downward as the positive direction with y = 0 and t = 0 at the top of the cliff. The freely falling pebble then has v0 = 0 and a = g = +9.8 m/s2. The displacement of the pebble at t = 1.0 s is given: y1 = 4.9 m.
The displacement of the pebble at t = 3.0 s is found from
y3 = v0t +
1 2
9.8 m/s2( ) 3.0 s( )2 = 44 m
The distance fallen in the 2.0-s interval from t = 1.0 s to t = 3.0 s is then
y = y3 − y1 = 44 m − 4.9 m = 39 m
and choice (c) is seen to be the correct answer.
*OQ2.7 (c) They are the same. After the first ball reaches its apex and falls back downward past the student, it will have a downward velocity of magnitude vi. This velocity is the same as the velocity of the second ball, so after they fall through equal heights their impact speeds will also be the same.
*OQ2.8 (b) Using v f 2 = vi
2 + 2a y, with vi = −12 m/s and y = −40 m:
v f 2 = vi
2 + 2aΔy
v2 = −12 m/s( )2 + 2 −9.80 m/s2( ) −40 m( ) v = −30 m/s
*OQ2.9 In a position vs. time graph, the velocity of the object at any point in time is the slope of the line tangent to the graph at that instant in time. The speed of the particle at this point in time is simply the magnitude (or absolute value) of the velocity at this instant in time. The displacement occurring during a time interval is equal to the difference in x coordinates at the final and initial times of the interval, x = xf − xi.
The average velocity during a time interval is the slope of the straight line connecting the points on the curve corresponding to the initial and final times of the interval,
v = Δx Δt
Thus, we see how the quantities in choices (a), (e), (c), and (d) can all be obtained from the graph. Only the acceleration, choice (b), cannot be obtained from the position vs. time graph.
*OQ2.10 (b) Above. Your ball has zero initial speed and smaller average speed during the time of flight to the passing point. So your ball must travel a smaller distance to the passing point than the ball your friend throws.
*OQ2.11 With original velocity zero, displacement is proportional to the square of time in (1/2)at2. Making the time one-third as large makes the displacement one-ninth as large, answer (c).
*OQ2.12 Once the ball has left the thrower’s hand, it is a freely falling body with a constant, nonzero, acceleration of a = −g. Since the acceleration of the ball is not zero at any point on its trajectory, choices (a) through (d) are all false and the correct response is (e).
*OQ2.13 (a) Its speed is zero at points B and D where the ball is reversing its direction of motion. Its speed is the same at A, C, and E because these points are at the same height. The assembled answer is A = C = E > B = D.
(b) The acceleration has a very large positive (upward) value at D. At all the other points it is −9.8 m/s2. The answer is D > A = B = C = E.
*OQ2.14 (i) (b) shows equal spacing, meaning constant nonzero velocity and constant zero acceleration. (ii) (c) shows positive acceleration throughout. (iii) (a) shows negative (leftward) acceleration in the first four images.
ANSWERS TO CONCEPTUAL QUESTIONS
*CQ2.1 Yes. If a car is travelling eastward and slowing down, its acceleration is opposite to the direction of travel: its acceleration is westward.
Chapter 2 33
*CQ2.2 Tramping hard on the brake at zero speed on a level road, you do not feel pushed around inside the car. The forces of rolling resistance and air resistance have dropped to zero as the car coasted to a stop, so the car’s acceleration is zero at this moment and afterward.
Tramping hard on the brake at zero speed on an uphill slope, you feel thrown backward against your seat. Before, during, and after the zero- speed moment, the car is moving with a downhill acceleration if you do not tramp on the brake.
*CQ2.3 (a) No. Constant acceleration only: the derivation of the equations assumes that d2x/dt2 is constant. (b) Yes. Zero is a constant.
*CQ2.4 Yes. If the speed of the object varies at all over the interval, the instantaneous velocity will sometimes be greater than the average velocity and will sometimes be less.
*CQ2.5 The net displacement must be zero. The object could have moved away from its starting point and back again, but it is at its initial position again at the end of the time interval.
*CQ2.6 Yes. Acceleration is the time rate of change of the velocity of a particle. If the velocity of a particle is zero at a given moment, and if the particle is not accelerating, the velocity will remain zero; if the particle is accelerating, the velocity will change from zero—the particle will begin to move. Velocity and acceleration are independent of each other.
*CQ2.7 Yes. Acceleration is the time rate of change of the velocity of a particle. If the velocity of a particle is nonzero at a given moment, and the particle is not accelerating, the velocity will remain the same; if the particle is accelerating, the velocity will change. The velocity of a particle at a given moment and how the velocity is changing at that moment are independent of each other.
*CQ2.8 Assuming no air resistance: (a) The ball reverses direction at its maximum altitude. For an object traveling along a straight line, its velocity is zero at the point of reversal. (b) Its acceleration is that of gravity: −9.80 m/s2 (9.80 m/s2, downward). (c) The velocity is −5.00 m/s2. (d) The acceleration of the ball remains −9.80 m/s2 as long as it does not touch anything. Its acceleration changes when the ball encounters the ground.
*CQ2.9 No: Car A might have greater acceleration than B, but they might both have zero acceleration, or otherwise equal accelerations; or the driver of B might have tramped hard on the gas pedal in the recent past to give car B greater acceleration just then.
SOLUTIONS TO END-OF-CHAPTER PROBLEMS
Section 2.1 Average Velocity
(d) vavg =
x2 − x1
t2 − t1
(e) vavg =
x2 − x1
t2 − t1
(a) vavg =
Δx Δt
(b) v=
Δx Δt
(c) v = Δx
Δt = 57.5 m − 0 m
5.00 s = 11.5 m s
P2.4 (a) Let d represent the distance between A and B. Let t1 be the time
for which the walker has the higher speed in 5.00 m/s =
d t1
. Let t2
represent the longer time for the return trip in −3.00 m/s = −
d t2
d 3.00 m/s( ) . The
Chapter 2 35
average speed is:
vavg = Total distance
8.00 m/s( )d / 15.0 m2/s2( ) vavg =
2 15.0 m2/s2( ) 8.00 m/s
= 3.75 m/s
(b) She starts and finishes at the same point A. With total displacement = 0, average velocity = 0 .
Section 2.2 Instantaneous Velocity P2.5 (a) At ti = 1.5 s, xi = 8.0 m (Point A)
At tf = 4.0 s, xf = 2.0 m (Point B)
vavg = x f − xi
= − 6.0 m 2.5 s
= −2.4 m/s
(b) The slope of the tangent line can be found from points C and D. (tC = 1.0 s, xC = 9.5 m) and (tD = 3.5 s, xD = 0),
v ≈ −3.8 m/s
(c) The velocity is zero when x is a minimum.
This is at t ≈ 4.0 s .
P2.6 (a) At any time, t, the position is given by x = (3.00 m/s2)t2.
Thus, at ti = 3.00 s: xi = (3.00 m/s2)(3.00 s)2 = 27.0 m .
(b) At tf = 3.00 s + t: xf = (3.00 m/s2)(3.00 s + t)2, or
x f = 27.0 m + 18.0 m/s( )Δt + 3.00 m/s2( ) Δt( )2
(c) The instantaneous velocity at t = 3.00 s is:
ANS. FIG. P2.5
lim Δt→0
Δt = lim
P2.7 (a) v =
= 5 m/s
(b) v =
(c) v =
(d) v =
2.8 (a)
ANS. FIG. P2.8
(b) At t = 5.0 s, the slope is v ≈ 58 m
2.5 s ≈ 23 m s .
At t = 4.0 s, the slope is v ≈ 54 m
3 s ≈ 18 m s .
3.4 s ≈ 14 m s .
4.0 s ≈ 9.0 m s .
(c) a = Δv
(d) Initial velocity of the car was zero .
ANS. FIG. P2.7
Chapter 2 37
Section 2.3 Analysis Model: Particle Under Constant Velocity P2.9 (a) The tortoise crawls through a distance D before the rabbit
resumes the race. When the rabbit resumes the race, the rabbit must run through 200 m at 8.00 m/s while the tortoise crawls through the distance (1 000 m – D) at 0.200 m/s. Each takes the same time interval to finish the race:
Δt = 200 m

→ 0.200 m/s( ) 200 m( ) = 8.00 m/s( ) 1 000 m − D( )
1 000 m − D = 0.200 m/s( ) 200 m( )
8.00 m/s → D = 995 m
So, the tortoise is 1 000 m – D = 5.00 m from the finish line when the rabbit resumes running.
(b) Both begin the race at the same time: t = 0. The rabbit reaches the 800-m position at time t = 800 m/(8.00 m/s) = 100 s. The tortoise has crawled through 995 m when t = 995 m/(0.200 m/s) = 4 975 s. The rabbit has waited for the time interval t = 4 975 s – 100 s =
4 875 s .
Section 2.4 Acceleration P2.10 (a) At t = 2.00 s, x = [3.00(2.00)2 – 2.00(2.00) + 3.00] m = 11.0 m.
At t = 3.00 s, x = [3.00(3.00)2 – 2.00(3.00) + 3.00] m = 24.0 m
so
vavg =
= 13.0 m/s
v =
At t = 2.00 s, v = [6.00(2.00) – 2.00] m/s = 10.0 m/s
At t = 3.00 s, v = [6.00(3.00) – 2.00] m/s = 16.0 m/s
(c) aavg =
Δv Δt
(d) At all times a =
d dt
= 2.00 s and t = 3.00 s.
(e) From (b), v = (6.00t – 2.00) = 0 → t = (2.00)/(6.00) = 0.333 s.
P2.11 x = 2.00 + 3.00t – t2, so v =
dx dt
= −2.00
(b) v = (3.00 – 6.00) m/s = –3.00 m/s
(c) a = –2.00 m/s2
P2.12 (a) See the x vs. t graph at the top right.
Choose x = 0 at t = 0.
At t = 3 s, x = 1 2
8 m/s( ) 3 s( ) = 12 m
At t = 5 s, x = 12 m + 8 m/s( ) 2 s( ) = 28 m
At t = 7 s, x = 28 m + 1 2
8 m/s( ) 2 s( ) = 36 m
(b) See the a vs. t graph at the bottom right.
For 0 < t < 3 s, a =
8 m/s 3 s
For 3 < t < 5 s, a = 0
At the points of inflection, t = 3 and 5 s, the slope of the velocity curve changes abruptly, so the acceleration is not defined.
(c) For 5 s < t < 9 s, a = −
16 m/s 4 s
= −4 m/s2
(d) The average velocity between t = 5 and 7 s is
vavg = (8 m/s +0)/2 = 4 m/s
At t = 6 s, x = 28 m + 4 m/s( ) 1 s( ) = 32 m
(e) The average velocity between t = 5 and 9 s is
vavg = [(8 m/s) + (−8 m/s)]/2 = 0 m/s
ANS. FIG. P2.12
Chapter 2 39
At t = 9 s, x = 28 m + (0 m/s)(1 s) = 28 m
P2.13 (a) The area under a graph of a vs. t is equal to the change in velocity, v. We can use Figure 2.16 to find the change in velocity during specific time intervals.
The area under the curve for the time interval 0 – 10 s has the shape of a rectangle. Its area is
v = (2 m/s2)(10 s) = 20 m/s
The particle starts from rest, v0 = 0, so its velocity at the end of the 10-s time interval is
v = v0 + v = 0 + 20 m/s = 20 m/s
Between t = 10 s and t = 15 s, the area is zero: v = 0 m/s.
Between t = 15 s and t = 20 s, the area is a rectangle: v = (−3 m/s2)(5 s) = −15 m/s.
So, between t = 0 s and t = 20 s, the total area is v = (20 m/s) + (0 m/s) + (−15 m/s) = 5 m/s, and the velocity at t = 20 s is
5 m/s.
(b) We can use the information we derived in part (a) to construct a graph of x vs. t; the area under such a graph is equal to the displacement, x, of the particle.
From (a), we have these points (t, v) = (0 s, 0 m/s), (10 s, 20 m/s), (15 s, 20 m/s), and (20 s, 5 m/s). The graph appears below.
The displacements are:
0 – 10 s (area of triangle): x = (1/2)(20 m/s)(10 s) = 100 m
10 – 15 s (area of rectangle): x = (20 m/s)(5 s) = 100 m
15 – 20 s (area of triangle and rectangle):
x = (1/2)[(20 – 5) m/s](5 s) + (5 m/s)(5 s)
= 37.5 m + 25 m = 62.5 m
Total displacement over the first 20.0 s:
x = 100 m + 100 m + 62.5 m = 262.5 m = 263 m
P2.14 Choose the positive direction to be the outward direction, perpendicular to the wall.
v f = vi + at: a =
Δv Δt
P2.15 (a) a =
6.00 s = 1.3 m/s2
(b) Maximum positive acceleration is at t = 3 s, and is the slope of the graph, approximately (6 − 2)/(4 − 2) =
2 m/s2 .
(c) a = 0 at t = 6 s , and also for
t > 10 s .
(d) Maximum negative acceleration is at t = 8 s, and is the slope of the
graph, approximately –1.5 m/s2 .
(b)
(c)
(d)
(e)
(f) One way of phrasing the answer: The spacing of the successive positions would change with
less regularity.
Another way: The object would move with some combination of the kinds of motion shown in (a) through (e). Within one drawing, the accelerations vectors would vary in magnitude and direction.
Chapter 2 41
Section 2.6 Analysis Model: Particle Under Constant Acceleration
P2.17 Given vi = 12.0 cm/s when xi = 3.00 cm (t = 0), and at t = 2.00 s, xf = –5.00 cm,
x f − xi = vit + 1 2
at2 : − 5.00 − 3.00 = 12.0 2.00( ) + 1 2
a 2.00( )2
2 = −16.0 cm/s2
P2.18 Method One
Suppose the unknown acceleration is constant as a car moving at
vi1 = 35.0 mi h comes to a v f = 0 stop in x f 1 − xi = 40.0 ft . We find its
acceleration from v f 1 2 = vi1
2 + 2a x f 1 − xi( ):
a =
2
2 40.0 ft( ) 5 280 ft
mi( )2 1 h 3 600 s( )2
= −32.9 ft s2
Now consider a car moving at vi2 = 70.0 mi h and stopping to v f = 0
with a = −32.9 ft s2 . From the same equation its stopping distance is
x f 2 − xi = v f 2
2 − vi 2
2 −32.9 ft s2( ) 5 280 ft
1 mi( )2 1 h 3 600 s( )2
= 160 ft
Method Two
For the process of stopping from the lower speed vi1 we have
v f 2 = vi1
2 + 2a x f 1 − xi( ) ; 0 = vi1 2 + 2ax f 1 ; vi1
2 = −2ax f 1 . For stopping from
vi2 = 2vi1 , similarly 0 = vi2 2 + 2a x f 2 − 0( ) ; vi2
2 = −2ax f 2 . Dividing gives
vi2 2
vi1 2 =
P2.19 (a) x f − xi =
1 2
1 2
(b) a =
= −0.448 m/s2
P2.20 (a) vi = 100 m/s, a = –5.00 m/s2, vf = vi + at, so 0 = 100 – 5t →
t = 20.0 s
(b) No. See reason in (c).
(c) Find the distance (displacement) in which the plane needs to stop:
v f 2 = vi
0 = (100)2 + 2(–5)Δx → Δx = 1000 m = 1.00 km.
The required distance is greater than 0.800 km. The plane would
overshoot the runway.
P2.21 We have vi = 2.00 × 104 m/s, v f = 6.00 × 106 m/s,
x f − xi = 1.50 × 10−2 m .
(a) x f − xi =
vi + v f
2.00 × 104 m s + 6.00 × 106 m s
= 4.98 × 10−9 s
ax = v f
2 − vi 2
2(x f − xi ) =
6.00 × 106 m s( )2 − 2.00 × 104 m s( )2
2(1.50 × 10−2 m)
= 1.20 × 1015 m s2
P2.22 (a) Choose the initial point where the pilot reduces the throttle and the final point where the boat passes the buoy: xi = 0, x f = 100 m,
vxi = 30 m/s, vxf = ?, ax = −3.5 m/s2 , t = ?
x f = xi + vxit +
2 −3.5 m s2( )t2
1.75 m s2( )t2 − 30 m s( )t + 100 m = 0
We use the quadratic formula:
t = −b ± b2 − 4ac
Chapter 2 43
t = 30 m s ± 900 m2 s2 − 4 1.75 m s2( ) 100 m( )
2 1.75 m s2( ) = 30 m s ± 14.1 m s
3.5 m s2 = 12.6 s or 4.53 s
The smaller value is the physical answer. If the boat kept moving with the same acceleration, it would stop and move backward, then gain speed, and pass the buoy again at 12.6 s.
(b) vxf = vxi + axt = 30 m s − 3.5 m s2( )4.53 s = 14.1 m s
P2.23 In the simultaneous equations:
vxf = vxi + axt
vxi + vxf( )t
62.4 m = 1 2
vxi + vxf( ) 4.20 s( )
62.4 m =
1 2
14.9 m/s = vxf +
vxf = 3.10 m/s
P2.24 (a) Take any two of the standard four equations, such as
vxf = vxi + axt
vxi + vxf( )t
vxi = vxf – axt
x f − xi =
1 2
axt 2
We note that the equation is dimensionally correct. The units are units of length in each term. Like the standard equation
x f − xi = vxit +
a quadratic function of time.
(b) Our newly derived equation gives us for the situation back in problem 23,
62.4 m = vxf 4.20 s( ) − 1 2
−5.60 m/s2( ) 4.20 s( )2
vxf = 62.4 m − 49.4 m
4.20 s = 3.10 m/s
P2.25 (a) The time it takes the truck to reach 20.0 m/s is found from
v f = vi + at. Solving for t yields
t =
2.00 m s2 = 10.0 s
The total time is thus 10.0 s + 20.0 s + 5.00 s = 35.0 s
(b) The average velocity is the total distance traveled divided by the total time taken. The distance traveled during the first 10.0 s is
x1 = vt = 0 + 20.0
2( ) 10.0( ) = 100 m
With a being 0 for this interval, the distance traveled during the next 20.0 s is
x2 = vit +
1 2
The distance traveled in the last 5.00 s is
x3 = vt = 20.0 + 0
2( ) 5.00( ) = 50.0 m
The total distance x = x1 + x2 + x3 = 100 + 400 + 50 = 550 m, and the
average velocity is given by v = x
t = 550
35.0 = 15.7 m s .
P2.26 (a) Compare the position equation x = 2.00+ 3.00t – 4.00t2 to the general form
x f = xi + vit +
at2
to recognize that xi = 2.00 m, vi = 3.00 m/s, and a = –8.00 m/s2. The velocity equation, vf = vi + at, is then
vf = 3.00 m/s – (8.00 m/s2)t
Chapter 2 45
The particle changes direction when vf = 0, which occurs at
t =
x = 2.00 m + 3.00 m/s( ) 3
8 s
1 2
given by t = −
position, the time is
3 4
4 s
= −3.00 m/s .
*P2.27 (a) We choose a coordinate system with the x axis positive to the right, in the direction of motion of the speedboat, as shown on the right.
(b) Since the speedboat is increasing its speed, the particle under constant acceleration model should be used here.
(c) Since the initial and final velocities are given along with the displacement of the speedboat, we use
vxf 2 = vxi
a =
2
2Δx
(e) We have vi = 20.0 m/s, vf = 30.0 m/s, and xf – xi = Δx = 200 m:
a =
(30.0 m/s)2 − (20.0 m/s)2
2(200 m) = 1.25 m/s2
(f) To find the time interval, we use vf = vi + at, which gives
t =
ANS. FIG. P2.27
Section 2.7 Freely Falling Objects P2.28 Choose the origin y = 0, t = 0( ) at the starting point of the cat and take
upward as positive. Then yi = 0 , vi = 0 , and a = −g = −9.80 m s2 . The position and the velocity at time t become:
y f − yi = vit +
and
v f = vi + at: v f = −gt = − 9.80 m s2( )t
(a) At t = 0.1 s:
y f = − 1
2 9.80 m s2( ) 0.1 s( )2 = −0.049 0 m
v f = − 9.80 m s2( ) 0.1 s( ) = −0.980 m s
(b) At t = 0.2 s:
y f = − 1
v f = − 9.80 m s2( ) 0.2 s( ) = −1.96 m s
(c) At t = 0.3 s:
y f = − 1
v f = − 9.80 m s2( ) 0.3 s( ) = −2.94 m s
*P2.29 The bill starts from rest vi = 0 and falls with a downward acceleration of 9.80 m/s2 (due to gravity). For an average human reaction time of about 0.20 s, we can find the distance the bill will fall:
y f = yi + vit + 1 2
at2 → Δy = vit − 1 2
gt2
9.80 m/s2( )(0.20 s)2 = −0.20 m
The bill falls about 20 cm—this distance is about twice the distance between the center of the bill and its top edge, about 8 cm. Thus
David could not respond fast enough to catch the bill.
Chapter 2 47
P2.30 (a) We know vf = 0 when t = 3.00 s, and a = −g = −9.80 m/s2.
v f = vi + at
vi = 9.80 m/s2( ) 3.00 s( ) = 29.4 m/s
(b) Find the vertical displacement y:
Δy = y f − yi = 1 2
vi + v f( )t Δy =
1 2
Δy = 44.1 m
P2.31 Time to fall 3.00 m is found from the equation describing position as a
function of time, with vi = 0, thus: 3.00 m =
1 2
t = 0.782 s.
(a) With the horse galloping at 10.0 m/s, the horizontal distance is vt = 7.82 m .
(b) From above, t = 0.782 s
P2.32 (a) Consider the upward flight of the arrow.
vyf 2 = vyi
2 + 2ay y f − yi( ) 0 = 100 m s( )2 + 2 −9.8 m s2( )Δy
Δy = 10 000 m2 s2
19.6 m s2 = 510 m
(b) Consider the whole flight of the arrow.
y f = yi + vyit + 1 2
ayt 2
−9.8 m s2( )t2
The root t = 0 refers to the starting point. The time of flight is given by
t = 100 m s
4.9 m s2 = 20.4 s
P2.33 (a) We know when t = 1.50 s, y = (yf – yi) = 4.00 m; also, a = –g.
y f − yi = vit − 1 2
gt2
4.00 m = vi 1.50 s( ) − 4.90 m/s2( ) 1.50 s( )2
vi = 10.0 m/s up
v f = vi + at
v f = 10.0 m/s − 9.80 m/s2( ) 1.50 s( ) v f = −4.7 m/s
The final velocity is 4.7 m/s, down.
*P2.34 (a) The keys, moving freely under the influence of gravity (a = −g), undergo a vertical displacement of y = +h in time t. We use
Δy = vit + 1 2 at2 to find the initial velocity as
Δy = vit + 1 2
gt2
h t +
gt 2
(b) We find the velocity of the keys just before they were caught (at time t) using v = vi + at:
v = vi + at
v = h t +
gt2
Solving for t,
Chapter 2 49
t = +8.00 ± −8.00( )2 − 4 −4.90( ) 30( )
2 −4.90( ) = +8.00 ± 64 + 588
−9.80
t = 1.79 s
Section 2.8 Context Connection: Acceleration Required by Consumers P2.36 (a) The Bugatti Veyron requires the shortest time to speed up from 0
to 60 mi/h, so it has the greatest average acceleration:
vxf = vxi + axt
ax = vxf − vxi
2.6 s = 23.1 mi h ⋅ s
in agreement with the value listed. For the Smart ForTwo,
ax =
13.3 s = 4.51 mi h ⋅ s
in agreement with the 4.5 mi h ⋅ s listed.
(b) Bugatti:

significant digits.
Smart ForTwo:

1 2

1 609 mi 1 mi

As the acceleration of 10 m/s2 calculated in part (b) shows, the Bugatti Veyron clearly accelerates at a rate faster than g from 0 to 60 mi/h.
P2.37 (a) Assuming a constant acceleration:
a =
8.00 s = 5.25 m s2
(b) Taking the origin at the original position of the car,
x f =
1 2
42.0 m s( ) 8.00 s( ) = 168 m
(c) From v f = vi + at, the velocity 10.0 s after the car starts from rest is:
v f = 0 + 5.25 m s2( ) 10.0 s( ) = 52.5 m s
P2.38 (a) Starting from rest and accelerating at ab = 13.0 mi/h · s, the bicycle reaches its maximum speed of vb,max = 20.0 mi/h in a time
tb ,1 =
13.0 mi/h ⋅ s = 1.54 s
Since the acceleration ac of the car is less than that of the bicycle, the car cannot catch the bicycle until some time t > tb,1 (that is, until the bicycle is at its maximum speed and coasting). The total displacement of the bicycle at time t is
Δxb = 1 2
= 1.47 ft/s 1 mi/h

The total displacement of the car at this time is
Δxc =
= 6.62 ft/s2( )t2
At the time the car catches the bicycle xc = xb. This gives
6.62 ft/s2( )t2 = 29.4 ft/s( )t − 22.6 ft
or t 2 − 4.44 s( )t + 3.42 s2 = 0
Chapter 2 51
that has only one physically meaningful solution t > tb,1. This solution gives the total time the bicycle leads the car and is
t = 3.45 s .
(b) The lead the bicycle has over the car continues to increase as long as the bicycle is moving faster than the car. This means until the car attains a speed of vc = vb,max = 20.0 mi/h. Thus, the elapsed time when the bicycle’s lead ceases to increase is
t =
At this time, the lead is
Δxb − Δxc( )max = Δxb − Δxc( ) t=2.22 s
= 29.4 ft/s( ) 2.22 s( ) − 22.6 ft[ ] − 6.62 ft/s2( ) 2.22 s( )2
or Δxb − Δxc( )max
ax = vxf − vxi
2.50 s = 70.0 mi h ⋅ s
= 70.0 mi h ⋅ s 1 609 mi 1 m

vxi + vxf( )t
= 1 2

= 97.8 m
*P2.40 (a) The velocity is constant between ti = 0 and t = 4 s. Its acceleration is 0 .
(b) a = (v9 − v4)/(9 s − 4 s) = (18 − [−12]) (m/s)/5 s = 6.0 m/s2
(c) a = (v18 − v13)/(18 s − 13 s) = (0 − 18) (m/s)/5 s = –3.6 m/s2
(d) We read from the graph that the speed is zero
at t = 6 s and at 18s .
(e) and (f) The object moves away from x = 0 into negative coordinates from t = 0 to t = 6 s, but then comes back again, crosses the origin and moves farther into positive coordinates until
t = 18 s , then attaining its maximum distance, which is the cumulative distance under the graph line:
x = (−12 m/s)(4 s) + (1/2)(−12 m/s)(2 s)
+ (1/2)(18 m/s)(3 s) + (18 m/s)(4 s) + (1/2)(18 m/s)(5 s)
x = −60 m + 144 m = 84 m
(g) We consider the total distance, rather than the resultant displacement, by counting the contributions computed in part (f) as all positive:
d = +60 m + 144 m = 204 m
P2.41 (a) a =
v f − vi
= −662 ft/s2 = −202 m/s2
at2 = 632 mi/h( ) 5 280 ft/mi 3 600 s/h

= 649 ft = 198 m
P2.42 Take downward as the positive y direction.
(a) While the woman was in free fall, Δy = 144 ft , vi = 0, and we take
a = g = 32.0 ft s2 . Thus,
Δy = vit +
1 2
giving tfall = 3.00 s. Her velocity just before impact is:
v f = vi + gt = 0 + 32.0 ft s2( ) 3.00 s( ) = 96.0 ft s
Chapter 2 53
(b) While crushing the box, vi = 96.0 ft s , v f = 0 , and
Δy = 18.0 in. = 1.50 ft. Therefore,
a =
or
(c) Time to crush box:
Δt = Δy
v = Δy
or Δt = 3.13 × 10−2 s .
P2.43 (a) Take initial and final points at top and bottom of the first incline. If the ball starts from rest, vi = 0 , a = 0.500 m/s2, xf – xi = 9.00 m. Then
v f 2 = vi
2 + 2a x f − xi( ) = 02 + 2 0.500 m/s2( ) 9.00 m( )
v f = 3.00 m/s
x f − xi = vit +
t = 6.00 s
(c) Take initial and final points at the bottom of the first plane and the top of the second plane, respectively: vi = 3.00 m/s, vf = 0, xf – xi = 15.0 m. We use
v f 2 = vi
which gives
2 15.0 m( ) = −0.300 m/s2
(d) Take the initial point at the bottom of the first plane and the final point 8.00 m along the second plane:
vi = 3.00 m/s, xf – xi = 8.00 m, a = –0.300 m/s2
v f 2 = vi
2 + 2a x f − xi( ) = 3.00 m/s( )2 + 2 −0.300 m/s2( ) 8.00 m( )
= 4.20 m2/s2
v f = 2.05 m/s
P2.44 Let the glider enter the photogate with velocity vi and move with constant acceleration a. For its motion from entry to exit,
x f = xi + vxit + 1 2
axt 2
aΔtd
(a) The speed halfway through the photogate in space is given by
vhs
not equal to vd unless a = 0 .
(b) The speed halfway through the photogate in time is given by
vht = vi + a
equal to vd as determined above.
P2.45 (a) Take the x axis along the tail section of the snake. The displacement from tail to head is
240 mi + (420 − 240) mcos(180° − 105°)i − 180 msin 75° j
= 287 mi − 174 mj
Its magnitude is 287( )2 + 174( )2 m = 335 m.
From v =
distance Δt
Inge: Δt = distance
v =
335 m h( ) 1 km( ) 3 600 s( ) 12 km( ) 1 000 m( ) 1 h( ) = 101 s
Olaf: Δt = 420 m ⋅ s 3.33 m
= 126 s
(b) Olaf must run the race in the same time:
Chapter 2 55
v =

= 15.0 km/h
P2.46 (a) As we see from the graph, from about –50 s to 50 s Acela is cruising at a constant positive velocity in the +x direction. From 50 s to 200 s, Acela accelerates in the +x direction reaching a top speed of about 170 mi/h. Around 200 s, the engineer applies the brakes, and the train, still traveling in the +x direction, slows down and then stops at 350 s. Just after 350 s, Acela reverses direction (v becomes negative) and steadily gains speed in the –x direction.
(b) The peak acceleration between 45 and 170 mi/h is given by the slope of the steepest tangent to the v versus t curve in this interval. From the tangent line shown, we find
a = slope = Δv Δt
100 − 50( ) s = 2.2 mi/h( )/s = 0.98 m/s2
(c) Let us use the fact that the area under the v versus t curve equals the displacement. The train’s displacement between 0 and 200 s is equal to the area of the gray shaded region, which we have approximated
with a series of triangles and rectangles.
Δx0→200 s = area1 + area2 + area3 + area4 + area5
≈ 50 mi/h( ) 50 s( ) + 50 mi/h( ) 50 s( ) + 160 mi/h( ) 100 s( )
+ 1 2
100 s( ) 170 mi/h − 160 mi/h( ) = 24 000 mi/h( ) s( )
Now, at the end of our calculation, we can find the displacement in miles by converting hours to seconds. As 1 h = 3 600 s,
ANS. FIG. P2.46(a)
ANS. FIG. P2.46(c)
Δx0→200 s =

s( ) = 6.7 mi .
P2.47 Average speed of every point on the train as the first car passes Liz:
Δx Δt
= 5.73 m s
The train has this as its instantaneous speed halfway through the 1.50-s time. Similarly, halfway through the next 1.10 s, the speed of the train
is
8.60 m 1.10 s
= 7.82 m s . The time required for the speed to change from
5.73 m/s to 7.82 m/s is
1 2
Δvx
1.30 s = 1.60 m s2
P2.48 The train accelerates with a1 = 0.100 m/s2 then decelerates with a2 = –0.500 m/s2. We can write the 1.00-km displacement of the train as
x = 1 000 m = 1
2 a1t1
1 000 m = 1
s ≈ 26 s
Total time = t = t1 + t2 = 129 s + 26 s = 155 s
P2.49 Let point 0 be at ground level and point 1 be at the end of the engine burn. Let point 2 be the highest point the rocket reaches and point 3 be just before impact. The data in the table below are found for each phase of the rocket’s motion.
(0 to 1): v f 2 – (80.0 m/s)2 = 2(4.00 m/s2 )(1 000 m) ANS. FIG. P2.49
Chapter 2 57
so vf = 120 m/s. Then, 120 m/s = 80.0 m/s + (4.00 m/s2)t
giving t = 10.0 s.
(1 to 2) 0 – (120 m/s)2 = 2(–9.80 m/s2)(xf – xi)
giving yf – yi = 735 m,
0 – 120 m/s = ( –9.80 m/s2)t
giving t = 12.2 s.
This is the time of maximum height of the rocket.
(2 to 3) v f 2 – 0 = 2(–9.80 m/s2 )(–1 735 m) or vf = –184 m/s
Then vf = –184 m/s = (–9.80 m/s2)t
giving t = 18.8 s.
(a) ttotal = 10 s + 12.2 s + 18.8 s = 41.0 s
(b)
#1 End Thrust 10.0 1 000 120 +4.00
#2 Rise Upwards 22.2 1 735 0 –9.80
#3 Fall to Earth 41.0 0 –184 –9.80
P2.50 Distance traveled by motorist = (15.0 m/s)t
Distance traveled by policeman = 1
2 2.00 m/s2( )t2
(a) intercept occurs when 15.0t = t2 , or t = 15.0 s
(b) v officer( ) = 2.00 m s2( )t = 30.0 m s
(c) x officer( ) = 1
2 2.00 m s2( )t2 = 225 m
P2.51 Consider the runners in general. Each completes the race in a total time interval T. Each runs at constant acceleration a for a time interval t, so
each covers a distance (displacement) xa = ½at2 where they eventually reach a final speed (velocity) v = at, after which they run at this constant speed for the remaining time (T – t) until the end of the race, covering distance xv = v(T – t) = at (T – t). The total distance (displacement) each covers is the same:
x = xa + xv
x = a[½t2 + t (T – t)]
so
where x = 100 m, and T = 10.4 s.
(a) For Laura (runner 1), t1 = 2.00 s:
a1 = (100 m)/(18.8 s2) = 5.32 m/s2
For Healan (runner 2), t2 = 3.00 s:
a2 = (100 m)/(26.7 s2) = 3.75 m/s2
(b) Laura (runner 1): v1 = a1t1 = 10.6 m/s
Healan (runner 2): v2 = a2t2 = 11.2 m/s
(c) The 6.00-s mark occurs after either time interval t. From the reasoning above, each has covered the distance
x = a[½t2 + t(t – t)]
where t = 6.00 s.
Laura (runner 1): x1 = 53.19 m
Healan (runner 2): x2 = 50.56 m
So, Laura is ahead by (53.19 m − 50.56 m) = 2.63 m.
(d) Laura accelerates at the greater rate, so she will be ahead of Healen at, and immediately after, the 2.00-s mark. After the 3.00-s mark, Healan is travelling faster than Laura, so the distance between them will shrink. In the time interval
Chapter 2 59
from the 2.00-s mark to the 3.00-s mark, the distance between them will be the greatest.
During that time interval, the distance between them (the position of Laura relative to Healan) is
D = x1 – x2 = a1[½t1 2 + t1(t – t1)] – ½a2t
2
because Laura has ceased to accelerate but Healan is still accelerating. Differentiating with respect to time, (and doing some simplification), we can solve for the time t when D is an maximum:
dD/dt = a1t1
t – a2t = 0 →
t = t1(a1/a2) = 2.84 s.
Substituting this time back into the expression for D, we find that D = 4.47 m, that is, Laura ahead of Healan by 4.47 m.
P2.52
5.00 0.58
TABLE P2.52
Chapter 2 61
The very convincing fit of a single straight line to the points in the graph of velocity versus time indicates that the rock does fall with constant acceleration. The acceleration is the slope of line:
aavg = –1.63 m/s2 = 1.63 m/s2 downward
P2.53 (a)
at2 = 50.0 m = 2.00 m/s( )t + 1 2
9.80 m/s2( )t2
t = 3.00 s after the first stone is thrown.
(b) y f = v2 it +
1 2
Substitute 50.0 = vi2 2.00( ) + 1
2 9.80( ) 2.00( )2 :
vi2 = 15.3 m/s downward
= 31.4 m/s downward
= 34.8 m/s downward
P2.54 Let the ball fall freely for 1.50 m after starting from rest. It strikes at speed given by
vxf
vxf
vxf = −5.42 m/s
If its acceleration were constant, its stopping would be described by
vxf 2 = vxi
2 + 2ax x f − xi( ) 0 = −5.42 m/s( )2 + 2ax −10−2 m( )
ax = −29.4 m2/s2
−2.00 × 10−2 m = +1.47 × 103 m/s2
Upward acceleration of this same order of magnitude will continue for some additional time after the dent is at its maximum depth, to give the ball the speed with which it rebounds from the pavement. The ball’s maximum acceleration will be larger than the average acceleration we estimate by imagining constant acceleration, but will
still be of order of magnitude ∼ 103 m/s2 .
P2.55 The rock falls a distance d for a time interval t1 and the sound of the splash travels upward through the same distance d for a time interval t2 before the man hears it. The total time interval t = t1 + t2 = 2.40 s.
(a) Relationship between distance the rock falls and time interval t1: d = ½ g t1
2
Relationship between distance the sound travels and time interval t2: d = vst2, where vs = 336 m/s.
d = vsΔt2 =
1 2
gΔt1 2
2
vsΔt2
9.80 m/s2
Δt2( )2 − 73.37( )Δt2 + 5.76 = 0
Solving the quadratic equation gives
t2 = 0.078 6 s → d = vst2 = 26.4 m
(b) Ignoring the sound travel time,
d =
1 2
9.80 m/s2( ) 2.40 s( )2 = 28.2 m, an error of 6.82% .
Chapter 2 63
*P2.56 We can write expressions of position to see whether the front of the passenger car (P) meets the back of the freight car (F) at some time.
Assume at t = 0, the coordinate of the front of the passenger car is xPi = 0; and the coordinate of the back of the freight car is xFi = 58.5 m. At later time t, the coordinate of the front of the passenger car is
xP = xPi + vPit + 1 2
aPt2
−3.00 m/s2( )t2
and the coordinate of the back of the freight car is
xF = xFi + vFit + 1 2
aFt2
Setting these expression equal to each other gives
xP = xF
−3.00 m/s2( )t2 = 58.5 m + 16.0 m/s( )t
or 1.50( )t2 + −24.0( )t + 58.5 = 0
after simplification and dropping units.
We do not have to solve this equation, we just want to check if a solution exists;
if a solution does exist, then the trains collide.
A solution does exist:
2 1.50( )
so the situation is impossible.
P2.57 The distance x and y are always related by
x 2 + y2 = L2 . Differentiating this equation
with respect to time, we have
2x
and
from differentiation, we have
vB = v
ANS. FIG. P2.57
Chapter 2 65
ANSWERS TO EVEN-NUMBERED PROBLEMS P2.2 (a) 50.0 m/s; (b) 41.0 m/s
P2.4 (a) 3.75 m/s; (b) 0
P2.6 (a) 27.0 m; (b) 27.0 m + (18.0 m/s) t + (3.00 m/s2)(t); (c) 18.0 m/s
P2.8 (a) See ANS. FIG. P2.8; (b) 23 m/s, 18 m/s, 14 m/s, and 9.0 m/s; (c) 4.6 m/s2; (d) zero
P2.10 (a) 13.0 m/s; (b) 10.0 m/s, 16.0 m/s; (c) 6.00 m/s2; (d) 6.00 m/s2; (e) 0.333 s
P2.12 (a) See Ans. Fig. P2.12, top; (b) See Ans. Fig. P2.12, bottom; (c) −4 m/s2; (d) 32 m; (e) 28 m
P2.14 1.34 × 104 m/s2
P2.16 (a–e) See graphs in P2.16; (f) with less regularity
P2.18 160 ft.
P2.20 (a) 20.0 s; (b) No; (c) The plane would overshoot the runway.
P2.22 4.53 s
1 2
P2.26 (a) 2.56 m; (b) −3.00 m/s
P2.28 (a) yf = –0.049 0 m, vf = –0.980 m/s; (b) yf = –0.196 m, vf = –1.96 m/s; (c) yf = –0.441 m, vf = –2.94 m/s
P2.30 (a) 29.4 m/s; (b) 44.1m
P2.32 (a) 510 m; (b) 20.4 s
P2.34 (a)
h t +
gt 2
h t −
gt 2
P2.36 (a) See proof in P2.36(a); (b) Bugatti: 10 m/s2, Smart: 2.0 m/s2; (c) Bugatti: 35 m, Smart: 180 m; (d) 2.74 s: the Bugatti is faster than this time from 0 to 60 mi/h
P2.38 (a) 3.45 s; (b) 10.0 ft.
P2.40 (a) 0; (b) 6.0 m/s2; (c) −3.6 m/s2; (d) at t = 6 s and at 18 s; (e and f) t = 18 s; (g) 204 m
P2.42 (a) 96.0 ft/s; (b) 3.07 × 103 ft s2 upward ; (c) 3.13 × 10−2 s
P2.44 (a) not equal to vd unless a = 0; (b) equal to vd
P2.46 (a) See complete description in P2.46(a); (b) 2.2 (mi/h)/s; (c) 6.7 mi
P2.48 155 s
P2.50 (a) 15.0 s; (b) 30.0 m/s; (c) 225 m
P2.52 (a and b) See Table P2.52; (c) 1.63 m/s2 downward and see graph in P2.52
P2.54 ~103 m/s2
P2.56 The trains do collide.
3 Motion in Two Dimensions
CHAPTER OUTLINE
3.2 Two-Dimensional Motion with Constant Acceleration
3.3 Projectile Motion
3.5 Tangential and Radial Acceleration
3.6 Relative Velocity and Relative Acceleration
3.7 Context Connection: Lateral Acceleration of Automobiles
*OQ3.1 (a) Yes (b) No: The escaping jet exhaust exerts an extra force on the plane. (c) No (d) Yes (e) No: The stone is only a few times more dense than water, so friction is a significant force on the stone. The answer is (a) and (d).
*OQ3.2 The centripetal acceleration a = v2/r becomes (3v)2/(3r) = 3v2/r, so it is 3 times larger. The answer is (b).
*OQ3.3 The car’s acceleration must have an inward component and a forward component: answer (e). Another argument: Draw a final velocity vector of two units west. Add to it a vector of one unit south. This represents subtracting the initial velocity from the final velocity, on the way to finding the acceleration. The direction of the resultant is that of vector (e).
*OQ3.4 (i) The 45° angle means that at point A the horizontal and vertical velocity components are equal. The horizontal velocity component is the same at A, B, and C. The vertical velocity component is zero at B and negative at C. The assembled answer is a = b = c > d = 0 > e.

Instructor's Solution Manuals to Principles of Physics: A Calculus-Based Text

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