Chapter 6 Position and Movement

Checkpoint (p.8)

1. time: second (s); length: metre (m)

2. (a) hour (h) / minute (min)

(b) metre (m) / feet (ft)

(c) metre (m) / kilometre (km) / feet (ft)

(d) minute (min) / second (s)

3. (a) The distance between the school and the nearest bus stop can be measured

by counting the number of footsteps needed to walk from one place to

another. (or any other reasonable answers)

(b) The time needed to travel to one place to another can be measured by a

wrist watch. (or any other reasonable answers)

Exercise (p.8)

1. 1 day = 24 hours = = 86 400 s

1 year = 365 days = = 31 536 000 s

2. (a) 1 inch = 2.54 cm = 0.0254 m

(b) 1 foot = 304.8 mm = 0.3048 m

(c) 1 yard = 914.4 mm = 0.9144 m

(d) 1 mile = 1.609 344 km = 1609.344 m

3. (a) minute (min) / hour (h)

(b) minute (min)

(c) millimetre (mm) / centimetre (cm)

(d) metre (m)

4. A sundial measures time by the position of the sun. As the position of the sun in

the sky varies with time, it casts a shadow of the rod onto a horizontal table with

hour-lines. The corresponding time can be read from the hour-line which aligns

with the shadow.

A sandglass consists of two bulbs connected vertically by a narrow tube. The

bottom bulb is usually filled with sand. When the sandglass is inverted, the sand

falls from one bulb to the other and the time needed is fixed. Therefore, by

defining the time needed as a fixed period, time intervals can be measured by

counting the number of such period passed.

In a Chinese water clock, water flows out from the upper vessel to the lowest

vessel at a fixed rate. As time passes, the ‘ruler’ floated on the lowest vessel will

rise due to the rise in water level. Time can be read from the markings on the

‘ruler’.

5. (a) percentage error of the atomic clock

(b) time taken for the atomic clock to miss 1 s

6. A stopwatch is more suitable because it is more convenient to use a stopwatch

for this measurement. The time taken for the Ferris wheel to complete a

revolution is quite long (about a few minutes) and a stopwatch is already

accurate enough (an error of a few seconds).

7. (a) Electronic timing is more accurate than the use of stopwatches.

(b) percentage error

(c) No. The error due to the time keeper’s reaction time is too large in this

situation because the finishing times are so close that the ranks determined

by the readings may be inaccurate. The tolerance of the error of the reading

and the practicality of usage are also factors needed to be considered.

8.

percentage error

Checkpoint (p.13)

1. (a) total distance travelled by the ball

= OA + AB + BC

= 120 + 360 + 82

= 562 cm

(b) displacement of the ball from O to A

displacement of the ball from A to B

displacement of the ball from B to C

total displacement =

(c) The sign of the total displacement is negative. It means that the final

position of the ball is on the left of the original position.

2. total displacement

total distance travelled

Exercise (p.14)

1. C

2. C

3. B

4. B

5. (a) The fare depends on the distance travelled.

(b) The distance between the road sign and the corresponding destination

6. (a) (i) A student goes to school from his home and returns home after school.

(or any other reasonable answers)

(ii) A person walks along a straight line.

(or any other reasonable answers)

(b) No. The shortest distance between two positions is the length of the straight

line joining the two positions, which is the magnitude of the displacement.

So, the distance travelled cannot be smaller than the magnitude of the

displacement.

7. (a) Take the downward direction as positive.

(i) displacement of the ball from A to B

displacement of the ball from B to C

displacement of the ball from C to D

(ii) total displacement

(iii) The sign of the total displacement is positive. It means that the ball is

below the initial position.

(b) Take the upward direction as positive.

total displacement

The negative sign suggests that the ball is below the initial position. It

agrees with the result in (a).

8. (a) distance travelled by the batter

distance travelled by the ball

(b) magnitude of displacement of the batter

The displacement of the batter is 38.7 m (NW).

magnitude of displacement of the ball

The direction in which the ball travels is NθW and θ can be determined by

The displacement of the ball is 57.2 m (N28.6°W).

9.

Applying the cosine law,

magnitude of displacement of the typhoon

Applying the sine law,

Therefore the typhoon travels towards the direction N(180°−106.3°)W

≈ N73.7°W.

The displacement of the typhoon is 589 km (N73.7°W).

10. (a) distance travelled

displacement

(b) Since his final position is still at B, his final displacement is 114 m.

Hong Kong (O)

400 m 800 m45°

B

A

11. (a)

(b) total displacement = displacement from A to D

12. (a) diameter of the Earth

(b) distance travelled

displacement = 12 740 km

Checkpoint (p.21)

1. B

2.average speed

Take the right as the positive direction.

average velocity

3. (a) Cockroach A has the highest average speed.

(b) The three cockroaches have the same average velocity.

Exercise (p.21)

1. (a) yes

(b) yes

(c) yes

(d) yes

2. (a) same

(b) different

(c) same

(d) different

3. (a) non-zero

(b) zero

(c) non-zero

(d) zero

4. A

5. (a) Student A walks towards the north for a certain distance and a certain time

interval while student B walks towards any other direction for the same

distance and time interval. (or any other reasonable answers)

(b) In a running race which starts and ends at the same position, the

displacements of the runners and hence their average velocities are zero but

not their average speeds during the race. (or any other reasonable answers)

6. average speed

7. (a) average speed

average velocity

(b) The winner is the one who has the highest average speed. In a 100 m

swimming contest, the swimmers return to their starting position and so

their average velocities are all zero.

8. (a) time of travel

average speed

Her average speed is lower than the speed limit.

(b) Although her average speed is lower than the speed limit, her instantaneous

speed at a certain instant may be higher than the limit. Hence, she may still

be fined for speeding.

9. (a) average speed

(b) Let h be the height of the orbit above the ground.

The orbit of the satellite is 4630 km above the ground.

10. (a) average speed of the plane

average speed of the train

ratio

(b) average speed

Take the north as positive direction.

average velocity

The average speed and average velocity of the train are 77.5 km h−1 and

22.5 km h−1 (to the north) respectively.

11. (a) Take the right as the positive direction.

total displacement

His total displacement is 4.8 m to the left of his original position.

(b) average speed

average velocity

12. (a) distance travelled

average speed

displacement (NE)

average velocity (NE)

(b) No. The instantaneous velocity of the truck is not constant as its direction

changes when the truck moves around the roundabout.

Checkpoint (p.26)

1. aeroplane: no

cable car: no

dog: yes

capsule: yes

2. average acceleration

final velocity

3. The object is speeding up because the magnitude of its velocity is increasing.

Exercise (p.27)

1. C

2. B

3. B

4. A

5. (a) A car accelerates along a straight road.

(b) A car moves around a corner at a constant speed.

(c) A ball rebounds off a racket.

6. At A, the cyclist has acceleration as he gains speed by cycling down the slope.

At B, he has no acceleration as he maintains his speed after reaching the ground

and there is no change in his direction of motion.

At C, although he moves at a constant speed, he has acceleration as his direction

of motion changes.

At D, he has no acceleration as there is no change in his speed and direction of

motion.

At E, he decelerates by applying a hard brake on the bicycle and hence he has

acceleration as there is a change in his velocity.

7. For Ferrari F430,

Applying ,

For Lamborghini Gallardo,

Hence, Ferrari F430 has a larger average acceleration.

8. (a) average acceleration

(b) final speed

9. Take the upward direction as positive.

acceleration during the first 2 s

acceleration during the next 10 s

acceleration during the last 2.5 s

The accelerations during the first 2 s, the next 10 s and the last 2.5 s are

0.75 m s−2 (upwards), 0 and 0.6 m s−2 (downwards).

10. (a) The skater moves to the right at a decreasing speed until her speed has

decreased to zero. Then she reverses her direction of motion and increases

her speed.

(b) Take the right as the positive direction.

Applying ,

She will reach a speed of 2 m s−1 again but in the opposite direction 5 s later.

11. (a) Take the direction down the inclined plane as positive.

initial velocity

final velocity

acceleration

As a is positive, it points to the direction down the inclined plane.

Applying , the velocity of the ball 1.2 s after the projection is

As v is positive, it also points to the direction down the inclined plane.

Hence, the physical results remain the same using the two methods of

calculation.

(b) (i) They are in the same direction.

(ii) They are in opposite directions.

12. (a) time taken for the water to move from A to B

(b) Take the right as the positive direction.

average acceleration

The average acceleration is 2 m s−2 to the left.

Checkpoint (p.31)

1.

time interval Simon’s motion velocity

0–50 s He walks towards the right at

a constant velocity.

50–110 s He stays at rest.

110–150 s He walks towards the left at a

constant velocity.

At t = 150 s, he is at a

distance of 20 m to the left of

O.

Checkpoint (p.36)

1. (a) During t = 0–5 s,

acceleration = slope of the v-t graph

During t = 5–10 s,

acceleration = slope of the v-t graph

(b) total displacement = total area under the v-t graph

2. D

Checkpoint (p.39)

1. (a) C

(b) D

Exercise (p.39)

1. (a) A

(b) D

2. C

3. A

4. D

5. A

6. (a) C < B < A

(b) C < B < A

(c) A < B < C

(d) A < B < C

7. (a) correct

(b) correct

(c) incorrect

(d) incorrect

8. (a) C

(b) D

(c) B

(d) A

9. (a) From the graph, the shop is located at a position 160 m east of Jeff’s home.

(b) During t = 0−3 min, he goes to the shop at a constant velocity. Then he stays

at the shop during t = 3−6 min. During t = 6−11 min, he returns home with a

constant velocity.

(c) total displacement = 0

It is because his motion starts and ends at the same position.

total distance travelled

(d) average speed

average velocity

10. (a)

(b) distance travelled between the stations

= area under the v-t graph

a / m s−2

t / m s−2

30 100 1400

−0.5

0.571

(c) Let tm be the time when the train is midway between the stations.

distance from a station to the position midway between the stations

distance travelled by the train

= area under the v-t graph from t = 0 to t = tm

= area under the v-t graph during t = 0 to t = 35 s

+ area under the v-t graph from t = 35 to t = tm

Hence,

(d) average velocity

11. (a) acceleration = slope of the v-t graph

initial acceleration of car A

initial acceleration of car B

Hence, car A has a larger acceleration.

(b) No. It only means that the cars are travelling at the same velocity at that

time.

(c) Car B overtakes car A at the time when the areas under their v-t graphs are

the same.

At t = 10 s,

distance travelled by car A

= area under its v-t graph during t = 0–10 s

distance travelled by car B

= area under its v-t graph during t = 0–10 s

Hence, car B overtakes car A at t = 10 s.

12. (a)

(b) During t = 0–5 s

acceleration of the car

During t = 5–7 s,

acceleration of the car

1 2 3 4 5 6 7t / s0

10

20

30

40

50

v / km h−1

5 7t / s0

2.78

a / m s−2

13. displacement = area under the v-t graph

For Figure Q13a,

time t / s displacement s / m

0 0

2 3

4 12

6 27

8 48

10 75

12 102

14 123

16 138

18 147

20 150

2 4 6 8 10 12 140

20

16 18 20t / s

s / m

40

60

80

100

120

140

160

For Figure Q13b,

time t / s displacement s / m

0 0

2 10

4 20

6 20

8 20

10 20

12 0

14 −20

16 −40

18 −60

20 −80

14. (a) Initially, the object moves with a constant acceleration from rest. It then

maintains its velocity after a certain time.

−80

s / m

−60

−40

−20

0

20

t / s2 4 6 8 10 12 14 16 18 20

0 0

s / m a / m s−2

t / s t / s

(b) The object decelerates until it stops.

(c) The object decelerates until it stops. Then it reverses its direction of motion.

The acceleration is constant throughout the motion.

(d) The object moves to the negative direction at a constant velocity.

0

0

s / m a / m s−2

t / s

t / s

0

0

s / m a / m s−2

t / s

t / s

0

v / m s−1 a / m s−2

t / s

t / s

0

15. The s-t, v-t and a-t graphs of the minibus are shown below.

Checkpoint (p.47)

1. (a) unsuitable

(b) suitable

(c) suitable

2. There are 12 ticks on the ticker-tape.

3. C

Exercise (p.47)

1. D

2. D

3. (a) The v-t graph represents a uniformly accelerated motion.

(b) The trolley moves away from the motion sensor.

(c) average acceleration of the trolley = slope of the v-t graph

4. (a) The car accelerates to the left.

(b) Plot an a-t graph and see whether the acceleration is non-zero.

[OR Plot a v-t graph and see whether the velocity is increasing.]

5. (a) Trolley A moves at a constant velocity.

Trolley B accelerates.

v

t

t

a

0

0

0

s

(b) Trolley B has a higher average velocity. The number of ticks on the tape of

trolley A and B are 9 and 7 respectively. Trolley B takes less time to travel

the same distance and hence it has a higher average velocity.

6. (a) The ball moves to the right and decelerates.

(b) (i) average velocity between A and B

(ii) average velocity between B and C

(iii) average velocity between C and D

(iv) average velocity between D and E

(c)

(d) average acceleration of the ball = slope of the v-t graph

v / cm s−1

0.1 0.2 0.3 0.40

10

t / s

20

30

40

Checkpoint (p.54)

1. We are given

, we have to find t and s.

Applying the equation ,

Applying the equation ,

Alternative method:

Applying the equation v2 − u2 = 2as,

2. D

Exercise (p.54)

1. B

2. D

3. A

4. C

5. C

6. B

7. The final velocity is

The distance it travel during the time interval is

Alternative method:

8. (a)

The time taken by the Shanghai Maglev train is

The time taken by the MTR train is

(b) The distance travelled by the maglev train when it accelerates is

The distance travelled by the MTR train when it accelerates is

The distance travelled by the maglev train is 10482.6 − 224.422 ≈ 10 300 m

longer.

9. The time taken for the runner to accelerate is

The distance she still has to run after acceleration is

The time taken for her to finish the race after acceleration is

Her finishing time is

10. (a) average acceleration

(b) average acceleration of the red ball

The distance travelled by the red ball after the collision is

Hence, the red ball will fall into the pocket.

11. Let t be the time needed to catch the train.

Applying ,

or (rejected)

Since t ≈ 3.16 s > 3 s, he will not be able to catch the train.

12. (a) Applying ,

The cheetah catches the deer at a position 27 m away from its initial

position.

(b) Applying ,

The velocity of the cheetah when it catches the deer is 18 m s−1.

(c) distance travelled by the deer during the chase

velocity of the deer

13. (a)

Applying ,

The velocity of the plane when it goes beyond the runway is 60.2 m s−1.

(b) Applying ,

The acceleration of the plane is −7.25 m s−2. It means the plane decelerates

at 7.25 m s−2 when it is on the field.

(c) Applying ,

The total time taken for the plane to stop on the ground is

14. (a)

reaction time

Applying ,

acceleration

The assumed reaction time and acceleration of the vehicle is 0.900 s and

−6.17 m s−2 respectively. It means the vehicle is assumed to be decelerated

at 6.17 m s−2.

(b)

thinking distance

braking distance

stopping distance

15. (a) From the graph, the reaction time of the driver is 0.8 s.

(b) acceleration

The acceleration of the car is −3.75 m s−2. It means the car decelerates at

3.75 m s−2.

(c) stopping distance

= area under the v-t graph

The car stops at a distance of before the traffic light.

(d) thinking distance

braking distance

acceleration

The acceleration of the car is −4.17 m s−2. It means the car decelerates at

4.71 m s−2.

Checkpoint (p.62)

1. Take the downward direction as positive. We are given , ,

, and we have to find s and v.

Applying the equation ,

Applying the equation ,

2. Take the upward direction as positive. We are given , and

, and we have to find s.

Applying the equation ,

The displacement is negative. This shows that the ball is 15 m below the point of

projection at t = 3 s.

(The mark should be drawn below the point of projection.)

3. (a) From the v-t graph, the stone is thrown at the speed of 15 m s−1.

(b) The stone takes 1.5 s to reach the highest point.

(c) maximum height = area under the v-t graph from t = 0 to t = 1.5 s

= 11.25 m

Exercise (p.63)

1. B

2. D

3. C

4. B

5. D

6. (a) correct

(b) incorrect

(c) incorrect

(d) correct

(e) correct

(f) correct

7. (a) correct

(b) incorrect

(c) correct

(d) correct

(e) correct

(f) correct

8. C

9. (a) Take the downward direction as positive.

The depth of the well is

(b) The velocity of the coin when it reaches the water surface is

The positive sign indicates the velocity is in the downward direction.

10. (a) Take the downward direction as positive.

Applying ,

The cat has 1.10 s to turn its body.

(b) Applying ,

The velocity of the cat is 11.0 m s−1 in the downward direction when it

reaches the ground.

11. (a) Take the upward direction as positive.

The velocity of the dolphin is 8.94 m s−1 in the upward direction when it

leaves the water surface.

(b) The displacement of the dolphin is zero when it returns to the water surface

(i.e. ).

Applying ,

or (rejected)

The spectators have 1.79 s to take pictures of the ‘flying dolphin’.

(c) Given that ,

The velocity of the dolphin has to be 11.0 m s−1 in the upward direction to

reach the ball if the ball is hung 2 m higher.

12. (a) (i) Take the downward direction as positive.

The displacement of the diver is +10 m when she enters the water.

or (rejected)

The time for her to fall to the water is 1.75 s.

(ii) Applying ,

Her velocity is 14.5 m s−1 when she enters the water.

(b) Applying , given that ,

Her displacement is 6.75 m (downwards) 1.5 s after she jumps.

13. (a) Take the upward direction as positive.

(rejected) or

The hang time of the player is 0.8 s.

(b) The player reaches the highest position when his velocity is zero (i.e. ).

Applying ,

The highest position of his feet during the jump is 0.8 m.

(c) From (b), the highest position of his feet during the jump is 0.8 m.

The maximum height he can reach when he jumps is

. Hence, he can reach the hoop.

14. (a) (i) From the graph, the fish reaches the highest point at t = 0.2 s.

(ii) displacement of the fish when it reaches the highest point

= area under the v-t graph from t = 0 s to t = 0.2 s

(iii) The time taken for the downward journey is .

(iv) From the graph, the initial velocity of the fish is 2 m s−1.

(v) The fish reaches the table at t = 0.6 s.

From the graph, the velocity of the fish at t = 0.6 s is −4 m s−1.

displacement of the fish when it reaches the table

= area under the v-t graph from t = 0 s to t = 0.6 s

The velocity and displacement of the fish when it reaches the table are

−4 m s−1 and −0.6 m respectively.

(vii) From the graph, the velocity of the fish at t = 0.32 s is −1.2 m s−1.

displacement of the fish at t = 0.32 s

= area under the v-t graph from t = 0 s to t = 0.32 s

The velocity and displacement of the fish 0.32 s after it jumps are

−1.2 m s−1 and 0.128 m respectively.

(b) The slope of the v-t graph represents the acceleration of the fish. In this

case, it is the acceleration due to gravity.

(i) The acceleration due to gravity always points downwards and it is

independent of the direction of motion of the fish.

(ii) The acceleration due to gravity is independent of the mass of an object.

15. Take the upward direction as positive.

m 3.7517.422

117.422.22 2

s

a / m s−2

−10

0

−10

0

10

0

5

1 2

1 2

1 2

v / m s−1

s / m

Chapter Exercise (p.69)

1. B

2. B

3. B

4. C

5. B

6. C

7. D

8. B

9. B

10. D

11. A

12. (a) A suggested route to the exit is shown below.

(1A)

(or any other reasonable answers)

(b) (i) The displacement is a measure of the change in position, which

includes the separation between the initial and final position

(magnitude) and the direction of the change (direction) (1A). The

distance travelled is the total length of the path (1A).

(ii) total displacement (1A)

total distance travelled (1A)

[must be consistent with the answer in (a)]

(iii) No (1A). The total displacement depends on the initial and final

positions only (1A).

(c) time of travel (1M+1A)

average velocity (1M+1A)

He takes 118 s to exit the maze and his average velocity is 0.449 m s−1.

[must be consistent with the answer in (a)]

13. (a) The displacement of the hare from the starting point is

The displacement of the tortoise from the starting point is

(1M)

The displacement of the hare from the tortoise is .

Hence, the hare is 450 m ahead of the tortoise at this time. (1A)

(b) The displacement of the tortoise is

(1M)

The displacement of the tortoise from the hare is .

Hence, the tortoise is 270 m ahead of the hare at this time. (1A)

(c) For the hare,

total time needed for it to finish the race

= time taken before it wakes up + time needed to finish the remaining race

The total time needed for the hare to finish the race is

(1M)

The total time needed for the tortoise to finish the race is

(1M)

The hare will win. (1A)

(d) No (1A). If the hare sleeps for 1 min longer, the total time needed for it to

finish the race becomes 26.67 min. It means that both the hare and tortoise

finish the race at the same time and hence it is a draw game (1A).

14. (a) The runner needs time to react after he hears the firing of the gun (1A). This

period of time is called reaction time (1A).

(b) (i) The acceleration of the runner when he speeds up is

(1M+1A)

(ii) displacement of the runner = area under the v-t graph

(1M)

His finishing time T is 14.65 s. (1A)

(iii) His average velocity during the race is (1M)

(1A)

(c) new acceleration

new maximum velocity

Applying , the time needed to reach the maximum velocity is

(1M)

As his reaction time remains unchanged, he reaches the maximum velocity

at time t = 0.3 + 3.7 = 4 s.

Let T′ be the time for him to finish the 100 m race. A new v-t graph is

shown below.

displacement of the runner = area under the v-t graph

(1M+1M)

His finishing time will be improved by . (1A)

15. (a)

The displacement of the drunken driver when his car is stopped is

(1M)

Hence, the stopping distance needed by the drunken driver is 25 m.

The displacement of the sober driver when his car is stopped is

Hence, the stopping distance needed by the sober driver is 12.5 m.

The minimum distance between the cars if they do not collide is

25 + 12.5 = 37.5 m > 30 m. Hence, the cars will collide. (1A)

(b) Let t be the time when the cars collide.

The initial displacement of the drunken driver from the position of the

collision is

(1M)

The initial displacement of the sober driver from the position of the

collision is

(1M)

Since the cars are 30 m apart initially,

or (rejected) (1M)

The distance travelled by the drunken driver before the collision is

30 mpositive direction

car driven by the drunken driver

car driven by the sober driver

s1 s2

position of collision

The collision occurs at a position 18.4 m away from the position where the

drunken driver applies the brake. (1A)

(c) The velocity of the drunken driver just before the collision is

(1M)

The velocity of the sober driver just before the collision is

The velocities of the car driven by the drunken driver and the sober driver

are 10.3 m s−1 and −4.05 m s−1. (1A)

16. (a) The initial velocity of the taxi is

The initial velocity of the van is

Applying , the time needed by the taxi to reach the merge point

is given by

(1M)

or (rejected)

The time needed by the van to reach the merge point is given by

(1M)

The taxi reaches the merge point earlier than the van. (1A)

(b) distance travelled by the van before it accelerates (1M)

The time needed by the van to reach the merge point after it accelerates is

given by

(1M)

or (rejected)

The total time needed by the van to reach the merge point is

The taxi still reaches the merge point line earlier than the van. (1A)

17. (a) Take the upward direction as positive.

Applying , when the ball reaches light-gate A, (1M)

...........................................(1)

When the ball reaches light-gate B,

........................(2) (1M)

Multiply (1) by a factor of , equation (1) becomes

.............................(3)

Subtract (2) by (3),

From (1),

The projection velocity u and the acceleration due to gravity g are 11.0 m s−1

(upwards) and 9.82 m s−2 (downwards) respectively. (1A+1A)

(b) Applying , the times that the ball passes light-gate A again is

given by (1M)

or (rejected)

The times that the ball passes light-gate B again is given by

or (rejected)

The ball will pass light-gate A and light-gate B again at the times of 2.04 s

and 1.78 s respectively. (1A+1A)

(c) (i) Applying , the velocity of the ball when it passes light-gate A

for the first time is

The velocity of the ball is 9.02 m s−1 in the upward direction when it

passes light-gate A for the first time. (1A)

(ii) Similarly, the velocity of the ball when it passes light-gate A for the

second time is

The velocity of the ball is 9.02 m s−1 in the downward direction when it

passes light-gate A for the second time. (1A)

18. (a) Take the upward direction as positive.

Applying , the value of u is given by (1M)

(1A)

(b) Ball A takes 3Δt to return to the juggler’s hand.

Applying , (1M)

or (rejected) (1A)

(c) From (b), the time when ball A returns to the juggler’s hand is

.

Since ball B is thrown at a time Δt after ball A is thrown, its time of flight is

(1M)

Applying , the displacement of ball B from the juggler’s hand is

(1M)

Applying , the velocity of ball B at that time is (1M)

Similarly, since ball C is thrown at a time 2Δt after ball A is thrown, its time

of flight is

The displacement of ball C from the juggler’s hand is

The velocity of ball C at that time is

At that time, ball B is 1.6 m above the juggler’s hand and moves at a

velocity of 2 m s−1 in the downward direction (1A). Ball C is also 1.6 m

above the juggler’s hand and moves upwards at a velocity of 2 m s−1 in the

upward direction (1A).

(d) From (c), ball B is at 1.6 m above the juggler’s hand when ball A is thrown

again, so its displacement is −1.6 m when it returns to his hand.

Applying , the time when ball B returns to the juggler’s hand is

given by (1M)

or (rejected) (1A)

Similarly, the displacement of ball C is −1.6 m when it returns to the

juggler’s hand.

Applying , the time when ball C returns to the juggler’s hand is

given by

or (rejected) (1A)

Ball B and C return to juggler’s hand at a time Δt and 2Δt after ball A is

thrown again respectively.

19. (Edexcel AS-level May 2002 Paper 6731/01 Unit Test PHY1 Q4)

20. (AQA AS-level Jan 2007 Paper PA02 Q2)

21. (HKCEE 2000 P1 Q7)

22. (HKCEE 2002 P1 Q8)

23. (a) The average speed of the Bar-tailed Godwit is given by

average speed (1A)

Its average velocity should have a magnitude smaller than its average speed

(1A) because the magnitude of the displacement is smaller than the distance

travelled (1A).

(b) (i) In Figure Q23b, 5 mm on the map represents 100 km, and hence 1 cm

on the map represents 200 km.

Applying ,

The average velocity of the Band-tailed Pigeon from15 Sep to 21 Sep

is

(1A)

Its average velocity of from 21 Sep to 03 Oct is

(1A)

Its average velocity of from 03 Oct to 31 Oct is

(1A)

Its average velocity of from 31 Oct to 25 Nov is

(1A)

(ii) From (b)(i), the average velocity of the Band-tailed Pigeon from

03 Oct to 31 Oct is 0.268 km h−1.

Applying , (1M)

its displacement on 25 Oct from its position on 03 Oct

Hence, the Band-tailed Pigeon is 141 km from its position on 3rd

October. (1A)