Active Physics Full Solutions to Textbook Exercises

Chapter 5

Motion

2. Average speed of A = 44 = 1 cm s1 = 0.01 m s1

Average speed of B = 44 = 1 cm s1 = 0.01 m s1
Average velocity of A = 44 = 1 cm s1 = 0.01 m s1 to
the right
1
Average velocity of B = 31
=
4 = 0.5 cm s
1
0.005 m s
to the right

Checkpoint
Checkpoint 1 (p.12)
1.

2.

3.

3. (a) Total distance = 160 + 160 + 240 =

560 m

(b) Keiths overall displacement is

240 m to the east.

4. (a) Total distance = 5 20 =

The ferries have different average velocities as

they travel towards different directions.
A

Checkpoint 3 (p.27)
1. Table:

100 cm

(b) dummy

2. (a)

(b)

Magnitude of the displacement

=

202 + 402

(c)

time / s

10

velocity / m s1

20

An object moving faster may not have a

larger change in velocity (i.e. higher acceleration). For example, an object moving at a
constant high speed along a straight line has
zero acceleration.
F

A braking car is moving forward, but since

its velocity is decreasing, its acceleration is
pointing backward.
F

Its velocity stays unchanged.

= 44.72 44.7 cm

Direction of the displacement:

20
tan =
40

= 26.57 26.6

Therefore the overall displacement of the bee

is 44.7 cm (N 26.6 E).

3. (a) No.
You are not undergoing acceleration when you
are moving at a constant speed along a straight
line.
(b) Yes.
You are undergoing acceleration whenever you
are speeding up.
(c) Yes.
You are undergoing acceleration whenever you
are changing your direction of motion.

Checkpoint 2 (p.21)
1. Table:
time / s

10

displacement / cm

10

15

50

(d) No.
You are not undergoing acceleration when you
are moving at a constant speed along a straight
line.

2|

Chapter 5 Motion Checkpoint

(e) Yes.
You are undergoing acceleration whenever you
are changing your direction of motion.

Checkpoint 4 (p.34)
1.
2.

Active Physics Full Solutions to Textbook Exercises

Checkpoint 6 (p.49)
1. Take the initial direction of motion of the motorcycle as positive.
Given: s = 500 m u = 50 m s1 v = 100 m s1
We have to ind t using the equation

The s t graph of car P has a larger slope.

From the graph, we can see that the car slows

down initially (i.e. v decreases in magnitude), then
reverses its direction of motion (i.e. v becomes
zero), and eventually speeds up (i.e. v increases in
magnitude but carries a negative sign). Hence, the
car is making a U-turn.

1
s = (u + v)t
2

3. (a) Acceleration
= slope of the v t graph
10 5
=
=
50

1 m s2

(b) Overall displacement

= total area under the v t graph
=

(5 + 10)(5)
=
2

Hence, we get
1
500 = (100 + 50)t
2
t = 6.667 s

It takes 6.67 s for the motorcycle to accelerate.

2. Take the initial direction of motion of the car as
positive.
Given: s = 50 m v = 0 a = 5 m s2
We have to ind u using the equation
v 2 u 2 = 2as

37.5 m

Hence, we get
Checkpoint 5 (p.40)
02 u 2 = 2(5)(50)

1. During time t = 05 s, Simon walks

away from the sensor at an average
velocity of 0.3 m s1 .
1
Slope = 20.5
50 = 0.3 m s

During time t = 510 s, Simon walks

towards the sensor at an average
velocity of 0.2 m s1 .
21
Slope = 105
= 0.2 m s1

2. Graphs:

u = 22.36 m s1

The speed of the car before the car braked is

22.4 m s1 .

Checkpoint 7 (p.61)
1. Table:
time / s

velocity / m s1

displacement / m

20

10

15

20

10

15

20

2. (a)
(b)

Both balls accelerate at the same rate under

gravity, regardless of what they are made of.
F

Its upward motion is slowing down.

Active Physics Full Solutions to Textbook Exercises

3. Take the upward direction as positive.
Given: u = 5 m s1 a = 9.81 m s2 t = 3 s
We have to ind s .
By the equation s = ut + 12 at 2 ,
1
s = (5)(3) + (9.81)(3)2
2
= 29.145 m

Hence, the cliff is 29.145 m high.

Exercise

Chapter 5 Motion Exercise

|3

Therefore, the smallest and the largest displacements of the marble are 2 cm and 6 cm
respectively.
6. Electronic timing is more accurate than using
stopwatches.
It is because the error due to the reaction times
of the timekeepers are signi icant in the results of
swimming competitions. The inishing times of the
swimmers are so close that the ranks determined by
the readings of stopwatches may be inaccurate due
to the reaction time.
7. (a) Displacements: dummy

Exercise 5.1 (p.12)

1. (a) 4.5 km = 4.5 1000 =

4500

(b) 3.2 cm = 3.2 102 =

0.032

(c) 6.5 h = 6.5 60 60 =

(d) 19 min = 19 60 =
2.
3.

23 400

1140

Only cars P and Q have the same initial and inal

positions.
A

4. It represents the distance.

The driver concerns mainly how far the car has to
travel to reach the destination (i.e. distance), but not
the length of the straight line path to the destination
(i.e. displacement).

(b) Adding the displacement vectors using the tipto-tail method:

5. The following shows the path of the marble when its

displacement is the smallest.

The following shows the path of the marble when its

displacement is the largest.

#
Magnitude of AD :
AD =

802 + 502 94.3 m

#
Direction of AD :
tan =

50
80

So the displacement is

32.0

94.3 m

(S 32.0 W).

4|

Chapter 5 Motion Exercise

Active Physics Full Solutions to Textbook Exercises

8. (a) The distance travelled = 120+360+82 =

562 cm

(b) The displacement

= (+120) + (360) + (+82) = 158 cm.

So the displacement is

158 cm

to the left.

(c) No.
The displacement of the ball does not depend
on the choice of the positive direction.

Magnitude of the displacement:

9. (a) The distance travelled by Linda

= 1.5 + 2.5 + 3 = 7 m .

AB =

(b) dummy

702 + 902 114 m

Direction of the displacement:

tan =

90
70

52.1

So the distance travelled by Joseph is 200 m ,

and his displacement is 114 m (S 52.1 E).
(b) The distance travelled by Joseph would
increase but his overall displacement would
remain unchanged.

Magnitude of the displacement:

AD =

(1.5 + 3)2 + 2.52 5.15 m

Direction of the displacement:

2.5
tan =
1.5 + 3

Exercise 5.2 (p.22)

1. (a) Yes

= 29.05

(b) Yes
(c) No

So, Lindas displacement is

(N 61.0 E).

5.15 m

10. (a) The distance travelled = (3 + 6 + 6)(1000) =

15 000 m .
(b) Ottos path:

2. (a) 110 km h1 = 110

3.6

30.6

(b) 55 cm s1 = 55 102 =
(c) 9 km min1 = 9000
60 =
3. (a)

(b)

(c)

150

m s1
0.55

m s1

m s1

4. (a) Time taken to go through the tunnel

=

(260)+10
6060

= 0.036 11 h

Average speed
distance
= total
total time =

Magnitude of the displacement:

s=

(6 3)2 + 62 = 6.7082 km 6710 m

63
6

Ottos displacement is

26.6

72 km h1

No.
Her average speed is lower than the speed
limit (80 km h1 ).

5. (a) Average speed

distance
= total
total time =

6710 m

(b) She is ined because her speed at a certain

instant is higher than the speed limit.

Direction of the displacement:

tan =

2.6
0.03611

(N 26.6 W).

11. (a) Distance travelled = 12 2( 70

2 ) + 90 200 m

(0.84)+0+(12)
4+6+2

0.433 m s1

Active Physics Full Solutions to Textbook Exercises

Chapter 5 Motion Exercise

(b) Take the direction to the right as positive.

Displacement from O to B :

Time required for Mandy to run from B to C

= 2(6/2)(1/2)
= 2.356 s.
4
Time taken for the whole journey
= 1.885 + 2.356 = 4.241 4.24 s .

s = (0.8 4) (1 2) = +1.2 m

(b) Average speed = 2(6/2)

4.241 =

Average velocity:

62
4.241

(c) Average velocity =

s
1.2
= +0.1 m s1
v= =
t 4+6+2

Hence, the average velocity is

the right.

0.1 m s1

|5

towards

(c) The time required to travel from B to O

= 1.2
1 = 1.2 s.
2(0.84)
The average speed = 4+6+2+1.2
0.485 m s1 .
The overall displacement = 0.
The average velocity = 0 .

4.44 m s1

2.83 m s1

due east.

8. Let d be the distance between X and Y .

The total time taken for the whole journey
= d2 + d1 = 3d
2 .
The average speed
distance
2d
4
1
= total
total time = 3d /2 = 3 1.33 m s .
Exercise 5.3 (p.28)

6. (a) The distance travelled = 50 (2 + 4) = 300 km.

1
The average speed = 3001000
.
63600 13.9 m s
Alternative Solution:
Since the speed of the car remains at 50 km h1
during the 6 hours, the average speed is
50 km h1 13.9 m s1 .

(b) dummy

1.

The acceleration of the car is the greatest when

the difference between u and v is the greatest.

2.

For a braking car, the direction of its acceleration is opposite to its velocity.

3. The cyclist undergoes acceleration at A , C and D

because he gains speed, changes his direction of
motion, and slows down respectively. He does not
undergo acceleration at B because he maintains his
speed.
4. Take the forward direction as positive.
By a =

vu
t

, we have

Overall displacement:
s=

2=

(50 2)2 + (50 4)2 = 223.6 km

Magnitude of the velocity:

v=

s 223.6 1000
=
10.4 m s1
t
6 3600

v 8
6

v = 20 m s1

The speed of the boat is

20 m s1

after 6 s.

5. Take the forward direction as positive. When the

minibus speeds up, by a = vu
t , we have
3=

15 0
t

t =5s

Direction of the velocity:

50 4
tan =
50 2

So the average velocity is

(N 63.4 E).

When it slows down, by a =

63.4
1.5 =

0 15
t

vu
t ,

we have

t = 10 s

10.4 m s1

(c) No.
The velocity of the car changes when it changes
its direction of motion.
7. (a) Time required for Mandy to run from A to B
= 1.885 s.
= 2(6/2)(1/2)
5

The total time of travel = 5 + 10 + 10 =

25 s

6. (a) Take the forward direction as positive.

By a =

vu
t ,

we have
a=

6 18
= 0.6 m s2
20

So the deceleration of the train is

0.6 m s2

6|

Chapter 5 Motion Exercise

(b) By a =

vu
t ,

Active Physics Full Solutions to Textbook Exercises

we have

0.6 =

06
t

The train takes a further

10 s

t = 10 s

to stop.

7. Take the direction away from the racket as positive.

By a = vu
t , we have
a=

8 (8)
= 32 m s2
0.5

The acceleration of the ball is

the racket.

32 m s2

away from

8. (a) The skater slides up the ramp with a decreasing speed until she reaches the highest point,
where she is momentarily at rest. Then she
slides down the ramp with an increasing speed.
(b) Take the direction up the ramp as the positive.
By a = vu
t , we have
0.8 =

2 2
t

(b) Velocity = slope of the s t graph =

0.04 m s1 .

0.80
200

0.04 m s1

The average velocity of the tank is

forward.

6. (a) Take the forward direction as positive.

Sketch of the v t graph:

t =5s

She will reach a speed of 2 m s1 again

later.

5s

Exercise 5.4 (p.41)

150
1. (a) Slope = 0.80
=

(b) Slope = 12.7

200 =

18.75 m s1
0.085 m s2

=
(c) Area = (90+120)60
2

6300 m

2.

The s t graph shows that the two cars have the

same displacement at t = T .

3.

Between t = 0 and t = T , the area under the

v t graph of car B has smaller area. So, car B has a
smaller displacement and lags behind car A at time
t =T.

4.

(b) The distance travelled = (0.5+2.5)15

= 22.5 m.
2
The car moves for a distance of
the driver sees the dog.
7. (a) From the graph, the shop is
Jeffs home.

22.5 m

after

due east of

160 m

(b) Velocity during time t = 05 min:

A Graphs B and D mismatch the decelerating and

accelerating stages. Graph C fails in the inal stage.

Option B could be correct if the backward direction of the bus

was taken as positive. But it is not natural. Besides, there are too
many types of motion that can produce such an a t graph.
5. (a) Plot of the s t graph: dummy

v=

160
0.533 m s1
5 60

Velocity during time t = 811 min:

v=

160
= 0.889 m s1
3 60

Jeff travels towards the east at a constant speed

of 0.533 m s1 during t = 05 min. He then
remains at rest during t = 58 min. Finally he
goes back to his starting position steadily at a
speed of 0.889 m s1 during t = 811 min.
(c) The total distance travelled by Jeff
= (160)(2) = 320 m .
The total displacement of Jeff =

Active Physics Full Solutions to Textbook Exercises

Chapter 5 Motion Exercise

8. (a) During t = 00.5 s, the acceleration =

6 m s2 away from the sensor.
During t = 0.52 s, the acceleration =

3
0.5

|7

(b) The trolley accelerates uniformly at 6 m s2

away from the sensor during t = 00.5 s. Then
it travels at a constant velocity of 3 m s1 away
from the sensor during t = 0.52 s.
(c) The overall displacement of the trolley
= (1.5+2)3
= 5.25 m away from the sensor.
2
9. (a) For time t = 035 s, the acceleration =
0.571 m s2 .

200
350

Consider the slope of the s t graph during

t = 10T , we have

For time t = 35100 s, the acceleration = 0.

For time t = 100140 s, the acceleration
020
= 140100
= 0.5 m s2 .
Therefore, the a t graph is

0 20
= 4
T 10

T=

15 s

(c) When she goes back, her velocity

=

change in displacement
time taken

+20
5

= 4 m s1 .

Therefore, the plot of the v t graph becomes

(b) The overall displacement of the train

= area under the v t graph = [(10035)+140]20
=
2
2050 m.
The average velocity =
forward.

2050
140

14.6 m s1

(c) Let T be the time when the train is midway

between the stations. Considering the displacement of that position from the departing
station, we have
[(T 35) + T ] 20)

2

2050

2

2T 35 = 102.5

T=

68.75 s

10. (a) The distance travelled = 5 4 + 4 (20 10) =

60 m .
The displacement = (5)(4) + (4)(20 10) =
20 m.

So the displacement is

20 m

backward.

(b) Let T be the required time. The following

shows the s t graph of Sara.

11. (a) The car travels at a uniform speed during

t =1220 s as its acceleration is zero and its
velocity is non-zero during that period.
(b) No.
At t = 1012 s, the car still speeds up (a > 0)
but the magnitude of its acceleration (i.e. the
change in velocity) decreases.
12. (a)

has a greater acceleration, because its

graph has a greater initial slope.
Car A

(b) The initial acceleration of car A

2
forward.
= 30
2 = 15 m s
The initial acceleration of car B
40
2
forward.
4 = 10 m s

8|

Chapter 5 Motion Exercise

Active Physics Full Solutions to Textbook Exercises

(c) Car B overtakes car A at the time when

their displacements (the areas under their
v t graphs) are the same. Let T be the
corresponding time.
[(T 2) + T ] 30

2

[(T 4) + T ] 40

2

3(2T 2) = 4(2T 4)

5. (a) Take the direction to the right as positive. By

v = u + at , we have
1 = 0 + a(0.02)

a = 50 m s2
50 m s2

The acceleration of the ball is

towards the right.
(b) By v 2 u 2 = 2as , we have

3T 3 = 4T 8
T=

02 0.42 = 2(0.1)s

5s

s = 0.8 m > 0.7 m

13. (a) The velocity of the car during t = 010 s

1
= 8010
10 = 7 m s .

Yes. The black ball will fall into the pocket.

The velocity of the car during t = 2030 s

120110
1
3020 = 1 m s .

First of all, the car starts moving from the

starting point at a constant speed of 7 m s1
during t = 010 s. And then it decelerates from
v = 7 m s1 to v = 1 m s1 during t = 1020 s. It
inally moves at at a constant speed of 1 m s1
during t = 2030 s.
(b) The average velocity =
forward.

12010
30

3.67 m s1

6. Take the forward direction as positive. Let t 1 and

t 2 be the time taken for the runner to accelerate
and move at a constant velocity respectively. By
v = u + at , we have
10 = 0 + (3.3)t 1

(b) The trolley in case B has a higher average

velocity. The number of ticks on the tapes of
A and B are 9 and 7 respectively. The trolley
in case B takes less time to travel the same
distance, and hence it has a higher average
velocity.

A Nancy measures the values of the time t and the

displacement s in order to ind the acceleration a .
Since we know that the initial velocity u = 0, the best
option is A.
B

s 2 = 100 15.15 = 84.85 m

Therefore, t 2 = vs22 = 84.85

10 = 8.485 s. The inishing
time of the runner = 3.030 + 8.485 11.5 s .
7. (a) By s = ut + 12 at 2 , we have

v 2 0 = 2ad

3.

v avg = st =

(u+v)t
2

v2 d

1t =

u+v
2

p
d

(0 + 5)t
2

3 m s2

forward.

1
s = 0 + (3)(4)2 = 24 m
2
24

8. (a) Take the direction to the left as positive. By

s = ut + 12 at 2 , we get
1
s = 0 + (6)(3)2 =
2

4. Take the forward direction as positive. By s =

we have
0.16 =

a = 3 m s2

(b) By s = ut + 12 at 2 , at t = 4 s, we have

Therefore, x =

Given that u = 0. By v 2 u 2 = 2as , we have

s 1 = 15.15 m

The displacement of the runner after the acceleration is

The acceleration is

2.

1
54 = 0 + a(6)2
2

Exercise 5.5 (p.50)

t 1 = 3.030 s

Let s 1 be the displacement of the runner during the

acceleration. By v 2 u 2 = 2as , we have
102 0 = 2(3.3)s

14. (a) The trolley in case A moves at a constant

velocity while that in case B accelerates.

1.

t=

0.064 s

(u+v)t
,
2

27 m

(b) By v = u + at , we have v = 0 + (6)(3) =

18 m s1

(c) The distance travelled by the deer during the

chase is s = 27 18 = 9 m.

So the speed of the deer v = st = 93 = 3 m s1 .

Active Physics Full Solutions to Textbook Exercises

Chapter 5 Motion Exercise

9. (a) Take the direction to the left as positive.

450 km h1 =

Hence, the total distance travelled by the car is

12 + 30 = 42 m > 40 m.
No. The car cannot stop before the traf ic light.

450
= 125 m s1
3.6

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

Assume that the boat passes A at time t = 0.
By s = ut + 21 at 2 , we have

By v 2 u 2 = 2as , we have
v 2 1252 = 2(5)(1200)

v = 2(5)(1200) + 1252

1
12 = 4t + (0.5)t 2
2

= 60.21 60.2 m s1

So the speed of the plane is

goes beyond the runway.

60.2 m s1

0 = t 2 16t + 48
t = 4 s or 12 s

when it

The boat reaches B for the irst time at t = 4 s

and the second time at t = 12 s. Therefore the
answer is 12 s .

(b) By v 2 u 2 = 2as , we have

0 60.212 = 2a(250)

(b) By v 2 u 2 = 2as , we have

0 60.212
a=
2(250)

v 2 42 = 2(0.5)(12)

= 7.25 m s2

The deceleration of the plane is

after it goes beyond the runway.

v2 = 4

7.25 m s2

v = 2 m s1 or 2 m s1

Therefore, the velocity of the boat when it

reaches B for the second time is 2 m s1
downstream.

(c) Let t1 and t 2 be the time intervals that the

plane moves on the runway and the ield
respectively.
As v = u + at

t=

vu
a ,

Positive v means the boat moves upstream and

negative v means the boat moves downstream.

we have

60.21 125
= 12.96 s
5
0 60.21
t2 =
= 8.305 s
7.25
t1 =

12. By v 2 u 2 = 2as , we have s =

10. (a) From the graph, his reaction time is

Hence,

202 402 1200

=
2a
2a
0 202 400
s2 =
=
2a
2a

21.3 s

0.8 s

012
(b) The acceleration = 40.8
= 3.75 m s2 .
So the deceleration of the car is 3.75 m s2 .

(c) The total distance travelled by the car

= area under the v t graph
= 28.8 m < 32 m.
= (0.8+4)(12)
2
Yes. The car can stop before the traf ic light.
(d) The distance travelled by the car before
braking
s 1 = v 1 t 1 = 15 0.8 = 12 m

By v 2 u 2 = 2as , the distance travelled by the

car during the braking process is

s 2 = 30 m

v 2 u 2
2a .

s1 =

So the total time taken = 12.96 + 8.305

0 152 = 2(3.75)s 2

|9

Combining the equations, we have

s 1 : s 2 = 1200 : 400 = 3 : 1 .
Exercise 5.6 (p.61)
1.

The acceleration due to gravity always points

downwards regardless of the motion of the ball.

2.

The graph given is a s t graph, whose area

under the curve has no physical meaning.

3.

4. Take the downward direction as positive.

By s = ut + 12 at 2 , we have
* 0 + 1 (9.81)t 2

112 = 
ut
2
t = 4.778 s or 4.778 s (rejected)

10 |

Chapter 5 Motion Exercise

Active Physics Full Solutions to Textbook Exercises

s = ut + 12 at 2 , we have

Average speed
v=

s
112
=

t 4.778

1
10 = (3)t + (9.81)t 2
2

23.4 m s1

0 = 4.905t 2 3t 10

Since the release takes place at t = 0, a negative time

indicates a time instant before the release. This solution is
physically impossible and is therefore rejected.

t = 1.766 1.77 s or 1.154 s (rejected)

So she takes
5. (a) Take the downward direction as positive.
By s = ut + 12 at 2 , we have

(b) By v = u + at , we have
v = 3 + (9.81)(1.766) 14.3 m s1


* 0 + 1 (9.81)(2)2 19.6 m
ut
s =
2

The depth of the well is

19.6 m

Her velocity when she reaches the water

surface is 14.3 m s1 downward.

(b) By v = u + at , we have

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

v = u + at , we have

v = 0 + (9.81)(2) = 19.62 m s1

0 = 2 + 9.81t

The velocity of the coin when it reaches the

water surface is 19.62 m s1 downward.
6. By s = ut +

1
2
2 at ,

t = 0.2039 0.204 s

The ish reaches the highest point at t =

we have

0.204 s

(b) Consider the motion of the ish before it

reaches the highest point. By s = ut + 12 at 2 ,
we have

* 0 + 1 (9.81)t 2

ut
1 =
2

2
t=
9.81

1
s = (2)(0.2039) + (9.81)(0.2039)2 0.2039 m
2

t 0.452 s or 0.452 s (rejected)

The minimum time required for the turn is

0.452 s

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

v 2 u 2 = 2as , we have
162 u 2 = 2(9.81)(13)

u = 162 2(9.81)(13)

0.970 m s1

1
s = 0 + (9.81)(0.6 0.2039)2 = 0.7697 m
2

The average speed of the ish =

1.62 m s1 .

0.9735
0.6

(c) By v 2 u 2 = 2as , we have

v 2 22 = 2(9.81)(0)

(b) By s = ut + 12 at 2 , we have
1
s = (0.9695)(1) + (9.81)(1)2 5.87 m
2

After throwing for 1 s, the ball is at

below the starting point.

Now consider the motion of the ish when it

falls from the highest point to the table. By
s = ut + 12 at 2 , we have

So the total distance travelled by the ish

= 0.2039 + 0.7697 = 0.9735 0.974 m .

u = 0.9695 0.970 m s1

The initial velocity of the ball is

downward.

to reach the water surface.

1.77 s

5.87 m

8. (a) Take the downward direction as positive. By

v2 = 4
v = 2 m s1 or 2 m s1

Since the downward direction is taken as

positive, v < 0 represents an upward motion
and v > 0 represents a downward motion.
So the velocity is

2 m s1

downward.

Active Physics Full Solutions to Textbook Exercises

Statement (3) is correct. As both cars have the same

initial speed and slow down at the same rate, they
travel the same distance when they decelerate.

Chapter Exercise
Multiple-choice Questions (p.65)
1.

2.

5.

The acceleration of the ball is equal to the

acceleration due to gravity g , which is constant
throughout the motion.

6.

If the object is thrown vertically downward,

the time required to reach the bottom of the cliff
decreases, but the acceleration (i.e. the slope of the
v t graph) remains unchanged.

D Statements (1) and (2) are correct as both cars

have the same displacement s at t = 4 s.

Statement (3) is correct. The s t graphs of both cars

have increasing slopes throughout.
From the v t graph, the acceleration of the
object from t = 48 s is
C

0 12
= 3 m s2
a=
84

By v = u + at , the velocity of the object at t = 5 s is

| 11

Chapter 5 Motion Chapter Exercise

7.

2
v 2 u
= 2as

v 2 = 2as .

The slope of the graph is 2a . If a , the slope .

8.

v = 12 + (3)(1) = 9 m s1

C Statement (1) is possible. A car turning a corner

at a constant speed is accelerating.

Statement (2) is possible. A ball being thrown

upward is momentarily at rest (i.e. zero velocity) at
the highest point, but it is accelerating downwards.

Therefore, we get

Statement (3) is impossible. Constant velocity

implies constant speed and ixed direction of
motion.
9.

u=

The speed of the ball during time t = 00.1 s is

0.0360
0.1

= 0.36 m s1

The speed of the ball during time t = 0.30.4 s is

v=

3.

Statement (1) is correct. At time t , the object is

momentarily at rest.
Statement (2) is incorrect. The acceleration of the
object at time t is determined by the slope of the
graph at that time, which takes a non-zero value.
Statement (3) is incorrect. The displacement of the
object at time t is determined by the area under the
graph during time = 0 to t , which takes a non-zero
value.

4.

0.2 = 0.36 + a(0.2)

(9 + 12)(1) (2 + 4)(12)
+
= 46.5 m
2
2

Statement (1) is correct. The graphs have the

same slope when the cars are braking, and so they
slow down at the same rate.
D

Statement (2) is incorrect. The reaction time of the

driver of car A is longer than that of car B .

= 0.2 m s1

By v = u + at , we have

The area under the v t graph is equal to the

displacement of the object. So,
x=

0.0840.064
0.1

a = 0.8 m s2

The deceleration of the ball is 0.8 m s2 .

10.

C Assuming that the reaction time of the student is

negligible. The vertical height h and the time t are
related by
0
1

>
h =
ut
+ at 2
2

p
h=

a
t
2

So, a straight line passing through the origin should

be obtained.
However, if the reaction time T is taken into the
account, the measured time t is given by
t = t T
p

t -intercept. Refer to the graph below:

a
2 t

a
2T.
p
So the graph of h against t is a graph with positive

Combining the equations, we have h =

12 |

Chapter 5 Motion Chapter Exercise

Active Physics Full Solutions to Textbook Exercises

Subtracting the equations, we have 18 9 = (u u) +
(5a 2a) a = 3 m s2 .
Structured Questions (p.67)

Consider option B. If air resistance is not negligible, the falling

object should take a longer time to fall the same vertical distance
than the case without air resistance. The slope of the graph would
become smaller, but the graph would not have a t intercept.
Consider option D. If the object falls with a downward velocity, we
get

p
h=

a t intercept.
For u = 0, we get s t 2 . Hence, the shorter the
distance between the adjacent images of the stone,
the shorter the time interval between which the
images are taken. So option A is the answer.
A

100 =
T=

[(T 4) + (T 0.3)](8)
2

By s = ut + 12 at 2 , we have s = 12 at 2
Therefore,
D

tX Y : tX Z =

XY :

XY +Y Z =

(1A)

6.83 m s1

Correct axis and labels: 1A

Correct curve and data: 1A

16. (a) Refer to the table below.

t 1 : t 2 = 3 : (5 3) = 3 : 2
13.

14.

d /m

v 2 / m2 s2

Statement (1) is incorrect. The sign of the

velocity of P does not change at t = 1 s.

0.2

4.0

0.4

7.8

Statement (2) is correct. At t = 2 s, the separation

[
]
between P and Q is sP Q = 21 (2)(2) 12 (2)(2) = 4 m.

0.6

11.8

0.8

15.8

Statement (3) is incorrect. At t = 4 s,

s P = 2 + 12 (2)(4) = 2 m and sQ = 12 (4)(4) = 8 m.
Therefore, the displacements of P and Q are not the
same, and they do not meet each other.

1.0

19.4

B Let u and v be the velocities of the particle at

time t = 0 and t = 4 s respectively.

By v = u + at , at t = 4 s, we have
1
36 = u(4) + a(4)2
2

9 = u + 2a

At t = 6 s, we have
1
36 = v(2)+ a(2)2
2

1
36 = (u+4a)(2)+ a(2)2
2

forward

(c) Let a be the acceleration of the runner during

t = 0.34 s. The acceleration at that time
period is equal to the slope of the graph:
80
a = 40.3
2.16 m s2 .

s t2

p p
9 : 9 + 16 = 3 : 5

(1M)

14.65

s
100
v avg = =

T
14.65

Note that the equations of uniformly accelerated

motion are independent of the mass of the object, so
both options C and D are incorrect.
12.

(1A)

(b) The displacement of the runner is equal to the

area under the graph. So

1
ut + at 2
2

p
When h = 0, t would be zero. Hence, the graph would not have

11.

15. (a) From the graph, the reaction time is

0.3 s .

18 = u+5a

(1M+1A)

Active Physics Full Solutions to Textbook Exercises

| 13

Chapter 5 Motion Chapter Exercise

v 2 u 2 = 2as , we have
0

2
>
v
v 2 = 2(0.85g )(12.8)


v =

2(0.85 9.8)(12.8)

= 14.60 14.6 m s

(1M)

Therefore the speed of the car before it

applies the brake is 14.6 m s1 .
(ii) Let t be the time required. By v =
have

s
t,

(1A)

we

29.3
t

14.60 =

(1M)

t 2.01 s

The required time is

2.01 s

(1A)

(b) The speed of the car is

v = 14.60 m s1 = 14.6 3.6 = 52.56 km h1

(1M)

The speed of the car does not exceed the speed

limit.
(1A)
But his reaction time is too long.

* + 1 at 2
ut
19. (a) s = 
2
2
s
.
a
0

Correct axis and labels: 1A

Correct scale: 1A
Correct data points: 1A
Correct best- it line: 1A
182
(b) The slope = 0.920.1
=

19.51 m s2

(1A)

2
Since v 2 u
= 2g s
of the graph is 2g .
Therefore, g = 19.51
2

v 2 = 2g s , the slope

9.76 m s2

(b) No.
The result would not be affected because all
rulers fall at the same acceleration under
gravity.
18. (a) (i) Take the forward direction as positive.
Let v be the inal velocity of the car. By

(1A)

s = 12 at 2

t2 =
(1M)

As the acceleration of the ball is a constant, the

slope a2 is also a constant.
(1M)
So he should have obtained a straight line
passing through the origin.
(b) (i)
Slope =

(1A)

17. (a) Ask the assistant to hold the ruler upright,

with the 0 cm mark at the bottom. Place your
ingers around the 0 cm mark.
(1A)
Ask the assistant to release the ruler without
warning. You need to catch the ruler as fast as
possible with your ingers once it falls.
(1A)
Find the height h fallen by the ruler.
(1A)
Use the equation h = 12 g t 2 to ind the reaction
time t , in which g is the acceleration due to
gravity.
(1A)

(1A)

(ii) By (i), slope =

a=

2
a

0.1 0.06
0.5 0.3

(1M)

0.2 m s2

(1A)

= 0.2, therefore

2
= 10 m s2
0.2

The acceleration is

10 m s2

downward.
(1A)

(c) Any

of the following:

Air resistance
Error in reading the ruler marks
The dimension of the ball which causes
error in measuring s

(1A)

(1M)

Time delay due to the trapdoor or the

electromagnet

(2A)

14 |

Chapter 5 Motion Chapter Exercise

20. (a) distance moved: scalar

speed: scalar
acceleration: scalar
(b) (i) (1) The velocity is represented by the
slope of the graph.
As the slope of the graph at t = 0 is
zero, the ball moves from rest.
(2) The slope of the graph becomes
constant after t = 0.8 s. So air
resistance cannot be neglected.

Active Physics Full Solutions to Textbook Exercises

(ii) Assume that car B catches up with car
A at time t = 20 + T , which satis ies the
condition

(1A)
(1A)
(1A)

(v B v A ) T = s

(1A)

Using the result in (c)(i),

(1A)

(20 15) T = 125

T = 25
(1A)

Car B catches up car A at

t = 20 + 25 = 45 s .

(ii) Sketch of the graph:

Shoot-the-stars Questions (p.69)

1.

The graph line is always above the given

line and and starts from zero: 1A
The graph line is continuous with an
increasing slope: 1A
The graph is curved: 1A

21. (a) During t = 010 s, car A moves with a constant

acceleration.
(1A)
During t = 1080 s, it moves with a constant
velocity.
(1A)
(b) (i) Car B .
(1A)
The greatest acceleration of car B appears
between t = 10 s and t = 20 s.
200
The acceleration = 2010
= 2 m s2 .
(1A)
(ii) Refer to the igure at the bottom on the
next page.
Correct acceleration and deceleration: 1A
Correct lines: 1A

(c) (i) The area under v t graph of car A by

t = 20 s t A = 10+20
15 = 225 m.
2
The area under v t graph of car B by
t = 20 s t B = 1020
= 100 m.
2
The separation between A and B
s = 225 100 = 125 m .

(1M)

(1M)

(1A)

Refer to the igures below.

(1A)

Active Physics Full Solutions to Textbook Exercises

Chapter 5 Motion Chapter Exercise

2. Take the downward direction as positive. Let

s A and s B be the displacement of balls A and B
respectively when they meet.

1
2

s A = uT + g T
2

s = (u)T + 1 g T 2
B
2

| 15

the lorry, we have

1
s T = 20T + (8)T 2
2
1
s L = 15T + (9)T 2
2

(1M)

Considering the magnitude of the displacements, we have

Since the downward direction is taken to be

positive, s B < 0.
(1M)
So s = s A + (s B ) = (uT + 12 g T 2 ) + (uT 12 g T 2 )
= 2uT .
(1A)

30 = s T + (s L )

30 = 20T 4T 2 + 15T 4.5T 2

0 = 8.5T 2 35T + 30
T = 1.217 1.22 s or 2.90 s (rejected)

3. (a) Take the direction to the left as positive. Let T

be the time when they collide.

So the collision takes place at time t =

By v = u + at , and considering the motion of the

lorry, we get
72
0 = ( 3.6
) + (8)t
t = 2.5 s.
The time required for the lorry to stop
completely is 2.5 s. As the collision happens
before the cars stop, we get T < 2.5 s.

(1M)

1.22 s

.
(1A)

s L is negative.

(b) By s = ut + 12 at 2 , we have
(1M)

Considering the displacement of the taxi and

1
s = (20)(1.217) + (8)(1.217)2 18.4 m
2

The position of the collision is 18.4 m away

from the position where the taxi applies the
brakes.

Chapter Exercise Q21(b)(ii)

(1A)