Active Physics Full Solutions to Textbook Exercises 05 Motion | p.

05 Motion For example, an object moving at a constant

high speed along a straight line has zero
Checkpoint
acceleration.
Checkpoint 1 (p.12)
(b) F
1 B A braking car is moving forward, but since its
2 B velocity is decreasing, its acceleration is
3. (a) Total distance =160 + 160 + 240 = 560 m pointing backward.

(b) Keith’s overall displacement is 240 m to the (c) T

east. Its velocity stays unchanged.

4. (a) Total distance = 5 × 20= 100 cm 3. (a) No.

You are not undergoing acceleration when
(b)
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.
(d) No.
Magnitude of the displacement You are not undergoing acceleration when
you are moving at a constant speed along a
= 44.72 = 44.7cm
straight line.
Direction of the displacement:
(e) Yes.
θ = 26.57° ≈ 26.6° You are undergoing acceleration whenever
Therefore the overall displacement of the you are changing your direction of motion.
bee is 44.7 cm (N 26.6° E).
Checkpoint 4 (p.34)
Checkpoint 2 (p.21) 1. A
1. Table: The s–t graph of car P has a larger slope.
2. C
time / s 0 1 2 3 10
From the graph, we can see that the car slows
displacement / cm 0 5 10 15 50
down initially (i.e. v decreases in magnitude), then
2. A’s average speed = 4/4 = 1 cm s−1 = 0.01 m s−1
reverses its direction of motion (i.e. v becomes
B’s average speed = 4/4 = 1 cm s−1 = 0.01 m s−1 zero), and eventually speeds up (i.e. v increases in
A’s average velocity magnitude but carries a negative sign). Hence, the
= 4/4 = 1 cm s−1 = 0.01 m s−1 to the right car is making a U-turn.
B’s Average velocity 3. (a) Acceleration
= (3−1)/4 = 0.5 cm s−1 = 0.005 m s−1 to the right = slope of the v–t graph
3. A = = 1 m s−2
The ferries have different average velocities as
(b) Overall displacement
they travel towards different directions.
= total area under the v–t graph
= = 37.5 m
Checkpoint 3 (p.27)
1. Table:
Checkpoint 5 (p.40)
time / s 0 1 2 3 10
velocity / m s−1 0 2 4 6 20 1. During time t =0 to 5 s , Simon walks away from
2. (a) F the sensor at an average velocity of 0.3 m s−1.
An object moving faster may not have a larger
 Slope = = 0.3 m s−1
change in velocity (i.e. higher acceleration).
Active Physics Full Solutions to Textbook Exercises 05 Motion | p.2

During time t =5 to 10 s , Simon walks towards the 3. Take the upward direction as positive.
sensor at an average velocity of 0.2 m s−1. Given: u =5 m s−1 a =−9.81 m s−2 t =3 s
We can have to find s.
 Slope = =0.2 m s−1
By the equation

2. Graphs:
= −29.145 m
Hence, the cliff is 29.145 m high.

Exercise

Exercise 5.1 (p.12)

1. (a) 4.5 km = 4.5 × 1000 = 4500 m
(b) 3.2 cm = 3.2 × 10−2 = 0.032 m
(c) 6.5 h = 6.5 × 60 × 60 = 23 400 s
Checkpoint 6 (p.49)
(d) 19 min = 19 × 60 = 1140 s
1. Take the initial direction of motion of the
2. C
motorcycle as positive.
Given: s =500 m u =50 m s−1 v =100 m s−1 3. A
We have to find t using the equation Only cars P and Q have the same initial and final
positions.
4. It represents the distance.
Hence, we get 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
t = 6.667 s
destination (i.e. displacement).
It takes 6.67 s for the motorcycle to accelerate.
5. The following shows the path of the marble when
2. Take the initial direction of motion of the car as its displacement is the smallest.
positive.
Given: s =50 m v=0 a = −5 m s−2
We have to find u using the equation
v2 - u2 = 2as
Hence, we get
v2 - u2 = 2(-5)(50)
u = 22.36 m s-1
The speed of the car before the car braked is The following shows the path of the marble when
22.4 m s−1. its displacement is the largest.

Checkpoint 7 (p.61)
1. Table:
time / s velocity / m s−1 displacement / m
0 20 0
1 10 15 Therefore, the smallest and the largest
2 0 20 displacements of the marble are 2 cm and 6 cm
3 −10 15 respectively.
4 −20 0 6. Electronic timing is more accurate than using
2. (a) F stopwatches.
Both balls accelerate at the same rate under It is because the finishing times of the swimmers
gravity, regardless of what they are made of. in competitions are very close. The error due to
(b) T the timekeeper’s reaction time would affect the
Its upward motion is slowing down. rankings if stopwatches were used.
Active Physics Full Solutions to Textbook Exercises 05 Motion | p.3

7. (a) Displacements: Magnitude of the displacement:

AD = ≈ 5.15 m

Direction of the displacement:

θ ≈ 29.05°

So, Linda’s displacement is 5.15 m

(N 61.0° E).
10. (a) The distance travelled
= (3 + 6 + 6)(1000) = 15 000 m.
(b) Otto’s path:

(b) Adding the displacement vectors using the

tip-to-tail method:

Magnitude of the displacement:

s= = 6.7082 km ≈ 6710 m

Direction of the displacement:

θ ≈ 26.6°
Magnitude of :
Otto’s displacement is 6710 m (N 26.6° W).
AD = ≈ 94.3 m
11. (a) Distance travelled
Direction of : = (1/2) × 2π(70/2) + 90 ≈ 200 m
θ ≈ 32.0°

So the displacement is 94.3 m (S 32.0° W).

8. (a) The distance travelled
= 120 + 360 + 82= 562 cm.
(b) The displacement
= (+120) + (−360) + (+82)= −158 cm. Magnitude of the displacement:
So the displacement is 158 cm to the left. AB = = 114 m
(c) No. Direction of the displacement:
The displacement of the ball does not
θ ≈ 52.1°
depend on the choice of the positive
direction. So the distance travelled by Joseph is 200 m,
9. (a) The distance travelled by Linda and his displacement is 114 m (S 52.1° E).
= 1.5 + 2.5 + 3 = 7 m. (b) The distance travelled by Joseph would
(b) increase but his overall displacement would
remain unchanged.

Exercise 5.2 (p.22)

1. (a) Yes (b) Yes (c) No

2. (a) 110 km h−1 = ≈ 30.6 m s−1

(b) 55 cm s−1 = 55 × 10−2 = 0.55 m s−1

Active Physics Full Solutions to Textbook Exercises 05 Motion | p.4

(c) 9 km min−1 = = 150 m s−1 Magnitude of the velocity:

v= = ≈ 10.4m s-1
3. (a) T (b) T (c) T
4. (a) Time taken to go through the tunnel Direction of the velocity:

= = 0.03611 h θ ≈ 63.4°

Average speed So the average velocity is 10.4 m s−1

(N 63.4° E).
(c) No.
= ≈ 72 km h-1 < 80 km h-1 The velocity of the car changes when it
No. changes its direction of motion.
Her average speed is lower than the speed 7. (a) Time required for Mandy to run from A to B
limit ( 80 km h−1).
= = 1.885 s
(b) She is fined because her speed at a certain
Time required for Mandy to run from B to C
instant is higher than the speed limit.
5. (a) Average speed = = 2.356 s
Time taken for the whole journey
= 1.885 + 2.356 = 4.241 ≈ 4.24 s
= ≈ 0.485 m s-1 (b) Average speed

(b) Take the direction to the right as positive. = = 4.44 m s-1

Displacement from O to B:
s = (0.8 × 4) − (1 × 2) = +1.2 m (c) Average velocity
Average velocity: = = 2.83 m s-1 due east
v= = = +0.1 m s-1
8. Let d be the distance between X and Y.
Hence, the average velocity is 0.1 m s−1 The total time taken for the whole journey
towards the right.
(c) The time required to travel from B to O The average speed
= = 1.2 s
The average speed
= = 0.485 m s-1
Exercise 5.3 (p.28)
The overall displacement = 0.
1. C
The average velocity = 0.
The acceleration of the car is the greatest when
6. (a) The distance travelled the difference between u and v is the greatest.
= 50 × (2 + 4) = 300 km.
2. B
The average speed
For a braking car, the direction of its acceleration
= = 13.9 m s-1 is opposite to its velocity.

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

Since the speed of the car remains at because he gains speed, changes his direction of
50 km h−1 during the 6 hours, the average motion, and slows down respectively. He does not
speed is 50 km h−1 ≈ 13.9 m s−1. undergo acceleration at B because he maintains
his speed.
(b)
4. Take the forward direction as positive.
By , we have

2= v = 20 m s-1
The speed of the boat is 20 m s−1 after 6 s.
Overall displacement: 5. Take the forward direction as positive. When the
s= = 223.6 km minibus speeds up, by , we have
Active Physics Full Solutions to Textbook Exercises 05 Motion | p.5

3= t=5s 4. A
Graphs B and D mismatch the decelerating and
When it slows down, by , we have accelerating stages. Graph C fails in the final stage.
−1.5 = t =10 s  Option B could be correct if the backward
The total time of travel =5+10+10= 25 s. direction of the bus was taken as positive. But it is
not natural. Besides, there are too many types of
6. (a) Take the forward direction as positive.
motion that can produce such an a–t graph.
By , we have
5. (a) Plot of the s–t graph:
a= = −0.6 m s-2
So the deceleration of the train is 0.6 m s−2.
(b) By , we have

−0.6 = t = 10 s
The train takes a further 10 s to stop.
7. Take the direction away from the racket as positive.
By , we have

a= = 32 m s-2
The acceleration of the ball is 32 m s−2 away from
the racket.
(b) Velocity = slope of the s–t graph
8. (a) The skater slides up the ramp with a
= = 0.04 m s-1
decreasing speed until she reaches the
highest point, where she is momentarily at The average velocity of the tank is 0.04 m s−1
rest. Then she slides down the ramp with an forward.
increasing speed. 6. (a) Take the forward direction as positive.
(b) Take the direction up the ramp as the Sketch of the v–t graph:
positive. By , we have

−0.8 = t=5s
She will reach a speed of 2 m s−1 again 5 s
later.

Exercise 5.4 (p.41)

(b) The distance travelled
1. (a) Slope = = 18.75 m s-1 = = 22.5 m
The car moves for a distance of 22.5 m after
(b) Slope = = −0.085 m s-2
the driver sees the dog.

(c) Area = = 6300 m2 7. (a) From the graph, the shop is 160 m due east
of Jeff’s home.
2. A
(b) Velocity during time t = 0 to 5 min:
The s–t graph shows that the two cars have the
same displacement at t = T. v= = 0.533 m s-1

3. C Velocity during time t = 8 to 11 min:

Between t =0 and t = T, the area under the v–t v= = −0.889 m s-1
graph of car B has smaller area. So, car B has a
Jeff travels towards the east at a constant
smaller displacement and lags behind car A at
speed of 0.533 m s−1 during t = 0 to 5 min. He
time t = T.
then remains at rest during t = 5 to 8 min.
Finally he goes back to his starting position
steadily at a speed of 0.889 m s−1 during t = 8
to 11 min.
Active Physics Full Solutions to Textbook Exercises 05 Motion | p.6

(c) The total distance travelled by Jeff (b) Let T be the required time. The following
= (160)(2) = 320 m. shows the s–t graph of Sara.
The total displacement of Jeff = 0.
8. (a) During t = 0 to 0.5 s, the acceleration = 3/0.5
= 6 m s−2 away from the sensor.
During t =0.5 to 2 s, the acceleration = 0.
(b) The trolley accelerates uniformly at 6 m s−2
away from the sensor during t = 0 to 0.5 s.
Then it travels at a constant velocity of
3 m s−1 away from the sensor during t = 0.5
to 2 s.
(c) The overall displacement of the trolley
Consider the slope of the s–t graph during
= t = 10 to T, we have
= 5.25 m away from the sensor. T = 15 s
9. (a) For time t = 0 to 35 s , the acceleration
(c) When she goes back, her velocity
= ≈ 0.571 m s−2.
For time t =35 to 100 s, the acceleration = 0.
= 4 m s−1.
For time t = 100 to 140 s, the acceleration
Therefore, the plot of the v–t graph becomes
= = −0.5 m s−2.
Therefore, the a–t graph is

(b) The overall displacement of the train

= area under the v–t graph
= = 2050 m.
11. (a) The car travels at a uniform speed during
The average velocity
t = 12 to 20 s as its acceleration is zero and
= 2050/140 ≈ 14.6 m s−1 forward.
its velocity is non-zero during that period.
(c) Let T be the time when the train is midway
(b) No.
between the stations. Considering the
At t =10 to 12 s , the car still speeds up
displacement of that position from the
(a > 0) but the magnitude of its acceleration
departing station, we have
(i.e. the change in velocity) decreases.
12. (a) Car A has a greater acceleration, because its
2T − 35 = 102.5 graph has a greater initial slope.
T = 68.75 s (b) The initial acceleration of car A = 30/2
10. (a) The distance travelled = 15 m s−2 forward.
= 5 × 4 + 4 × (20 − 10) = 60 m. The initial acceleration of car B = 40/4
The displacement = 10 m s−2 forward.
= (5)(4) + (−4)(20 − 10) = −20 m (c) Car B overtakes car A at the time when their
So the displacement is 20 m backward. displacements (the areas under their v–t
graphs) are the same. Let T be the
corresponding time.

3(2T − 2) = 4(2T − 4)
Active Physics Full Solutions to Textbook Exercises 05 Motion | p.7

3T − 3 = 4T − 8 s
T=5s Let s1 be the displacement of the runner during
13. (a) The velocity of the car during t = 0 to 10 s the acceleration. By v2 − u2 = 2as, we have
= = 7 m s−1. 102 − 0 = 2(3.3)s or s1 =15.15 m.
The displacement of the runner after the
The velocity of the car during t = 20 to 30 s
acceleration is s2 = 100 − 15.15 = 84.85 m.
= = 1 m s−1.
Therefore, t2 = s2 / v2 = 84.85/10 = 8.485 s. The
First of all, the car starts moving from the
finishing time of the runner = 3.030 + 8.485
starting point at a constant speed of 7 m s−1
≈ 11.5 s.
during t = 0 to10 s. And then it decelerates
from v = 7 to 1 m s−1 during t = 10 to 20 s. It 7. (a) By , we have
finally moves at a constant speed of 1 m s−1 a = 3 m s-2
during t = 20 to 30 s.
The acceleration is 3 m s−2 forward.
(b) The average velocity
(b) At t =4 s, we have
= ≈ 3.67 m s−1 forward.
14. (a) The trolley in case A moves at a constant
velocity while that in case B accelerates. Therefore, x = 24.

(b) The trolley in case B has a higher average 8. (a) Take the direction to the left as positive.
velocity. The number of ticks on the tapes of At t = 2 s, we have
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 (b) By v = u + at, we have
velocity. v = 0 + (6)(3) = 18 m s−1.
(c) The distance travelled by the deer during the
chase is s' = 27 − 18 = 9 m. So the speed of
Exercise 5.5 (p.50)
the deer v' = s'/t' = 9/3 = 3 m s−1.
1. A
9. (a) Take the direction to the left as positive.
Nancy measures the values of the time t and the
displacement s in order to find the acceleration a. 450 km h-1 = = 125 m s-1
Since we know that the initial velocity u = 0, the By v2 − u2 = 2as, we have
best option is A. v2 − 1252 = 2(−5)(1200)
2. B Therefore,
Given that u = 0. By v2 − u2 = 2as, we have v=
v2 − 0 = 2ad = 60.21 ≈ 60.2 m s-1
3. C So the speed of the plane is 60.2 m s−1 when
vavg = it goes beyond the runway.
(b) By v2 − u2 = 2as, we have
4. Take the forward direction as positive. 0 − 60.212 = 2a(250)
By , we have
a=
t = 0.064 s
= −7.25 m s-2
5. (a) Take the direction to the right as positive. By The deceleration of the plane is 7.25 m s−2
v = u + at, we have 1 = 0 + a(0.02) or after it goes beyond the runway.
a =50 m s−2. The acceleration of the ball is
(c) Let t1 and t2 be the time intervals that the
50 m s−2 towards the right.
plane moves on the runway and the field
(b) Yes. respectively. Since
The black ball will fall into the pocket v = u + at
because by v2 − u2 = 2as, we have
we have
02 − 0.42 = 2(−0.1) s or s =0.8 m > 0.7 m.
t1 = = 12.96 s
6. Take the forward direction as positive. Let t1 and t2
be the time taken for the runner to accelerate and t2 = = 8.305 s
move at a constant velocity respectively.
So the total time taken
By v = u + at, we have 10 = 0 + (3.3)t1 or t1 =3.030
= 12.96 + 8.305 ≈ 21.3 s.
Active Physics Full Solutions to Textbook Exercises 05 Motion | p.8

10. (a) From the graph, his reaction time is 0.8 s. Exercise 5.6 (p.61)
1. D
(b) The acceleration = = −3.75 m s-2
The acceleration due to gravity always points
So the deceleration of the car is 3.75 m s−2. downwards regardless of the motion of the ball.
(c) Yes. 2. A
The car can stop before the traffic light. The graph given is a s–t graph, whose area under
The total distance travelled by the car the curve has no physical meaning.
= area under the v–t graph
3. C
= = 28.8 m < 32 m.
4. Take the downward direction as positive.
(d) No.
The car cannot stop before the traffic light.
The distance travelled by the car before
braking is s1 = v1 t1 = 15 × 0.8 =12 m. t = 4.778 s or −4.778 s (rejected)
By v2 − u2 = 2as, the distance travelled by the Average speed
car during the braking process is
0 − 152 = 2 (−3.75) s2 v= ≈ 23.4 m s-1
s2 = 30 m
 Since the release takes place at t = 0, a negative
Hence, the total distance travelled by the car
time indicates a time instant before the release.
is 12 + 30 = 42 m > 40 m.
This solution is physically impossible and is
11. (a) Take the upstream direction as positive. therefore rejected.
Assume that the boat passes A at time t = 0.
5. (a) Take the downward direction as positive.
By , we have

The depth of the well is 19.6 m.

0 = t2 − 16t + 48
(b) v = u + at = 0 + (9.81)(2) = 19.62 m s−1
t=4s or 12 s
The velocity of the coin when it reaches the
The boat reaches B for the first time at t = 4 s
water surface is 19.62 m s−1 downward.
and the second time at t = 12 s. Therefore the
answer is 12 s. 6. By , we have
(b) By v2 − u2 = 2as, we have
v2 − 42 = 2 ( −0.5)(12)
v2 = 4
v = 2 m s−1 or −2 m s−1 t ≈ 0.452 s or −0.452 s (rejected)
Therefore, the velocity of the boat when it The minimum time required for the turn is 0.452 s.
reaches B for the second time is 2 m s−1
7. (a) Take the downward direction as positive. By
downstream.
v2 − u2 = 2as, we have
 Positive v means the boat moves upstream 162 − u2 = 2(9.81)(13)
and negative v means the boat moves
u=
downstream.
u = 0.9695 ≈ 0.970 m s-1
The initial velocity of the ball is 0.970 m s−1
12. By v2 − u2 = 2as, we have downward.
(b) After throwing for 1 s,

Hence,

s1 = s= ≈ 5.87 m
the ball is at 5.87 m below the starting
s2 = point.
Combining the equations, we have 8. (a) Take the downward direction as positive.
s1 : s2 = −1200 : −400 = 3 : 1.
Active Physics Full Solutions to Textbook Exercises 05 Motion | p.9

0 = 4.905t2 − 3t − 10
t = 1.766 ≈ 1.77 s or −1.154 s (rejected)
So she takes 1.77 s to reach the water
surface.
(b) v =u + at = −3+(9.81)(1.766) ≈ 14.3 m s−1
Her velocity when she reaches the water
surface is 14.3 m s−1 downward.
9. (a) Take the downward direction as positive.
By v = u + at , we have 0 = −2 + (9.81)(t) The area under the v–t graph is equal to the
⇒ t = 0.2039 s displacement of the object. So,
The fish reaches the highest point at t =
x= = 46.5 m
0.204 s.
(b) Consider the motion of the fish before it 3. C
reaches the highest point. Statement (1) is correct. At time t , the object is
momentarily at rest.
Statement (2) is incorrect. The acceleration of the
s= object at time t is determined by the slope of the
≈ −0.2039 m graph at that time, which takes a non-zero value.
Now consider the motion of the fish when it Statement (3) is incorrect. The displacement of
falls from the highest point to the table. the object at time t is determined by the area
under the graph during time = 0 to t, which takes a
non-zero value.
s= = 0.7697 m
4. D
So the total distance travelled by the fish Statement (1) is correct. The graphs have the same
= 0.2039 + 0.7697 = 0.9735 ≈ 0.974 m. slope when the cars are braking, and so they slow
The average speed of the fish down at the same rate.
= 0.9735/0.6 ≈ 1.62 m s−1. Statement (2) is incorrect. The reaction time of
(c) By v2 − u2 = 2as, we have the driver of car A is longer than that of car B .
v2 −22 = 2 (9.81)(0) Statement (3) is correct. As both cars have the
v2 = 4 same initial speed and slow down at the same rate,
v = 2 m s−1 or −2 m s−1 they travel the same distance when they
Since the downward direction is taken as decelerate.
positive, v < 0 represents an upward motion
5. C
and v >0 represents a downward motion. So
The acceleration of the ball is equal to the
the velocity is 2 m s−1 downward.
acceleration due to gravity g, which is constant
throughout the motion.
Chapter Exercise
6. A
Multiple-choice Questions (p.65)
If the object is thrown vertically downward, the
1. D time required to reach the bottom of the cliff
Statements (1) and (2) are correct as both cars decreases, but the acceleration (i.e. the slope of the
have the same displacement s at t =4 s. v–t graph) remains unchanged.
Statement (3) is correct. The s–t graphs of both 7. D
cars have increasing slopes throughout.
2. C
The slope of the graph is 2a. If a increases, the
From the v–t graph, the acceleration of the object
slope increases.
from t = 4 to 8 s is
8. C
a= = −3 m s-2
Statement (1) is possible. A car turning a corner at
By v = u + at, the velocity of the object at t = 5 s is a constant speed is accelerating.
v =12 + (−3)(1) = 9 m s−1. Statement (2) is possible. A ball being thrown
Therefore, we get upward is momentarily at rest (i.e. zero velocity)
Active Physics Full Solutions to Textbook Exercises 05 Motion | p.10

at the highest point, but it is accelerating the shorter the time interval between which the
downwards. images are taken. So option A is the answer.
Statement (3) is impossible. Constant velocity Note that the equations of uniformly accelerated
implies constant speed and fixed direction of motion are independent of the mass of the object,
motion. so both options C and D are incorrect.
9. B 12. D
The speed of the ball during time t = 0 to 0.1 s is
u= = 0.36 m s-1
Therefore,
The speed of the ball during time t = 0.3 to 0.4 s is
v= = 0.2 m s-1
13. B
By v = u + at, we have
Statement (1) is incorrect. The sign of the velocity
0.2 = 0.36 + a (0.2) ⇒ a = −0.8 m s−2
of P does not change at t =1 s.
The deceleration of the ball is 0.8 m s−2.
Statement (2) is correct. At t = 2 s, the separation
10. C
between P and Q is
Assuming that the reaction time of the student is
negligible. The vertical height h and the time t' are sPQ = =4m
related by
Statement (3) is incorrect. At t = 4 s ,
sP = =2m

So, a straight line passing through the origin sQ = =8m

should be obtained.
Therefore, the displacements of P and Q are not
However, if the reaction time T is taken into the
the same, and they do not meet each other.
account, the measured time t is given by
t' = t − T. 14. B
Let u and v be the velocities of the particle at time t
Combining the equations, we have
=0 and t =4 s respectively. Note that v = u + at.
At t =4 s, 36 = 9 = u + 2a
So the graph of against t is a graph with At t =6 s,
positive t–intercept. Refer to the graph below:
36 =

18 = u + 5a
Subtracting the equations, we have
18 − 9 = (u − u) + (5a − 2a) ⇒ a =3 m s−2.

 Consider option B. If air resistance is not Structured Questions (p.67)

negligible, the falling object should take a longer
15. (a) From the graph, the reaction time is 0.3 s.
time to fall the same vertical distance than the (1A)
case without air resistance. The slope of the graph
(b) The displacement of the runner is equal to
would become smaller, but the graph would not
the area under the graph. So
have a t–intercept. Consider option D. If the object
falls with a downward velocity, we get 100 = (1M)

T = 14.65 (1A)

∴ vavg = ≈ 6.83 m s-1（forward）

When =0, t would be zero. Hence, the graph
(1M+1A)
would not have a t–intercept.
(c) Let a be the acceleration of the runner
11. A during t = 0.3 to 4 s . The acceleration at that
For u = 0, we get s ∝ t2. Hence, the shorter the
distance between the adjacent images of the stone,
Active Physics Full Solutions to Textbook Exercises 05 Motion | p.11

time period is equal to the slope of the graph: reaction time t, in which g is the acceleration
due to gravity. (1A)
(b) No. (1A)

The result would not be affected because all

rulers fall at the same acceleration under
gravity. (1A)
18. (a) (i) Take the forward direction as positive.
a= ≈ 2.16 m s-2
Let v' be the final velocity of the car. By
 Correct axis and labels: 1A , we have
 Correct curve and data: 1A
16. (a) Refer to the table below.
(1M)
d /m 0.2 0.4 0.6 0.8 1.0 = 14.60 ≈ 14.6 m s-1
v2 / m2 s−2 4.0 7.8 11.8 15.8 19.4
Therefore the speed of the car before it
applies the brake is 14.6 m s−1. (1A)
(ii) Let t be the time required.
By v = st, the required time is
t = v/s = 14.60/29.3 ≈ 2.01 s. (1M+1A)

(b) The speed of the car is

v = 14.60 m s−1 = 14.6 × 3.6 = 52.56 km h−1
(1M)

The speed of the car does not exceed the

speed limit. (1A)
But his reaction time is too long. (1A)
19. (a) The distance

(1M)

As the acceleration of the ball is a constant,

the slope of t2–s graph 2/a is also a constant.
(1M)

So he should have obtained a straight line

passing through the origin.
(b) (i) The slope is
= 0.2 m s-2 (1M+1A)

 Correct axis and labels: 1A (ii) Slope = 2/a = 0.2

 Correct scale: 1A ∴ a = 2/0.2 =10 m s−2 (1M)

 Correct data points: 1A

The acceleration is 10 m s−2
 Correct best-fit line: 1A
downward. (1A)
(b) The slope = = 19.51 m s−2. (1A)

Since ⇒ v2 =2gs, the slope (c) Any two of the following: (2A)

of the v2–s graph is 2g. • Air resistance

Therefore, g = 19.51/2 ≈ 9.76 m s−2. (1A)
• Error in reading the ruler marks
17. (a) Ask the assistant to hold the ruler upright,
• The dimension of the ball which causes
with the 0 cm mark at the bottom. Place your
error in measuring s
fingers around the 0 cm mark. (1A)
Ask the assistant to release the ruler without • Time delay due to the trapdoor or the
warning. You need to catch the ruler as fast electromagnet
as possible with your fingers once it falls. (1A) 20. (a) distance moved: scalar (1A)

Find the height h fallen by the ruler. (1A) speed: scalar (1A)
Use the equation to find the
acceleration: scalar (1A)
Active Physics Full Solutions to Textbook Exercises 05 Motion | p.12

(b) (i) (1) The velocity is represented by the (ii) Assume that car B catches up with car
slope of the graph. (1A) A at time t = 20 + T, which satisfies the
As the slope of the graph at t = 0 condition
is zero, the ball moves from rest. (vB − vA) × T = ∆s
(1A)
Using the result in (c)(i),
(2) The slope of the graph becomes (20 − 15) × T = 125 ⇒ T = 25 s (1M)
constant after t = 0.8 s. So air Car B catches up car A at
resistance cannot be neglected. t =20 + 25= 45 s. (1A)
(1A)

(ii) Sketch of the graph:

Shoot-the-stars Questions (p.69)
1. C

• The graph line is always above

the given line and starts from
zero: 1A
• The graph line is continuous with
an increasing slope: 1A
• The graph is curved: 1A 2. Take the downward direction as positive. Let sA
21. (a) During t = 0 to 10 s , car A moves with a and sB be the displacement of balls A and B
constant acceleration. (1A) respectively when they meet.
During t = 10 to 80 s , it moves with a
constant velocity. (1A)
(b) (i) Car B. (1A)
The greatest acceleration of car B Since the downward direction is taken to be
appears between t =10 s and t =20 s. positive, sB < 0. (1M)
Acceleration = = 2 m s−2 (1A)
(ii) The graph required: = 2uT。 (1A)

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

T be the time when they collide.
By v = u + at and considering the motion of
the lorry, we get

∴ t = 2.5 s
• Correct values: 1A The time required for the lorry to stop
• Correct lines: 1A completely is 2.5 s . As the collision happens
before the cars stop, we get T < 2.5 s. (1M)
(c) (i) The area under v–t graph of car A
Considering the displacement of the taxi and
tA = = 225 m
the lorry, we have
The area under v–t graph of car B
tB = = 100 m. (1M)
The separation between A and B
∆s =225 − 100= 125 m. (1A)
Considering the magnitude of the
displacements, we have
Active Physics Full Solutions to Textbook Exercises 05 Motion | p.13

30 = sT + (− sL)
∴ 30 = 20T − 4T2 + 15T − 4.5T2
0 = 8.5T2 − 35T + 30 (1M)
T = 1.217 ≈ 1.22 s or 2.90 s (rejected)
So the collision takes place at t = 1.22 s. (1A)
 The displacement sL is negative.
(b) By , we have

s= ≈ 18.4 m
The position of the collision is 18.4 m away
from the position where the taxi applies the
brakes. (1A)