Chapter 5 MCQs

For Full Scale Course: Written and Compiled By Sir Hunain Zia (AYLOTI), World Record Holder With 154 Total Personal A Grades, 11 World Records and 7 Distinctions, Educate A Change.

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1

A car travels at 20 m/s. The driver’s reaction time is 0.75 s.

What is the thinking distance?

A 15 m
B 19 m
C 20.75 m
D 26.7 m

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2

A driver has a reaction time of 0.60 s. During this time, the car travels 18 m before the brakes are applied.

What is the speed of the car?

A 0.033 m/s
B 10.8 m/s
C 30 m/s
D 300 m/s

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3

A car travelling at 24 m/s has a thinking distance of 18 m and a braking distance of 54 m.

What is the stopping distance?

A 36 m
B 54 m
C 72 m
D 96 m

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4

A car travels at 30 m/s. The driver’s reaction time is 0.80 s. After the brakes are applied, the car decelerates uniformly to rest in 5.0 s.

What is the stopping distance?

A 24 m
B 75 m
C 99 m
D 174 m

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5

A car travels at 18 m/s. The driver reacts in 0.50 s. The car then brakes uniformly to rest in 3.0 s.

What is the stopping distance?

A 9.0 m
B 27 m
C 36 m
D 63 m

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6

A car travels at 25 m/s. The thinking distance is 20 m.

What is the driver’s reaction time?

A 0.80 s
B 1.25 s
C 5.0 s
D 45 s

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7

A car travels at 28 m/s. Its braking distance is 56 m.

Assuming uniform deceleration, what is the magnitude of the deceleration?

A 0.25 m/s²
B 7.0 m/s²
C 14 m/s²
D 28 m/s²

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8

A car travelling at 20 m/s brakes uniformly to rest with a deceleration of 5.0 m/s².

What is the braking distance?

A 2.0 m
B 4.0 m
C 40 m
D 100 m

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9

A car travelling at 16 m/s has a reaction time of 0.75 s and then brakes uniformly at 4.0 m/s².

What is the stopping distance?

A 12 m
B 32 m
C 44 m
D 76 m

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10

A car’s speed doubles. The driver’s reaction time and the braking deceleration remain unchanged.

Which row shows how the thinking distance and braking distance change?

	thinking distance	braking distance
A	doubles	doubles
B	doubles	quadruples
C	quadruples	doubles
D	quadruples	quadruples

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For Full Scale Course: Written and Compiled By Sir Hunain Zia (AYLOTI), World Record Holder With 154 Total Personal A Grades, 11 World Records and 7 Distinctions, Educate A Change.

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11

A car travels at speed v. The driver’s reaction time is t. The car then brakes uniformly with deceleration a.

Which expression gives the stopping distance?

A vt + v/a
B vt + v²/2a
C vt + 2av²
D v/t + v²/a

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12

A car travels at 12 m/s. The driver reacts in 0.50 s. The braking distance is 18 m.

What is the stopping distance?

A 6 m
B 18 m
C 24 m
D 30 m

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13

A car travels at 36 km/h. The driver’s reaction time is 0.70 s.

What is the thinking distance?

A 2.5 m
B 7.0 m
C 25 m
D 50 m

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14

A car travels at 72 km/h. The driver reacts in 0.90 s. The car then brakes uniformly to rest in 4.0 s.

What is the stopping distance?

A 18 m
B 40 m
C 58 m
D 98 m

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15

A car travels at 90 km/h. The driver reacts in 0.80 s. The braking distance is 62.5 m.

What is the stopping distance?

A 20 m
B 62.5 m
C 82.5 m
D 134.5 m

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16

Which factor directly increases thinking distance but does not directly increase braking distance?

A icy road
B worn tyres
C alcohol affecting the driver
D greater mass of passengers

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17

Which change increases braking distance but does not directly increase thinking distance?

A tiredness
B alcohol
C wet road
D drug use affecting reaction time

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18

A driver is tired but the road and tyres are unchanged.

Which distances are directly affected?

A thinking distance only
B braking distance only
C both thinking and braking distance
D neither thinking nor braking distance

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19

A car carries a heavy load. The driver’s reaction time and the speed before braking are unchanged.

Which statement is most likely correct?

A Thinking distance increases because reaction time increases.
B Thinking distance decreases because the car is heavier.
C Braking distance may increase because more kinetic energy must be removed.
D Braking distance is unchanged because speed is unchanged.

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20

A car’s tyres are worn smooth. The driver’s reaction time and speed are unchanged.

Which distance is most directly increased?

A thinking distance
B braking distance
C distance travelled before seeing the hazard
D total journey distance only

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21

A driver sees a hazard at time t = 0. The graph of speed against time is horizontal for 0.8 s, then slopes uniformly down to zero over the next 3.2 s.

The initial speed is 20 m/s.

What is the stopping distance?

A 16 m
B 32 m
C 48 m
D 64 m

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22

A speed–time graph for stopping has:

• horizontal section: 15 m/s for 0.60 s

• straight-line braking section: 15 m/s to 0 in 2.4 s

What is the stopping distance?

A 9 m
B 18 m
C 27 m
D 45 m

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23

A driver’s reaction time doubles, but the car’s initial speed and braking deceleration remain unchanged.

Which statement is correct?

A Thinking distance doubles and braking distance is unchanged.
B Thinking distance is unchanged and braking distance doubles.
C Both thinking and braking distance double.
D Stopping distance is unchanged.

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24

A car travels at 15 m/s. The driver’s reaction time is 0.40 s. The braking distance is 22.5 m.

The driver becomes distracted and reaction time increases to 1.2 s. The braking distance is unchanged.

What is the new stopping distance?

A 28.5 m
B 34.5 m
C 40.5 m
D 49.5 m

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25

A car travels at 10 m/s and stops with a braking distance of 8.0 m.

If the same car travels at 30 m/s and brakes with the same deceleration, what is the braking distance?

A 24 m
B 48 m
C 72 m
D 96 m

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For Full Scale Course: Written and Compiled By Sir Hunain Zia (AYLOTI), World Record Holder With 154 Total Personal A Grades, 11 World Records and 7 Distinctions, Educate A Change.

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26

A car travels at speed v and has braking distance d on a dry road.

On an icy road, the maximum frictional force is half as large. The speed is unchanged.

What is the new braking distance?

A d/4
B d/2
C 2d
D 4d

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27

A car has mass m and speed v. During braking, a constant frictional force F acts.

Which expression gives the braking distance?

A F / mv²
B mv² / 2F
C 2F / mv²
D mv / F

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28

A car of mass 1000 kg travels at 20 m/s. During braking, the average resistive force is 5000 N.

What is the braking distance?

A 20 m
B 40 m
C 80 m
D 200 m

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29

A car of mass 1200 kg travels at 15 m/s. During braking, a constant frictional force of 4500 N acts.

What is the braking distance?

A 15 m
B 30 m
C 45 m
D 60 m

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30

A lorry and a car travel at the same speed. The lorry has twice the mass of the car. The maximum braking force on the lorry is also twice the maximum braking force on the car.

How do their braking distances compare?

A lorry has half the braking distance
B lorry has the same braking distance
C lorry has twice the braking distance
D lorry has four times the braking distance

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31

An object falls through air from rest.

Which row correctly shows how weight and air resistance compare at the instant it is released?

	weight	air resistance
A	zero	zero
B	non-zero	zero
C	zero	non-zero
D	non-zero	equal to weight

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32

A skydiver falls from rest before opening the parachute.

Which statement correctly describes the motion at first?

A Air resistance is greater than weight, so acceleration is upwards.
B Weight is greater than air resistance, so acceleration is downwards.
C Weight equals air resistance, so speed is constant immediately.
D Both weight and air resistance are zero.

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33

A falling object reaches terminal velocity.

Which statement is correct?

A air resistance is zero
B weight is zero
C air resistance equals weight
D air resistance is greater than weight

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34

A falling object is moving downwards and slowing down.

Which statement must be correct?

A weight is greater than air resistance
B air resistance is greater than weight
C weight equals air resistance
D no resultant force acts

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35

A skydiver opens a parachute while falling.

Immediately after the parachute opens, what happens to the air resistance and acceleration?

	air resistance	acceleration
A	increases greatly	upwards
B	increases greatly	zero
C	decreases greatly	downwards
D	becomes zero	downwards

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36

A skydiver falls at terminal velocity before opening the parachute. The parachute then opens.

Which sequence is correct?

A speed increases, then becomes constant at a higher terminal velocity
B speed decreases, then becomes constant at a lower terminal velocity
C speed remains constant throughout
D speed decreases to zero and remains zero

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37

An object falls through a liquid. The resistive force increases as the speed increases.

Which graph best describes resistive force against speed?

A horizontal line through zero force
B straight line or curve increasing from the origin
C vertical line at one speed
D line decreasing as speed increases

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38

A metal sphere falls through oil. It reaches terminal velocity.

Which statement explains why it no longer accelerates?

A The oil removes the sphere’s weight.
B The sphere has no mass in oil.
C The upward drag equals the downward weight.
D The sphere stops moving.

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39

A small ball falls through oil. Its speed–time graph starts steep, then becomes less steep, then becomes horizontal.

Which statement is correct?

A acceleration increases then becomes constant
B acceleration decreases to zero
C speed decreases to zero
D weight decreases to zero

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40

Two identical balls fall through the same liquid. Ball X is released from rest. Ball Y is thrown downwards at a speed greater than terminal velocity.

What happens to ball Y initially?

A It speeds up because weight is greater than drag.
B It slows down because drag is greater than weight.
C It moves at constant speed because it is already moving.
D It accelerates upwards until it moves upwards.

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For Full Scale Course: Written and Compiled By Sir Hunain Zia (AYLOTI), World Record Holder With 154 Total Personal A Grades, 11 World Records and 7 Distinctions, Educate A Change.

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41

A car moves along a level road at constant speed. The driving force is 1200 N.

What is the total resistive force?

A 0 N
B 600 N
C 1200 N
D greater than 1200 N

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42

A car of mass 900 kg has a driving force of 1800 N and total resistive force of 900 N.

What is the acceleration?

A 1.0 m/s²
B 2.0 m/s²
C 3.0 m/s²
D 2700 m/s²

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43

A car accelerates from rest. The driving force is constant, but air resistance increases with speed.

Which statement is correct?

A acceleration remains constant forever
B acceleration decreases as speed increases
C acceleration increases as speed increases
D acceleration is zero from the start

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44

A vehicle travels with constant driving force. As speed increases, resistive force increases.

When does the vehicle reach maximum speed?

A when driving force becomes zero
B when resistive force becomes zero
C when driving force equals resistive force
D when weight equals driving force

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45

A cyclist moves down a hill at constant speed.

Which statement is correct?

A No forces act on the cyclist.
B The downhill component of weight equals resistive forces.
C Weight equals normal contact force only.
D Air resistance is zero.

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46

A block slides across a rough horizontal surface and eventually stops.

Which energy transfer occurs mainly because of friction?

A kinetic store to thermal store
B thermal store to kinetic store
C gravitational store to chemical store
D elastic store to nuclear store

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47

A student rubs her hands together quickly.

Why do her hands become warmer?

A friction transfers energy to thermal stores
B weight increases between the hands
C air resistance becomes zero
D mass is converted into heat

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48

A box is pulled at constant speed across a rough floor by a force of 60 N. It moves 4.0 m.

How much energy is transferred to thermal stores by friction?

A 15 J
B 60 J
C 120 J
D 240 J

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49

A 2.0 kg block is sliding on a rough horizontal surface at 6.0 m/s. Friction brings it to rest.

How much energy is transferred to thermal stores?

A 12 J
B 36 J
C 72 J
D 144 J

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50

A car brakes to rest. The braking force is doubled while the initial speed is unchanged.

What happens to the braking distance?

A it becomes one quarter
B it becomes half
C it doubles
D it becomes four times larger

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For Full Scale Course: Written and Compiled By Sir Hunain Zia (AYLOTI), World Record Holder With 154 Total Personal A Grades, 11 World Records and 7 Distinctions, Educate A Change.

Chapter 5 Answer Key

Q	Ans	Q	Ans	Q	Ans	Q	Ans	Q	Ans
1	A	11	B	21	C	31	B	41	C
2	C	12	C	22	C	32	B	42	A
3	C	13	B	23	A	33	C	43	B
4	C	14	C	24	C	34	B	44	C
5	C	15	C	25	C	35	A	45	B
6	A	16	C	26	C	36	B	46	A
7	B	17	C	27	B	37	B	47	A
8	C	18	A	28	B	38	C	48	D
9	C	19	C	29	B	39	B	49	B
10	B	20	B	30	B	40	B	50	B

For Full Scale Course: Written and Compiled By Sir Hunain Zia (AYLOTI), World Record Holder With 154 Total Personal A Grades, 11 World Records and 7 Distinctions, Educate A Change.

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Detailed Explanations

1. A

• Thinking distance = speed × reaction time

• Thinking distance = 20 × 0.75

• Thinking distance = 15 m

• B and C are rough addition traps.

• D divides 20 by 0.75, which is not thinking distance.

2. C

• Thinking distance = speed × reaction time

• speed = thinking distance / reaction time

• speed = 18 / 0.60

• speed = 30 m/s

• B multiplies 18 × 0.60 instead of dividing.

3. C

• Stopping distance = thinking distance + braking distance

• Stopping distance = 18 + 54

• Stopping distance = 72 m

4. C

• Thinking distance = 30 × 0.80 = 24 m

• Braking distance = area under speed–time graph during braking

• Average speed during braking = (30 + 0) / 2 = 15 m/s

• Braking distance = 15 × 5.0 = 75 m

• Stopping distance = 24 + 75 = 99 m

5. C

• Thinking distance = 18 × 0.50 = 9.0 m

• During braking:

  • average speed = (18 + 0) / 2 = 9.0 m/s

  • braking distance = 9.0 × 3.0 = 27 m

• Stopping distance = 9.0 + 27 = 36 m

6. A

• Thinking distance = speed × reaction time

• reaction time = thinking distance / speed

• reaction time = 20 / 25

• reaction time = 0.80 s

7. B

• Use: v² = 2as

• a = v² / 2s

• a = 28² / (2 × 56)

• a = 784 / 112

• a = 7.0 m/s²

8. C

• Braking distance = v² / 2a

• Braking distance = 20² / (2 × 5.0)

• Braking distance = 400 / 10

• Braking distance = 40 m

9. C

• Thinking distance = 16 × 0.75 = 12 m

• Braking distance = v² / 2a

• Braking distance = 16² / (2 × 4.0)

• Braking distance = 256 / 8

• Braking distance = 32 m

• Stopping distance = 12 + 32 = 44 m

10. B

• Thinking distance = speed × reaction time

• If speed doubles and reaction time is unchanged:

  • thinking distance doubles

• Braking distance depends on v²:

  • if speed doubles, braking distance becomes 2² = 4 times larger

• So thinking distance doubles and braking distance quadruples.

For Full Scale Course: Written and Compiled By Sir Hunain Zia (AYLOTI), World Record Holder With 154 Total Personal A Grades, 11 World Records and 7 Distinctions, Educate A Change.

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11. B

• Thinking distance = vt

• Braking distance = v² / 2a

• Stopping distance = thinking distance + braking distance

• So stopping distance = vt + v²/2a

• A is wrong because v/a gives time, not distance.

• C is dimensionally wrong. Physics does not forgive bad units.

12. C

• Thinking distance = 12 × 0.50 = 6 m

• Stopping distance = thinking distance + braking distance

• Stopping distance = 6 + 18

• Stopping distance = 24 m

13. B

• Convert 36 km/h to m/s:

• 36 / 3.6 = 10 m/s

• Thinking distance = 10 × 0.70

• Thinking distance = 7.0 m

• C is the trap from forgetting the km/h to m/s conversion.

14. C

• Convert 72 km/h to m/s:

• 72 / 3.6 = 20 m/s

• Thinking distance = 20 × 0.90 = 18 m

• Braking distance:

  • average speed = (20 + 0) / 2 = 10 m/s

  • braking distance = 10 × 4.0 = 40 m

• Stopping distance = 18 + 40 = 58 m

15. C

• Convert 90 km/h to m/s:

• 90 / 3.6 = 25 m/s

• Thinking distance = 25 × 0.80 = 20 m

• Stopping distance = 20 + 62.5

• Stopping distance = 82.5 m

16. C

• Alcohol affects the driver’s reaction time.

• Greater reaction time directly increases thinking distance.

• It does not directly change the braking force between tyre and road.

• Icy road and worn tyres affect braking distance.

17. C

• A wet road reduces friction between tyre and road.

• Lower friction means smaller braking force.

• Smaller braking force means longer braking distance.

• Tiredness, alcohol and drugs mainly increase reaction time, so they affect thinking distance.

18. A

• Tiredness increases reaction time.

• Thinking distance = speed × reaction time.

• So thinking distance increases.

• Road and tyre conditions are unchanged, so braking distance is not directly affected.

19. C

• A heavier car has more kinetic energy at the same speed:

• Ek = 1/2 mv²

• More kinetic energy must be removed during braking.

• So braking distance may increase.

• Thinking distance is unchanged because reaction time and speed are unchanged.

20. B

• Worn tyres reduce friction with the road.

• Reduced friction reduces braking force.

• Therefore braking distance increases.

• Thinking distance is about reaction time, not tyre grip.

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21. C

• First 0.8 s: car moves at constant speed during reaction time.

• Thinking distance = 20 × 0.8 = 16 m

• Braking distance = triangle area

• Braking distance = 1/2 × 3.2 × 20 = 32 m

• Stopping distance = 16 + 32 = 48 m

22. C

• Horizontal section distance:

  • 15 × 0.60 = 9 m

• Braking section distance:

  • 1/2 × 2.4 × 15 = 18 m

• Stopping distance = 9 + 18 = 27 m

23. A

• Thinking distance = speed × reaction time

• If reaction time doubles, thinking distance doubles.

• Braking distance depends on speed and braking deceleration.

• Since speed and braking deceleration are unchanged, braking distance is unchanged.

24. C

• New thinking distance = 15 × 1.2 = 18 m

• Braking distance remains 22.5 m

• New stopping distance = 18 + 22.5

• New stopping distance = 40.5 m

25. C

• Braking distance is proportional to speed² if deceleration is unchanged.

• Speed changes from 10 m/s to 30 m/s.

• Speed becomes 3 times larger.

• Braking distance becomes 3² = 9 times larger.

• New braking distance = 8.0 × 9 = 72 m

For Full Scale Course: Written and Compiled By Sir Hunain Zia (AYLOTI), World Record Holder With 154 Total Personal A Grades, 11 World Records and 7 Distinctions, Educate A Change.

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26. C

• Braking force is due to friction.

• If frictional force is halved, deceleration is halved.

• Braking distance = v² / 2a

• If a is halved, braking distance doubles.

• New braking distance = 2d

27. B

• Initial kinetic energy = 1/2 mv²

• Work done by braking force = Fd

• Fd = 1/2 mv²

• d = mv² / 2F

• So answer is mv² / 2F

28. B

• Kinetic energy = 1/2 mv²

• Ek = 1/2 × 1000 × 20²

• Ek = 500 × 400

• Ek = 200 000 J

• Work done by braking force = force × distance

• distance = energy / force

• distance = 200 000 / 5000

• distance = 40 m

29. B

• Ek = 1/2 mv²

• Ek = 1/2 × 1200 × 15²

• Ek = 600 × 225

• Ek = 135 000 J

• Braking distance = 135 000 / 4500

• Braking distance = 30 m

30. B

• Braking distance = mv² / 2F

• Lorry has twice the mass, so kinetic energy is twice as large.

• But lorry also has twice the braking force.

• Both effects cancel.

• Therefore braking distance is the same.

31. B

• At the instant the object is released:

  • weight is non-zero because gravity acts immediately

  • air resistance is zero because speed is initially zero

• Air resistance depends on speed.

• No speed, no drag.

32. B

• At first, air resistance is small.

• Weight is greater than air resistance.

• Resultant force is downwards.

• So acceleration is downwards.

33. C

• At terminal velocity:

  • speed is constant

  • acceleration is zero

  • resultant force is zero

• Therefore air resistance equals weight.

34. B

• Object is moving downwards but slowing down.

• This means acceleration/resultant force is upwards.

• Upward force is air resistance.

• So air resistance must be greater than weight.

35. A

• Opening the parachute greatly increases surface area.

• Air resistance increases greatly.

• It becomes greater than weight.

• Resultant force is upwards, so acceleration is upwards.

• The skydiver is still moving down, but slowing down. That’s the killer trap.

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36. B

• Before parachute opens: skydiver is at high terminal velocity.

• Parachute opens: air resistance increases greatly.

• Skydiver slows down.

• Eventually air resistance equals weight again.

• New terminal velocity is lower.

• So speed decreases, then becomes constant at a lower terminal velocity.

37. B

• Resistive force increases as speed increases.

• At zero speed, resistive force is zero.

• So the graph should increase from the origin.

• It may be a straight line or curve depending on the situation.

38. C

• Terminal velocity means constant speed.

• Constant speed means zero resultant force.

• Downward weight equals upward drag.

• The sphere still has weight and is still moving.

39. B

• Gradient of speed–time graph = acceleration.

• At first, graph is steep, so acceleration is large.

• As drag increases, resultant force decreases.

• So acceleration decreases.

• When graph becomes horizontal, acceleration = 0.

40. B

• Ball Y is thrown downwards faster than terminal velocity.

• At a speed greater than terminal velocity, drag is greater than weight.

• Resultant force is upwards.

• Since the ball is moving downwards, this upward resultant force slows it down.

• It does not keep accelerating until it moves upwards; it slows towards terminal velocity.

For Full Scale Course: Written and Compiled By Sir Hunain Zia (AYLOTI), World Record Holder With 154 Total Personal A Grades, 11 World Records and 7 Distinctions, Educate A Change.

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41. C

• Constant speed means zero acceleration.

• Zero acceleration means zero resultant force.

• Driving force = resistive force.

• Resistive force = 1200 N

42. A

• Driving force = 1800 N

• Resistive force = 900 N

• Resultant force = 1800 − 900 = 900 N

• a = F / m

• a = 900 / 900

• a = 1.0 m/s²

43. B

• Driving force is constant.

• As speed increases, air resistance increases.

• Resultant force = driving force − resistive force.

• Resultant force decreases.

• Since F = ma, acceleration decreases.

44. C

• Maximum speed occurs when acceleration becomes zero.

• Acceleration becomes zero when resultant force is zero.

• Resultant force is zero when driving force equals resistive force.

• So maximum speed occurs when driving force = resistive force.

45. B

• Cyclist moves down the hill at constant speed.

• Constant speed means zero resultant force.

• The downhill component of weight is balanced by resistive forces.

• Forces still act; they are just balanced.

46. A

• Friction opposes motion.

• It transfers energy from the kinetic store to thermal stores.

• The block and surface become warmer.

47. A

• Rubbing hands creates friction.

• Work is done against friction.

• Energy is transferred to thermal stores.

• That is why the hands become warmer.

48. D

• Box moves at constant speed, so pulling force = frictional force.

• Friction = 60 N

• Energy transferred = work done against friction

• Work done = force × distance

• Work done = 60 × 4.0

• Work done = 240 J

49. B

• Energy transferred to thermal stores = initial kinetic energy

• Ek = 1/2 mv²

• Ek = 1/2 × 2.0 × 6.0²

• Ek = 1.0 × 36

• Ek = 36 J

50. B

• Braking distance = mv² / 2F

• Initial speed is unchanged, so kinetic energy is unchanged.

• If braking force is doubled, braking distance is halved.

• New braking distance = half

For Full Scale Course: Written and Compiled By Sir Hunain Zia (AYLOTI), World Record Holder With 154 Total Personal A Grades, 11 World Records and 7 Distinctions, Educate A Change.

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Common Traps From This Chapter

Trap	Correct Rule
Thinking distance	speed × reaction time
Braking distance	distance travelled after brakes are applied
Stopping distance	thinking distance + braking distance
Speed doubles	thinking distance doubles
Speed doubles	braking distance quadruples
Reaction time doubles	thinking distance doubles only
Wet/icy road	increases braking distance
Tiredness/alcohol/drugs	increase thinking distance
Worn tyres	increase braking distance
Heavy load	may increase braking distance
Horizontal part of stopping graph	thinking distance
Sloping part of stopping graph	braking distance
Area under speed–time graph	distance
Terminal velocity	weight = drag
Just released falling object	weight non-zero, drag zero
Falling and speeding up	weight > drag
Falling and slowing down	drag > weight
Parachute opens	drag suddenly increases
New parachute terminal velocity	lower than before
Constant speed vehicle	driving force = resistive force
Braking force doubled	braking distance halves
Friction	transfers kinetic energy to thermal stores