IGCSE Physics Motion Questions and Worked Solutions (1.2) | 0625

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IGCSE Physics 0625 · Topic 1.2

IGCSE Physics Motion Questions and Worked Solutions

50 topic-wise practice questions on Motion (Unit 1: Motion, Forces and Energy), written to the Cambridge IGCSE Physics 0625 syllabus for 2026–2028, each with a step-by-step worked solution.

50 questions23 Core17 Extended10 Challenge127 marksTextbook §1.2 · pp. 12–24
Questions tagged Core are for all candidates (Papers 1 & 3). Questions tagged Extended are Supplement content (Papers 2 & 4). Questions 41–50 are harder Challenge questions. Try each question first, then tap Show solution to check your working.

Topic guide: Motion

This topic describes how objects move. Speed is distance travelled per unit time, velocity adds direction, and acceleration is the rate of change of velocity. Distance–time and speed–time graphs show motion visually: gradients and areas carry meaning.

Key formulas

v = s ÷ tspeed v (m/s), distance s (m), time t (s)Core
average speed = total distance ÷ total timeinclude stationary time in total timeCore
a = Δv ÷ Δtacceleration a (m/s²), change in velocity Δv (m/s), time Δt (s)Extended
distance = area under speed–time graphtriangles, rectangles, trapeziaCore
g ≈ 9.8 m/s²acceleration of free fall near the Earth's surfaceCore

Prerequisite knowledge

  • Rearranging simple equations
  • Area of a triangle and a rectangle
  • Plotting and reading graphs; finding a gradient

Exam tips & common mistakes

  • Convert km/h to m/s by dividing by 3.6
  • Do not confuse distance–time and speed–time graphs — read the axis labels first
  • Average speed is NOT the average of two speeds when the times are different

Key facts from the syllabus (1.2)

  • Distance–time graph: gradient = speed; horizontal line = at rest; straight line = constant speed; curve = changing speed
  • Speed–time graph: gradient = acceleration; horizontal line = constant speed; area = distance travelled
  • A negative acceleration (deceleration) means the object is slowing down
  • Objects falling freely near the Earth with no air resistance accelerate at about 9.8 m/s²
  • With air resistance, a falling object reaches terminal velocity when air resistance equals weight
  • Extended the gradient of a curved speed–time graph at a point is found using a tangent

Textbook: §1.2 Motion, pp. 12–24

1
Core[1 mark]

Define speed.

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Method: Speed is distance travelled per unit time.

Speed = distance ÷ time (distance travelled per unit time)
2
Core[1 mark]

Define velocity.

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Method: Velocity is speed in a given direction.

Speed in a given direction
3
Core[2 marks]

A cyclist travels 1800 m in 240 s. Calculate the average speed.

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Method: v = s ÷ t.

v = 1800 ÷ 240 = 7.50 m/s
4
Core[2 marks]

A car travels at 25 m/s for 40 s. Calculate the distance travelled.

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Method: s = v × t.

s = 25 × 40 = 1000 m
5
Core[3 marks]

A bus travels 12 km in 20 minutes and then 18 km in 25 minutes. Calculate its average speed in km/h.

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Method: Average speed = total distance ÷ total time. Total time = 45 min = 0.75 h.

Average speed = 30 ÷ 0.75 = 40.0 km/h
6
Core[1 mark]

State what a horizontal line on a distance–time graph shows.

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Method: Distance is not changing.

The object is at rest (stationary)
7
Core[1 mark]

State what a straight line with a positive gradient on a speed–time graph shows.

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Method: Speed increases at a steady rate.

Constant acceleration
8
Core[2 marks]

A distance–time graph is a straight line from (0 s, 0 m) to (8.0 s, 56 m). Calculate the speed.

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Method: Speed = gradient of the distance–time graph.

v = 56 ÷ 8.0 = 7.00 m/s
9
Core[3 marks]

A car accelerates uniformly from rest to 20 m/s in 10 s. Use the area under the speed–time graph to calculate the distance travelled.

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Method: Area of triangle = ½ × base × height.

s = ½ × 10 × 20 = 100 m
10
Core[3 marks]

A train moves at 15 m/s for 60 s, then decelerates uniformly to rest in 20 s. Calculate the total distance travelled.

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Method: Rectangle + triangle under the speed–time graph.

s = (15 × 60) + (½ × 20 × 15) = 900 + 150 = 1050 m
11
Core[1 mark]

State the approximate value of the acceleration of free fall near the Earth's surface.

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Method: g is approximately constant near the surface.

9.8 m/s²
12
Core[2 marks]

A car travels 150 km in 2.5 h. Calculate its average speed in km/h.

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Method: v = d ÷ t.

v = 150 ÷ 2.5 = 60.0 km/h
13
Core[2 marks]

Convert 72 km/h into m/s.

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Method: × 1000 ÷ 3600.

72 × 1000 ÷ 3600 = 20.0 m/s
14
Core[2 marks]

A sound travels 1700 m in 5.0 s. Calculate its speed.

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Method: v = d ÷ t.

v = 1700 ÷ 5.0 = 340 m/s
15
Core[2 marks]

A snail moves at 1.2 mm/s. Calculate the time it takes to move 30 cm.

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Method: t = d ÷ v; 30 cm = 300 mm.

t = 300 ÷ 1.2 = 250 s
16
Core[1 mark]

State what a straight line with a positive gradient on a distance–time graph shows.

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Method: Steady speed.

Constant speed
17
Core[1 mark]

State what a horizontal line on a speed–time graph shows.

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Method: Speed unchanged.

Constant speed
18
Core[1 mark]

State what a curve getting steeper on a distance–time graph shows.

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Method: Speed increasing.

The object is accelerating
19
Core[3 marks]

A distance–time graph shows a cyclist travelling 600 m in 120 s, stopping for 60 s, then travelling a further 400 m in 80 s. Calculate the average speed for the whole journey.

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Method: Total distance ÷ total time.

v = 1000 ÷ 260 = 3.85 m/s
20
Core[3 marks]

A lift accelerates uniformly from rest to 3.0 m/s in 2.0 s, moves at 3.0 m/s for 8.0 s, then decelerates uniformly to rest in 2.0 s. Calculate the distance travelled.

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Method: Area under graph (trapezium).

s = 3.0 + 24 + 3.0 = 30.0 m
21
Core[2 marks]

State how the distance travelled is found from a speed–time graph.

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Method: Area.

It is the area under the graph
22
Core[2 marks]

State how the speed is found from a distance–time graph.

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Method: Gradient.

It is the gradient of the graph
23
Core[2 marks]

A ball rolls 4.5 m in 1.5 s at constant speed. Calculate its speed.

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Method: v = d ÷ t.

v = 4.5 ÷ 1.5 = 3.00 m/s
24
Extended[1 mark]

Define acceleration.

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Method: Acceleration is change in velocity per unit time.

a = Δv ÷ Δt (change in velocity per unit time)
25
Extended[2 marks]

A sprinter accelerates from rest to 9.0 m/s in 2.5 s. Calculate the acceleration.

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Method: a = Δv ÷ Δt.

a = 9.0 ÷ 2.5 = 3.60 m/s²
26
Extended[3 marks]

A car slows from 28 m/s to 12 m/s in 4.0 s. Calculate the acceleration and state what the sign means.

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Method: a = (v − u) ÷ t.

a = (12 − 28) ÷ 4.0 = -4.00 m/s²; the negative sign shows a deceleration of 4.00 m/s²
27
Extended[2 marks]

A speed–time graph is a curve whose gradient gets smaller with time. Describe the motion.

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Method: Gradient of a speed–time graph = acceleration.

The object is speeding up with decreasing acceleration (changing acceleration)
28
Extended[3 marks]

A tangent drawn to a speed–time curve passes through (2.0 s, 3.0 m/s) and (6.0 s, 11.0 m/s). Calculate the acceleration at that point.

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Method: Acceleration = gradient of the tangent.

a = (11.0 − 3.0) ÷ (6.0 − 2.0) = 2.00 m/s²
29
Extended[3 marks]

A ball falls from rest with no air resistance. Calculate its speed after 2.5 s (g = 9.8 m/s²).

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Method: v = u + at with u = 0 (from a = Δv/Δt).

v = 9.8 × 2.5 = 24.5 m/s
30
Extended[3 marks]

Describe the motion of a skydiver from leaving the plane until reaching terminal velocity.

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Method: At first weight > air resistance so she accelerates at about 9.8 m/s². As speed increases, air resistance increases, so the resultant force and the acceleration decrease. When air resistance = weight, resultant force is zero.

Accelerates with decreasing acceleration, then falls at constant (terminal) velocity when air resistance equals weight
31
Extended[2 marks]

Explain why an object dropped on the Moon falls with constant acceleration all the way down.

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Method: The Moon has no atmosphere, so there is no air resistance; the only force is weight.

No air resistance, so the only force is constant weight, giving constant acceleration
32
Extended[3 marks]

A cyclist decelerates uniformly from 12 m/s to rest in 6.0 s. Calculate the deceleration and the distance travelled.

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Method: a = Δv/Δt; distance = area of triangle.

Deceleration = 12 ÷ 6.0 = 2.00 m/s²; s = ½ × 6.0 × 12 = 36 m
33
Extended[2 marks]

A car accelerates at 2.5 m/s² for 6.0 s from 10 m/s. Calculate its final speed.

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Method: Δv = a × t.

v = 10 + 2.5 × 6.0 = 25.0 m/s
34
Extended[2 marks]

A train decelerates at 0.50 m/s² from 30 m/s. Calculate the time it takes to stop.

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Method: t = Δv ÷ a.

t = 30 ÷ 0.50 = 60.0 s
35
Extended[2 marks]

State how acceleration is found from a speed–time graph.

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Method: Gradient.

It is the gradient of the speed–time graph
36
Extended[3 marks]

A speed–time graph shows a straight line from (0, 4.0 m/s) to (5.0 s, 14 m/s). Calculate the acceleration and the distance travelled.

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Method: Gradient; trapezium area.

a = 10 ÷ 5.0 = 2.00 m/s²; s = ½(4.0 + 14) × 5.0 = 45.0 m
37
Extended[2 marks]

A negative gradient on a speed–time graph is described as deceleration. Calculate the deceleration if speed falls from 18 m/s to 6.0 m/s in 3.0 s.

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Method: Gradient.

Deceleration = 12 ÷ 3.0 = 4.00 m/s²
38
Extended[3 marks]

A parachutist opens the parachute while falling at terminal velocity. Describe and explain the motion until a new terminal velocity is reached.

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Method: Air resistance increases suddenly.

Air resistance becomes greater than weight, so there is an upward resultant force and the parachutist decelerates; as speed falls air resistance decreases until it equals weight, giving a new, lower terminal velocity
39
Extended[2 marks]

An object falls freely from rest for 1.5 s (g = 9.8 m/s²). Calculate its speed.

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Method: v = gt.

v = 9.8 × 1.5 = 14.7 m/s
40
Extended[3 marks]

A stone falls from rest and hits the ground at 19.6 m/s (no air resistance). Calculate the time of fall and the height using the speed–time graph.

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Method: t = v/g; area of triangle.

t = 19.6 ÷ 9.8 = 2.00 s; h = ½ × 2.0 × 19.6 = 19.6 m

Challenge questions

Harder, multi-step questions at the top end of the exam. Attempt these once you are confident with the questions above.

41
ChallengeCore[5 marks]

A car accelerates uniformly from 5.0 m/s to 25 m/s in 8.0 s, travels at 25 m/s for 12 s, then brakes uniformly to rest in 5.0 s. Calculate (a) the total distance, (b) the average speed.

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Method: Area: trapezium + rectangle + triangle.

(a) ½(5+25)×8 + 25×12 + ½×25×5 = 120 + 300 + 62.5 = 482.5 m (b) 19.3 m/s
42
ChallengeCore[4 marks]

A cyclist travels 6.0 km at 20 km/h and then 6.0 km at 12 km/h. Calculate the average speed for the whole journey. Explain why it is not 16 km/h.

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Method: Find each time.

t = 0.30 h + 0.50 h = 0.80 h; v = 12 ÷ 0.80 = 15.0 km/h; more time is spent at the slower speed
43
ChallengeCore[4 marks]

Car A leaves a town at 60 km/h. Car B leaves the same town 30 minutes later along the same road at 90 km/h. Calculate how long after B leaves it catches A.

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Method: A has a 30 km head start; closing speed 30 km/h.

1.0 h after B leaves
44
ChallengeCore[4 marks]

A distance–time graph is a straight line from (0, 0) to (10 s, 40 m), horizontal to 15 s, then a straight line to (25 s, 100 m). Calculate the speed in each section and the average speed.

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Method: Gradients.

4.0 m/s, 0, 6.00 m/s; average = 100 ÷ 25 = 4.00 m/s
45
ChallengeCore[4 marks]

A train must travel 4.5 km between stations. It accelerates uniformly from rest to 30 m/s in 50 s and decelerates uniformly to rest in 40 s. Calculate the time spent at constant speed.

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Method: Distance while changing speed = ½×30×50 + ½×30×40.

Accelerating and braking: 750 + 600 = 1350 m; remaining 3150 m at 30 m/s takes 105 s
46
ChallengeExtended[5 marks]

A ball is dropped from rest from a height of 20 m (g = 9.8 m/s², no air resistance). Using a speed–time graph, calculate (a) the time to hit the ground, (b) the speed at impact.

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Method: h = ½ × t × gt = ½gt².

(a) t = √(2 × 20 ÷ 9.8) = 2.02 s (b) v = 9.8 × 2.02 = 19.8 m/s
47
ChallengeExtended[4 marks]

A car moving at 30 m/s brakes with a deceleration of 6.0 m/s². The driver's reaction time is 0.70 s. Calculate the total stopping distance.

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Method: Thinking distance + braking distance (area of triangle).

Thinking = 21 m; braking time 5.0 s, distance = 75 m; total = 96.0 m
48
ChallengeExtended[4 marks]

A ball is thrown vertically upwards at 14.7 m/s (g = 9.8 m/s²). Calculate (a) the time to reach the top, (b) the maximum height, (c) the total time before it returns to the hand.

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Method: Deceleration at 9.8 m/s².

(a) 1.50 s (b) ½ × 1.5 × 14.7 = 11.0 m (c) 3.00 s
49
ChallengeExtended[4 marks]

A speed–time graph for a sprinter is a curve. Tangents drawn at 1.0 s and 4.0 s have gradients 6.0 m/s² and 1.5 m/s². Describe the sprinter's motion and explain how the resultant force changes.

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Method: Decreasing acceleration.

The sprinter speeds up with decreasing acceleration; as acceleration falls, the resultant force falls (air resistance grows while driving force is limited)
50
ChallengeExtended[5 marks]

A skydiver reaches a terminal velocity of 55 m/s, opens the parachute, and 4.0 s later has a new terminal velocity of 6.0 m/s. (a) Calculate the average deceleration. (b) Explain, in terms of forces, why the speed falls and then becomes constant.

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Method: Δv/Δt.

(a) 12.2 m/s² (b) Air resistance suddenly exceeds weight, giving an upward resultant force; as speed falls, air resistance falls until it equals weight, so resultant force and acceleration become zero
All IGCSE Physics topics← 1.1 Physical Quantities and Measurement Techniques1.3 Mass and Weight →

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