Speed, velocity, acceleration and motion graphs – GCSE Physics exam questions with mark scheme

1Grade 4–5Higher4 marks

The velocity–time graph shows a car's journey between two sets of traffic lights.

(a) Calculate the acceleration of the car during the first 4 seconds. [2]

(b) Determine the total distance travelled by the car. [2]

(a) a = 12 ÷ 4 (change in velocity ÷ time; allow gradient of the first section)(1)
(a) 3.0 m/s²(1)
(b) area under the graph used, e.g. ½ × 4 × 12 + 6 × 12 + ½ × 3 × 12 (allow trapezium: ½ × (13 + 6) × 12)(1)
(b) 114 m(1)
Answer: (a) 3.0 m/s² (b) 114 m
Examiner tip: Distance is the AREA under a velocity–time graph, not the gradient.

2Grade 6–75 marks

A train travelling at 30 m/s brakes with a constant deceleration. It stops in a distance of 450 m.

(a) Calculate the deceleration of the train. Use the Physics Equations Sheet. [3]

(b) The mass of the train is $4.0 \times 10^{5}$ kg. Calculate the braking force on the train. [2]

(a) $0^2 - 30^2 = 2 \times a \times 450$(1)
(a) $a = \dfrac{-900}{900}$ (rearrangement)(1)
(a) deceleration = 1.0 m/s² (allow −1.0 m/s² as acceleration)(1)
(b) $F = 4.0 \times 10^{5} \times 1.0$ (allow ecf from (a))(1)
(b) $4.0 \times 10^{5}$ N(1)
Answer: (a) 1.0 m/s² (b) $4.0 \times 10^5$ N
Examiner tip: Use v² − u² = 2as with v = 0; a negative answer just means the train is slowing down.

3Grade 1–34 marks

Distance is a scalar quantity. Displacement is a vector quantity.

(a) State what is meant by a vector quantity. [1]

(b) An athlete runs exactly one lap of a 400 m running track. She finishes at the same point that she started from.

Give the distance she ran and her displacement. [2]

☐ 0.15 m/s ☐ 1.5 m/s ☐ 15 m/s ☐ 150 m/s

(a) (a quantity that has) both magnitude and direction (allow size for magnitude; both needed)(1)
(b) distance = 400 m(1)
(b) displacement = 0 (m) (allow zero / none; do not accept 400 m)(1)
(c) 1.5 m/s (only one box ticked)(1)
Answer: (a) magnitude and direction (b) distance 400 m, displacement 0 m (c) 1.5 m/s
Examiner tip: A closed lap has zero displacement because the athlete ends where she started; the distance is still the full 400 m.

4Grade 1–34 marks

A student sees a flash of lightning. She hears the thunder 4.5 s later.

Speed of sound in air = 330 m/s

(a) Write down the equation that links distance travelled, speed and time. [1]

(b) Calculate the distance between the student and the lightning. [2]

(c) The student assumes that she sees the flash at the same moment as the lightning strikes. Explain why this is a reasonable assumption. [1]

(a) distance travelled = speed × time (allow $s = vt$ or any correct rearrangement)(1)
(b) distance = 330 × 4.5(1)
(b) 1485 m (allow 1500 m; allow 1.485 km if unit given)(1)
(c) light travels much faster than sound (so the light takes a negligible time to reach her) (allow light travels (almost) instantly; ignore 'light is faster' without comparison to sound or the time being negligible)(1)
Answer: (a) distance = speed × time (b) 1485 m (c) light is far faster than sound, so its travel time is negligible
Examiner tip: Use the time the SOUND takes; the light's journey time is so tiny it can be ignored.

5Grade 6–7Higher4 marks

A car pulls away from traffic lights. The graph shows how the distance travelled by the car changes during the first 8.0 s.

(a) How does the graph show that the speed of the car is increasing? [1]

(b) Determine the speed of the car at a time of 4.0 s. Show clearly on the graph how you work out your answer. [3]

(a) the gradient (of the line) increases (allow the line gets steeper; ignore 'the line goes up')(1)
(b) tangent drawn to the curve at t = 4.0 s(1)
(b) gradient of the tangent calculated, e.g. (36 − 0) ÷ (8.0 − 2.0) (allow any correct Δdistance ÷ Δtime from their tangent)(1)
(b) 6.0 m/s (allow 5.5 – 6.5 m/s)(1)
Answer: (a) gradient increases (b) 6.0 m/s (allow 5.5–6.5)
Examiner tip: Draw a long, straight tangent touching the curve only at 4.0 s and use a big triangle – a short tangent gives a gradient outside the allowed range.

6Grade 4–55 marks

A stone is dropped from rest from a bridge. It falls 18 m into the river below. Air resistance can be ignored.

acceleration due to gravity = 9.8 m/s²

(a) Calculate the speed of the stone just before it hits the water. Use the Physics Equations Sheet. [3]

(b) Calculate the time taken for the stone to fall to the water. [2]

(a) $v^2 - 0^2 = 2 \times 9.8 \times 18$ (substitution)(1)
(a) $v^2 = 352.8$ (allow $v = \sqrt{352.8}$)(1)
(a) $v$ = 19 m/s (allow 18.8 m/s or 18.78 m/s)(1)
(b) time = 18.8 ÷ 9.8 (allow their (a) ÷ 9.8; rearrangement of $a = \Delta v / t$)(1)
(b) 1.9 s (allow 1.92 s; allow 1.94 s from 19 m/s; allow ecf from (a))(1)
Answer: (a) 19 m/s (18.8 m/s) (b) 1.9 s
Examiner tip: Remember the square root: a final answer of 352.8 m/s (v² not v) scores only the first two marks.

7Grade 6–7Higher6 marks

A child sits on the edge of a playground roundabout. The roundabout turns at a steady rate so the child moves in a circle at a constant speed.

(a) Explain why the child is accelerating, even though her speed is constant. [3]

(b) The child is 1.8 m from the centre of the roundabout. The roundabout makes one complete turn every 4.5 s.

Calculate the speed of the child. [3]

(a) velocity is a vector / velocity has (magnitude and) direction(1)
(a) the direction of the child's motion is (continuously) changing(1)
(a) so her velocity is changing and acceleration is a change in velocity (with time) (allow so there is a change in velocity)(1)
(b) distance in one turn = 2 × π × 1.8 (= 11.3 m) (allow π × 3.6)(1)
(b) speed = 11.3 ÷ 4.5(1)
(b) 2.5 m/s (allow 2.51 m/s; 0.40 m/s from 1.8 ÷ 4.5 scores 0)(1)
Answer: (a) direction changes so velocity changes, so she accelerates (b) 2.5 m/s
Examiner tip: Say the word 'direction': constant speed with changing direction means changing velocity, which is acceleration.

8Grade 6–77 marks

A student investigates how the mass of a trolley affects its acceleration when the force pulling it is kept constant.

The trolley is pulled along a horizontal bench by a string that passes over a pulley to a falling mass hanger. Two light gates, connected to a data logger, measure the velocity of the trolley at two points and the time taken to travel between them.

(a) Give one variable the student must keep the same. [1]

(b) In one trial the velocity at the first light gate was 0.24 m/s and the velocity at the second light gate was 0.84 m/s. The time taken between the light gates was 0.40 s.

Calculate the acceleration of the trolley. [2]

The table shows the student's results.

Mass of trolley in kg	Trial 1 acceleration in m/s²	Trial 2 acceleration in m/s²	Trial 3 acceleration in m/s²	Mean acceleration in m/s²
0.40	3.02	2.96	3.05	3.01
0.60	2.04	1.97	1.99	2.00
0.80	1.52	1.91	1.48	X
1.00	1.18	1.23	1.19	1.20
1.20	0.98	1.01	1.02	1.00

(d) The student concludes that the acceleration of the trolley is inversely proportional to its mass.

Use data from the table to show that the student is correct. [2]

(a) the (pulling) force / the weight of the mass hanger (and masses) (allow the falling mass; allow the slope/angle of the bench; ignore 'same trolley', 'same equipment')(1)
(b) acceleration = (0.84 − 0.24) ÷ 0.40(1)
(b) 1.5 m/s²(1)
(c) 1.91 (m/s²) identified as anomalous(1)
(c) X = (1.52 + 1.48) ÷ 2 = 1.50 m/s² (allow 1.5; do not accept 1.64, which includes the anomaly)(1)
(d) suitable data quoted, e.g. mass doubles from 0.40 kg to 0.80 kg and mean acceleration goes from 3.01 to 1.50 m/s² (allow 0.60 → 1.20 kg: 2.00 → 1.00 m/s²; allow mass × acceleration calculated for at least two masses)(1)
(d) acceleration halves (approximately) when mass doubles / mass × acceleration is constant (≈ 1.2) (allow 'inversely proportional' only if linked to data)(1)
Answer: (a) pulling force (weight on hanger) (b) 1.5 m/s² (c) 1.91 is anomalous; X = 1.50 m/s² (d) doubling mass halves acceleration (m × a ≈ 1.2 N each time)
Examiner tip: To show inverse proportion you must use numbers: double the mass and show the acceleration halves, or show mass × acceleration stays constant.

9Grade 8–9HigherTriple only6 marks

A skydiver jumps from a plane. She opens her parachute part of the way down.

The graph shows how her velocity changes from the moment she leaves the plane (A) until just before she lands (E).

Explain the motion of the skydiver between A and E.

In your answer you should refer to the forces acting on the skydiver. [6]

at A the only force is her weight (no air resistance yet), so her acceleration is greatest (≈ 9.8 m/s²) – steepest gradient(1)
as her velocity increases, the air resistance (drag) increases(1)
so the resultant force (weight − air resistance) decreases and her acceleration decreases – the gradient gets smaller(1)
between B and C air resistance equals weight, so the resultant force is zero(1)
so she moves at a constant (terminal) velocity(1)
at C the parachute opens, increasing her surface area, so air resistance increases (suddenly)(1)
air resistance is now greater than weight, so there is a resultant force upwards and she decelerates(1)
as her velocity decreases, air resistance decreases until it equals her weight again (at D)(1)
between D and E the forces are balanced again so she falls at a new, lower terminal velocity(1)

Level	Marks	What it looks like
Level 3	5–6	A clear, logically ordered explanation of the motion in every section (A–B, B–C, C–D and D–E), consistently linking each change in velocity to the size and direction of the resultant force and to how air resistance changes with speed and parachute area.
Level 2	3–4	The motion in most sections is explained, with some correct links between the forces and the change in velocity; the reasoning may have gaps or one section may be missing or only described.
Level 1	1–2	Simple, separate statements about the motion (e.g. speeds up, then steady, then slows down) or about the forces, with little or no link made between them.
No relevant content	0	Nothing worthy of credit.

Answer: Accelerates (decreasing acceleration as drag rises) to terminal velocity at B; constant velocity B–C (drag = weight); parachute opens at C so drag > weight and she decelerates; drag falls as she slows until drag = weight at D; lower terminal velocity D–E.
Examiner tip: Link every change in the graph's gradient to the resultant force – describing the shape without explaining the forces stays in Level 1.

10Grade 4–56 marks

The distance–time graph shows part of a cyclist's journey along a straight road.

(a) Describe the motion of the cyclist between Q and R. [1]

(b) Calculate the speed of the cyclist between P and Q. [2]

(d) Calculate the average speed of the cyclist for the whole journey from P to S. [2]

(a) stationary / not moving (allow stopped; allow speed is zero)(1)
(b) speed = 240 ÷ 40 (allow gradient of PQ correctly read from graph)(1)
(b) 6.0 m/s (allow 6 m/s)(1)
(c) the line (gradient) is steeper between R and S (allow gradient greater; ignore 'line goes up more')(1)
(d) average speed = 480 ÷ 100(1)
(d) 4.8 m/s (do not accept (6 + 0 + 8) ÷ 3 = 4.7 m/s, which averages the speeds)(1)
Answer: (a) stationary (b) 6.0 m/s (c) steeper gradient (d) 4.8 m/s
Examiner tip: Average speed is total distance ÷ total time (including the time stopped) – never average the three section speeds.

Still losing marks on this topic?

Book a free introductory session and we'll find exactly which step is costing you.

Book a free intro