Why is the area under a velocity-time graph the displacement?

Area is velocity times time, which is exactly displacement.

Can two speeds simply be averaged?

Only when the two time intervals are equal; otherwise divide total distance by total time.

How do the two graph types differ?

On a distance-time graph the gradient is speed; on a velocity-time graph it is acceleration and the area is displacement.

S4 Physics: Motion Graphs Practice Paper 4

2026-11-01

Tutor Prep Guide · Secondary Physics

Paper four opens with an acceleration from 8 to 28 m/s and the area of a flat 12 m/s section, then adds a rectangle to a trapezium, sizes a six-second free fall, works a negative acceleration from 45 to 15 m/s, and closes with a three-part graph, a split average speed and the rising-line trap.

Quick answer　4 m/s^2, 72 m, 6 m/s, 4 m/s^2, 96 m at 10.67 m/s, 60 m/s over 180 m, -6 m/s^2 over 150 m, 119 m at 9.92 m/s, 12 m/s, and a rising line means acceleration.

01　Paper details (check answers afterwards)

Item	Detail
Questions	10 questions
Suggested time	25 minutes
Difficulty mix	Foundation 4 · Intermediate 3 · Challenge 2 · Error redo 1
Chapter	Chapter: S4 Physics: Motion Graphs – Tutor Lesson Plan
Level	S4

02　Questions (10 items)

FoundationA car goes from 8 m/s to 28 m/s in 5 s. Find its acceleration.
FoundationA velocity-time graph shows a flat section at 12 m/s lasting 6 s. How far is travelled in that time?
FoundationA straight line on a distance-time graph rises from 0 m to 18 m in 3 s. Find the speed.
FoundationA bus starts from rest and reaches 12 m/s in 3 s. What is its acceleration?
IntermediateA car runs 5 s at a steady 8 m/s, then speeds up from 8 m/s to 20 m/s in 4 s. Find the total displacement and the average speed.
IntermediateAn object falls freely for 6 s (take g as 10 m/s^2). Find its final velocity and the height fallen.
IntermediateA car slows from 45 m/s to 15 m/s in 5 s. Find the acceleration and the distance covered.
ChallengeA velocity-time graph rises from 0 to 14 m/s in 3 s, holds 14 m/s for 5 s, then falls to 0 in 4 s. Find the total displacement and average speed.
ChallengeA 60 m stretch is covered at 15 m/s and a further 60 m at 10 m/s. Find the average speed for the whole trip.
Error redoA student sees a rising straight line on a velocity-time graph and answers that the object moves at steady speed. What is wrong?

03　Answers and one-line notes

#	Answer	One-line note
1	4 m/s^2 ((28 – 8) / 5)	Acceleration is the change in velocity divided by the time taken.
2	72 m (12 x 6)	The rectangle under a velocity-time graph is the displacement.
3	6 m/s (18 / 3)	The gradient of a distance-time graph is the speed.
4	4 m/s^2 (12 / 3)	Starting from rest means the initial velocity is zero.
5	96 m and an average speed of about 10.67 m/s	A 40 m rectangle plus a 56 m trapezium, divided by the 9 s total.
6	60 m/s and 180 m	v is gt, which is 60, and s is half g t squared, which is 180.
7	-6 m/s^2, covering 150 m	Deceleration is negative; use the mean velocity of 30 m/s times the time.
8	119 m and an average speed of about 9.92 m/s	The three areas are 21, 70 and 28, giving 119 over 12 s.
9	12 m/s (120 m in 10 s)	The two legs take 4 s and 6 s, so divide the total distance by the total time.
10	A rising line means the velocity is increasing, so it accelerates; steady speed would be a flat line	First identify which graph it is, then read what the gradient means.

04　What to do after marking

After marking, redo only the wrong items and write down which step went wrong (formula, unit, or misread number).
Repeat the same item two days later; move to the next paper only when both attempts are correct.
Parents only need the answer and note columns; there is no need to explain every item.

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