Why is the area under a velocity graph the displacement?

The area is speed times time, and that product is exactly the displacement.

Why can a downward sloping graph still mean forward motion?

As long as the velocity value stays positive the direction has not changed.

When are average velocity and average speed the same?

When the motion never reverses direction.

S4 Physics: Motion Graphs Practice Paper 1

2026-11-01

Tutor Prep Guide · Secondary Physics

Paper one opens with a 5 s acceleration and a 12 m/s run, then reads flat distance lines and the area under a velocity graph, moves on to an 8 to 20 m/s step, a three-stage graph and an average speed with a rest, then free fall for 3 s, ending with the backward-slope mix-up.

Quick answer　Check: 4 m/s^2, 60 m, displacement, 2 m/s^2 with 84 m, 160 m, 5 m/s with 2.9 m/s, 30 m/s with 45 m, 216 m with 15.4 m/s, and slowing down only.

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 speeds up from rest to 20 m/s in 5 s. Find its average acceleration.
FoundationWhat does a horizontal section on a distance-time graph show?
FoundationA car travels at a steady 12 m/s for 5 s. Find the distance covered.
FoundationOn a velocity-time graph, what does the area under the line represent?
IntermediateA car accelerates uniformly from 8 m/s to 20 m/s in 6 s. Find the acceleration and the displacement.
IntermediateA car goes from 0 to 16 m/s in 4 s, holds 16 m/s from 4 s to 10 s, then stops in 4 s. Find the total displacement.
IntermediateA walker covers 20 m in 4 s then rests for 3 s. Find the walking speed and the average speed for the whole trip.
ChallengeA ball falls from rest for 3 s with g = 10 m/s^2. Find its speed and the height fallen.
ChallengeA car reaches 24 m/s in 6 s, holds it for 4 s, then stops in 4 s. Find the displacement and the average speed.
Error redoA student sees a downward slope on a velocity-time graph and answers that the car moves backwards. Why is that wrong?

03　Answers and one-line notes

#	Answer	One-line note
1	4 m/s^2 (20 / 5)	Acceleration is the change in velocity divided by the time taken.
2	The object is at rest, so its speed is zero	A flat line means the distance stops growing, so the slope is zero.
3	60 m (12 x 5)	At steady speed the distance is speed times time.
4	The displacement travelled	The unit of the area is m/s times s, which is metres.
5	Acceleration 2 m/s^2 and displacement 84 m	Use the mean speed (8 + 20) / 2 times the time for the displacement.
6	160 m (32 + 96 + 32)	Add the triangle and rectangle areas section by section.
7	Walking 5 m/s and about 2.9 m/s overall (20 / 7)	Average speed uses total displacement over total time, rest included.
8	30 m/s and 45 m	Free fall adds 10 m/s each second, and the height is 1/2 g t^2.
9	216 m and about 15.4 m/s (216 / 14)	Sum the areas first, then divide by the total time.
10	A downward slope only means slowing down; the velocity stays positive so the car still moves forward	Read the sign of the velocity, not the sign of the slope.

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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