Why use the mean speed for braking distance?

The speed falls evenly from 25 to 0, so the mean is 12.5 m/s.

What is the slope of a distance-time graph?

It is distance over time, which is the speed.

Why is the second second not 20 m?

20 m is the total in two seconds, so the first second of 5 m must come off.

S4 Physics: Motion Graphs Practice Paper 3

2026-11-01

Tutor Prep Guide · Secondary Physics

Paper three warms up with a 15 to 35 m/s acceleration, flat lines and slope meaning, then works six seconds from rest, a 25 m/s braking distance and a two-stage average, before a three-stage graph and a four-second fall, ending with the negative-velocity misread.

Quick answer　5 m/s^2, at rest, 216 m, acceleration, 5 m/s^2 with 90 m, 5 m/s^2 with 62.5 m, 136 m at 13.6 m/s, 102 m at 7.85 m/s, 40 m/s with 80 m and 15 m, and the opposite direction.

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 15 m/s to 35 m/s in 4 s. Find the acceleration.
FoundationWhat does a horizontal section on a distance-time graph show?
FoundationA lorry travels at a steady 18 m/s for 12 s. How far does it go?
FoundationWhat does the slope of a velocity-time graph represent?
IntermediateA car starts from rest and reaches 30 m/s in 6 s at constant acceleration. Find the acceleration and the displacement.
IntermediateA car at 25 m/s brakes and stops in 5 s. Find the deceleration and the braking distance.
IntermediateA car runs at 10 m/s for 4 s, then speeds up from 10 m/s to 22 m/s in 6 s. Find the total displacement and the average speed.
ChallengeA velocity-time graph rises from 0 to 12 m/s over 5 s, stays at 12 m/s for 4 s, then falls to 0 over the next 4 s. Find the total displacement and the average speed.
ChallengeA stone falls freely from a tower (g = 10 m/s^2) and lands after 4 s. Find the landing speed, the height and the distance covered during the second second.
Error redoA student sees the velocity-time line drop below the time axis and answers that the object is at rest. What is wrong?

03　Answers and one-line notes

#	Answer	One-line note
1	5 m/s^2 ((35 – 15) / 4)	Acceleration is the change in velocity divided by the time taken.
2	The object is at rest (the distance does not change, so the speed is zero)	A flat line has zero slope, and the slope is the speed.
3	216 m (18 x 12)	At steady speed the distance is speed times time.
4	The acceleration	Slope is change in velocity over change in time, which is the definition of acceleration.
5	5 m/s^2 and 90 m	a is 30 / 6 = 5 and s is 1/2 x 5 x 6^2 = 90 m.
6	5 m/s^2 and 62.5 m	The deceleration is 25 / 5 = 5, and the distance is the mean speed 12.5 times 5 s.
7	136 m, average 13.6 m/s	The first part is 40 m and the second 16 m/s x 6 s = 96 m, so 136 m over 10 s.
8	102 m, average 7.85 m/s	The three areas give 30 + 48 + 24 = 102 m, divided by 13 s.
9	40 m/s, 80 m and 15 m	v is 10 x 4 = 40 and h is 1/2 x 10 x 4^2 = 80; the second second is 20 m minus 5 m.
10	A line below the time axis means negative velocity, so the object moves in the opposite direction, not that it stops	Rest means zero velocity, which sits on the axis; a negative value only shows the opposite direction.

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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