What topics does this third paper cover?

Constant-speed distance, net force and acceleration, uniform deceleration and braking distance, displacement from uniform acceleration, apparent weight in a lift, vertical throw, and a catch-up problem between two cars.

Which value of g is used throughout?

Every question takes g = 10 m/s^2, and the answers keep accelerations and forces in m/s^2 and N.

S4 Physics Force and Motion Practice Paper (Set 3): 10 Questions with Answers on Braking Distance, Lift Scale Readings and Catch-Up Problems

2026-11-01

Tutor Prep Guide · Secondary Physics

The third force and motion paper drills braking distance, displacement under uniform acceleration, lift scale readings, vertical throws and catch-up problems, each with an answer and a one-line explanation.

Direct answers This paper has 10 questions, from basic constant speed to braking distance and catch-up problems, all with g = 10 m/s^2; suggested time is 25 minutes.

01 Paper details (check answers afterwards)

Item	Detail
Questions	10
Suggested time	25 minutes
Difficulty mix	Foundation 3 · Intermediate 4 · Challenge 2 · Error redo 1
Chapter	S4 Physics: Force and Motion
Level	S4 (DSE)

02 Questions (10 items)

FoundationA car travels at a constant 8 m/s for 5 s. How far does it go?
FoundationA body of mass 4 kg experiences a net force of 24 N. What is its acceleration?
FoundationA car travelling at 12 m/s decelerates uniformly and stops in 4 s. What is its acceleration?
IntermediateA car brakes from 30 m/s and stops after sliding 45 m. Find the deceleration and the braking time.
IntermediateA body starts from rest and accelerates uniformly at 3 m/s^2 for 6 s. How far does it travel?
IntermediateA 2 kg box is pulled horizontally with 20 N while friction is 8 N. What is its acceleration?
IntermediateA 50 kg person stands on a scale inside a lift accelerating downwards at 2 m/s^2. What does the scale read?
ChallengeA body is thrown upwards at 20 m/s (g = 10 m/s^2). What are its velocity and its height above the ground after 3 s?
ChallengeCar A travels at a constant 15 m/s. Car B starts from rest behind A and accelerates at 3 m/s^2 to catch up. How long does B take to catch A, and how far has it travelled when it does?
Error redoA body accelerates uniformly from 10 m/s to 30 m/s in 4 s. How far does it travel? (A common wrong answer is 120 m.)

03 Answers and one-line notes

#	Answer	One-line note
1	s = v x t = 8 x 5 = 40 m.	There is only one formula for constant speed: distance = speed x time.
2	a = F / m = 24 / 4 = 6 metres per second squared.	The larger the net force and the smaller the mass, the larger the acceleration.
3	a = (v – u) / t = (0 – 12) / 4 = -3 metres per second squared.	A final velocity of zero means the car decelerates throughout; the minus sign shows the acceleration opposes the motion.
4	a = -(30 squared) / (2 x 45) = -10 metres per second squared; t = u / \|a\| = 30 / 10 = 3 s.	With no time given, use v squared = u squared + 2as, then recover the time from a.
5	s = half of a t squared = half of 3 x 6 squared = 54 m.	From rest, the displacement depends only on the acceleration and the square of the time.
6	Net force = 20 – 8 = 12 N; a = 12 / 2 = 6 metres per second squared.	Work out the net force, then divide by the mass; never divide the pull alone.
7	N = m(g – a) = 50 x (10 – 2) = 400 N (that is, a 40 kg weight).	The normal force drops when the lift accelerates downwards, so the reading is lighter than the true weight.
8	v = u – gt = 20 – 10 x 3 = -10 m/s (10 m/s downwards); h = ut – half g t squared = 60 – 45 = 15 m.	The negative velocity shows the body has passed the highest point and is falling; the height uses the same displacement formula.
9	half of 3 x t squared = 15t, so 1.5 t squared = 15t and t = 10 s; s = 15 x 10 = 150 m.	A catch-up problem sets the two displacements equal; take the time you solve for and substitute it back into either expression.
10	s = (u + v) / 2 x t = (10 + 30) / 2 x 4 = 80 m.	Accelerating motion needs the average velocity; multiplying by the final velocity overestimates the distance.

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.

Do the catch-up question last without looking at the algebra shown, then compare your line of working with the answer.Tutor Registration →

Questions and answers are written originally by the TutorZone team and are not taken from any school, publisher or examination authority paper. Every value is checked by computer.