What is this second paper for?

It practices uniform deceleration, the time-free formula v^2 = u^2 + 2as, net force and acceleration, apparent weight in a lift, and vertical throw and free-fall timing.

When should I use v^2 = u^2 + 2as instead of v = u + at?

Use v^2 = u^2 + 2as when the time is not given, and v = u + at when the acceleration and time are known; both hold only for uniform acceleration.

S4 Physics Force and Motion Practice Paper (Set 2): 10 Questions with Answers on Deceleration, v^2 = u^2 + 2as and Vertical Throws

2026-11-01

Tutor Prep Guide · Secondary Physics

The second force and motion paper drills uniform deceleration, v^2 = u^2 + 2as, net force and acceleration, vertical throws and free fall, each with an answer and a one-line explanation.

Direct answers This paper has 10 questions, from uniform deceleration to the top of a vertical throw, each solved in one substitution; suggested time is 20 minutes.

01 Paper details (check answers afterwards)

Item	Detail
Questions	10
Suggested time	20 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 accelerates from 8 m/s to 20 m/s in 4 s. What is its acceleration?
FoundationA body of mass 6 kg experiences a net force of 18 N. What is its acceleration?
FoundationA body starts from rest and accelerates uniformly at 4 m/s^2 for 8 s. How far does it travel?
IntermediateA car travelling at 20 m/s decelerates uniformly for 4 s and stops. Find its acceleration and its braking distance.
IntermediateA body starts from rest and accelerates at 2 m/s^2 over 25 m. What is its final velocity?
IntermediateA 60 kg person stands on a scale in a lift and the reading is 720 N. (a) What is the net force? (b) What is the acceleration?
IntermediateA 5 kg crate is pushed horizontally with 30 N while friction is 10 N. Find its acceleration.
ChallengeA ball is thrown vertically upwards at 30 m/s (take g = 10 m/s^2). Find the height of the highest point and the total time for the return trip.
ChallengeA stone is dropped from a height of 45 m. Find the time to fall and the velocity on landing.
Error redoXiao Ming reads a velocity-time graph running from (0, 0) to (5 s, 10 m/s) and answers that the displacement is 2 m/s^2. What is wrong?

03 Answers and one-line notes

#	Answer	One-line note
1	a = (v – u) / t = (20 – 8) / 4 = 3 metres per second squared.	Find the change in velocity first, then divide by the time.
2	a = F / m = 18 / 6 = 3 metres per second squared.	The greater the mass, the smaller the acceleration for the same net force.
3	s = half of a t squared = half of 4 x 8 squared = 128 m.	When the initial velocity is zero, s = half a t squared is the quickest route.
4	a = (0 – 20) / 4 = -5 metres per second squared; s = (20 + 0) / 2 x 4 = 40 m.	Average velocity multiplied by time gives the distance.
5	v squared = u squared + 2as = 0 + 2 x 2 x 25 = 100, so v = 10 m/s.	When time is not involved, use v squared = u squared + 2as.
6	(a) 720 – 60 x 10 = 120 N; (b) a = 120 / 60 = 2 metres per second squared, upwards.	The scale reading is the normal force, not the weight.
7	Net force = 30 – 10 = 20 N, a = 20 / 5 = 4 metres per second squared.	Work out the net force first, then divide by the mass.
8	Height = u squared / 2g = 900 / 20 = 45 m; time = 2u / g = 60 / 10 = 6 s.	The velocity is zero at the highest point of a vertical throw.
9	t = square root of (2s / g) = square root of (90 / 10) = 3 s; v = gt = 10 x 3 = 30 m/s.	A freely falling body starts from rest.
10	He took the gradient as the answer; the area under the graph gives the displacement = half of 5 x 10 = 25 m, while 2 metres per second squared is the acceleration.	On a v-t graph the gradient is the acceleration and the area is the displacement.

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.

Answer all ten before you look at the solutions, then rework any question where your number differs from the one shown.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.