Kinematics of Linear Motion Practice

How to use these practice questions

The point of practice is not to read solutions but to produce them. Work each of the six Kinematics of Linear Motion questions below with pen and paper first, under something close to exam conditions, write the formula, then the substitution, then the answer, one clear line at a time.

Only then open the worked solution and mark your own script the way an examiner would, awarding yourself a mark for the right method, a mark for correct substitution, and a mark for the final answer. Where your working differs from ours, find the exact line where it went wrong rather than simply copying the correct version.

Because SPM marking is analytic, clear, ordered working protects most of the marks even when a final answer slips, so treat neat layout as part of the practice, not an afterthought.

Six practice questions

Q1[3 marks]

A particle moves along a straight line so that its displacement from a fixed point O, $t$ seconds after passing O, is $s=t^{3}-6t^{2}+9t$ metres. Find expressions for the velocity $v$ and the acceleration $a$ .

Show worked solution

Velocity is the rate of change of displacement, and acceleration the rate of change of velocity. Differentiate once, then again.

v=\frac{ds}{dt}=3t^{2}-12t+9

a=\frac{dv}{dt}=6t-12

Q2[3 marks]

Using $v=3t^{2}-12t+9$ , find the times when the particle is instantaneously at rest.

Show worked solution

The particle is at rest when its velocity is zero. Set $v=0$ and solve.

3t^{2}-12t+9=0\;\Rightarrow\;t^{2}-4t+3=0\;\Rightarrow\;(t-1)(t-3)=0

So $t=1$ s and $t=3$ s.

Q3[4 marks]

For the same particle, find the displacement at the instant when the acceleration is zero.

Show worked solution

First find when $a=0$ , then substitute that time into $s$ .

6t-12=0\;\Rightarrow\;t=2

s(2)=2^{3}-6(2)^{2}+9(2)=8-24+18=2

The displacement is $2$ m from O.

Q4[3 marks]

A second particle has velocity $v=2t-6$ m s $^{-1}$ and is at O when $t=0$ . Find its displacement when $t=4$ s.

Show worked solution

Displacement is the integral of velocity. Integrate, then use $s=0$ at $t=0$ to find the constant.

s=\int (2t-6)\,dt=t^{2}-6t+c,\quad s(0)=0\Rightarrow c=0

s(4)=16-24=-8

The displacement is $-8$ m, that is, 8 m on the negative side of O.

Q5[5 marks]

A particle has velocity $v=t^{2}-4t+3$ m s $^{-1}$ and starts at O when $t=0$ . Find the total distance travelled in the first 3 seconds.

Show worked solution

Total distance is not the same as final displacement when the particle changes direction. First find when $v=0$ : $(t-1)(t-3)=0$ , so $t=1$ and $t=3$ , the particle reverses at $t=1$ .

s=\int v\,dt=\frac{t^{3}}{3}-2t^{2}+3t\ (\,c=0\,)

Evaluate the displacement at the key times:

s(0)=0,\quad s(1)=\tfrac{1}{3}-2+3=\tfrac{4}{3},\quad s(3)=9-18+9=0

From $t=0$ to $1$ it moves $\tfrac{4}{3}$ m forward; from $t=1$ to $3$ it moves from $\tfrac{4}{3}$ back to $0$ , a distance of $\tfrac{4}{3}$ m. Total distance $=\tfrac{4}{3}+\tfrac{4}{3}=\tfrac{8}{3}$ m.

Q6[4 marks]

A particle moves with displacement $s=t^{3}-3t^{2}$ m. Find its minimum velocity and the time at which this occurs.

Show worked solution

Velocity is $v=\frac{ds}{dt}$ ; its minimum occurs where its own rate of change, the acceleration, is zero.

v=3t^{2}-6t,\qquad a=\frac{dv}{dt}=6t-6

Set $a=0$ : $6t-6=0$ , so $t=1$ . Since $a$ changes from negative to positive here, this is a minimum of $v$ .

v(1)=3(1)^{2}-6(1)=-3

The minimum velocity is $-3$ m s $^{-1}$ , at $t=1$ s.

How to mark yourself like an examiner

An examiner does not just look at your final box, they follow your working and award marks step by step. So when you mark yourself, resist the temptation to tick only right answers.

Ask, for each question: did I state the correct method or formula? Did I substitute the right values?

Is my final answer in the exact form the question asked for, with units where needed? Give yourself the method marks you genuinely earned even when the final number is wrong, and, just as importantly, deny yourself marks where your working was too cramped or jumped a step, because a real marker would.

Keep a short list of the exact slips you repeat across these questions; that list, not the score, is what tells you where to spend your next practice session.

How a teacher helps

Marking your own work is powerful, but a student cannot always see why a method keeps going wrong, that is where a second pair of eyes changes things. Working one-to-one, a teacher watches your Kinematics of Linear Motion working as it happens and catches the exact step that costs the marks, then teaches straight to it, often tracing the real gap back to an earlier idea the question quietly assumes.

Our teachers are experienced; lessons are online and in English, and the first lesson is a one-hour paid class at the teacher's rate, from RM50 an hour. Bring the questions here that you could not fully solve, and an hour aimed at exactly those is worth far more than an hour spent on what you already know.

Get 1-to-1 help.

Book a Trial Class

Frequently asked questions

Are these real past-year SPM questions?

No. Every question here is original, written in the style of the exam.

We never reproduce real past-year SPM questions.

How should I use the marks shown on each question?

Treat them as a guide to how much working is expected, and mark your own solution method by method, because SPM marking is analytic, you earn marks step by step, not only for the final answer.

What if I get an answer wrong?

Find the exact line where your working and ours diverge, and note the slip. Repeated slips are the fastest thing to fix, and exactly what a one-to-one lesson can target.

Source:SRC-DSKP-EN

Written by the spmaddmath.com.my editorial team.· Last updated 5 September 2026

Kinematics of Linear Motion, Practice Questions

How to use these practice questions

Six practice questions

How to mark yourself like an examiner

How a teacher helps

Frequently asked questions

Ready to get started?