Kinematics Worked Examples (Easy)

Q: What does "momentarily at rest" mean?

It means the velocity is zero at that instant, v=0v=0v=0, even though the acceleration may not be. Solve v=0v=0v=0 for ttt; a quadratic velocity can give two separate times at rest.

What these examples cover

These easy Kinematics examples build the four moves the whole chapter rests on. Two use differentiation: velocity is the rate of change of displacement, $v=\frac{ds}{dt}$ , and acceleration is the rate of change of velocity, $a=\frac{dv}{dt}$ .

One uses the condition for a particle to be momentarily at rest, $v=0$ . Two use integration to reverse those steps, recovering velocity from acceleration and displacement from velocity, where the given initial value fixes the constant.

Every number here is small and clean, so you can follow each line without a calculator. Cover the solution, attempt the question in full on paper, then check line by line and find the exact step where any difference begins.

Worked examples

Work through all four. Attempt each fully before you read the matching solution, and notice how the same discipline, write down the relationship, differentiate or integrate, then substitute carefully, runs through every one.

Q1[4 marks]

A particle moves along a straight line and passes a fixed point $O$ . Its displacement $s$ m from $O$ , $t$ s after passing $O$ , is $s=t^{3}-3t^{2}+5t$ .

Find (a) the velocity when $t=2$ , and (b) the acceleration when $t=3$ .

Show worked solution

Velocity is the rate of change of displacement, so differentiate $s$ with respect to $t$ .

v=\frac{ds}{dt}=3t^{2}-6t+5

Substitute $t=2$ for part (a):

v|_{t=2}=3(2)^{2}-6(2)+5=12-12+5=5

Acceleration is the rate of change of velocity, so differentiate once more.

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

Substitute $t=3$ for part (b):

a|_{t=3}=6(3)-6=18-6=12

Answer

The velocity when $t=2$ is $5\ \text{m s}^{-1}$ , and the acceleration when $t=3$ is $12\ \text{m s}^{-2}$ .

Q2[4 marks]

A particle moves along a straight line. Its velocity $v$ , in $\text{m s}^{-1}$ , at time $t$ s is $v=t^{2}-5t+6$ .

Find (a) the times when the particle is momentarily at rest, and (b) the acceleration when $t=1$ .

Show worked solution

A particle is momentarily at rest when its velocity is zero, so solve $v=0$ .

t^{2}-5t+6=0

Factorise the quadratic.

(t-2)(t-3)=0

So $t=2$ or $t=3$ ; both are valid because $t\ge 0$ .

For part (b), acceleration is $\frac{dv}{dt}$ , so differentiate the velocity.

a=\frac{dv}{dt}=2t-5

Substitute $t=1$ :

a|_{t=1}=2(1)-5=-3

Answer

The particle is momentarily at rest at $t=2$ s and $t=3$ s. The acceleration when $t=1$ is $-3\ \text{m s}^{-2}$ ; the negative sign shows it is decelerating at that instant.

Q3[4 marks]

A particle moves along a straight line with acceleration $a=6t-4$ , in $\text{m s}^{-2}$ , at time $t$ s. When $t=0$ its velocity is $5\ \text{m s}^{-1}$ .

Find the velocity when $t=2$ .

Show worked solution

Velocity is the integral of acceleration with respect to time, so integrate and include a constant $c$ .

v=\int a\,dt=\int (6t-4)\,dt=3t^{2}-4t+c

Use the initial condition $v=5$ when $t=0$ to find $c$ .

5=3(0)^{2}-4(0)+c\ \Rightarrow\ c=5

So the velocity function is:

v=3t^{2}-4t+5

Substitute $t=2$ :

v|_{t=2}=3(2)^{2}-4(2)+5=12-8+5=9

Answer

The velocity when $t=2$ is $9\ \text{m s}^{-1}$ .

Q4[3 marks]

A particle moves along a straight line with velocity $v=3t^{2}-2t$ , in $\text{m s}^{-1}$ , at time $t$ s. Its displacement from a fixed point $O$ is $2$ m when $t=0$ .

Find its displacement from $O$ when $t=3$ .

Show worked solution

Displacement is the integral of velocity, so integrate and include a constant $c$ .

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

Use $s=2$ when $t=0$ to find $c$ .

2=(0)^{3}-(0)^{2}+c\ \Rightarrow\ c=2

So the displacement function is:

s=t^{3}-t^{2}+2

Substitute $t=3$ :

s|_{t=3}=(3)^{3}-(3)^{2}+2=27-9+2=20

Answer

The displacement from $O$ when $t=3$ is $20$ m.

Q5[3 marks]

A particle moves along a straight line with velocity $v=12t-3t^{2}$ , in $\text{m s}^{-1}$ , at time $t$ s, where $t\ge 0$ . Find (a) the time when the acceleration of the particle is zero, and (b) the velocity of the particle at that instant.

Show worked solution

Acceleration is the rate of change of velocity, so differentiate $v$ with respect to $t$ .

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

For part (a), set $a=0$ and solve for $t$ .

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

For part (b), substitute $t=2$ into the velocity function.

v|_{t=2}=12(2)-3(2)^{2}=24-12=12

Answer

The acceleration is zero when $t=2$ s, and the velocity at that instant is $12\ \text{m s}^{-1}$ . Because the acceleration changes from positive to negative at $t=2$ , this is the particle's maximum velocity.

Q6[2 marks]

A particle moves along a straight line and passes through a fixed point $O$ where $t=0$ . Its displacement from $O$ , $s$ m, at time $t$ s, is $s=t^{2}-6t$ .

Find the value of $t$ , other than $t=0$ , at which the particle passes through $O$ again.

Show worked solution

The particle is at $O$ when $s=0$ , so solve this equation for $t$ .

t^{2}-6t=0

Factorise.

t(t-6)=0

So $t=0$ or $t=6$ ; since $t=0$ is the starting point, the particle returns to $O$ at the other value.

Answer

The particle passes through $O$ again when $t=6$ s.

Q7[3 marks]

A particle moves along a straight line with velocity $v=4t-2$ , in $\text{m s}^{-1}$ , at time $t$ s. Find the displacement of the particle between $t=1$ s and $t=3$ s.

Show worked solution

The displacement between two times is given by the definite integral of velocity over that interval.

\Delta s=\int_{1}^{3}(4t-2)\,dt=\Big[2t^{2}-2t\Big]_{1}^{3}

Evaluate at the upper and lower limits.

\Delta s=(2(3)^{2}-2(3))-(2(1)^{2}-2(1))=(18-6)-(2-2)=12

Answer

The displacement between $t=1$ s and $t=3$ s is $12$ m. Since $v>0$ throughout this interval, this is also the total distance travelled.

Q8[3 marks]

A particle moves along a straight line and passes through a fixed point $O$ where $t=0$ . Its displacement from $O$ , $s$ m, at time $t$ s, is $s=5t-t^{2}$ .

Find the velocity of the particle when $t=4$ , and state whether the particle is moving in the positive or negative direction at that instant.

Show worked solution

Velocity is the rate of change of displacement, so differentiate $s$ with respect to $t$ .

v=\frac{ds}{dt}=5-2t

Substitute $t=4$ .

v|_{t=4}=5-2(4)=5-8=-3

Answer

The velocity when $t=4$ is $-3\ \text{m s}^{-1}$ . Since the velocity is negative, the particle is moving in the negative direction (back towards $O$ ) at that instant.

Key method points

These four examples rehearse the skills that open almost every Kinematics question in Add Math. Keep the following points in mind as you practise more.

Velocity is the derivative of displacement, $v=\frac{ds}{dt}$ ; acceleration is the derivative of velocity, $a=\frac{dv}{dt}=\frac{d^{2}s}{dt^{2}}$ .
A particle is momentarily at rest when $v=0$ ; solve that equation for $t$ .
To reverse the process, integrate: $v=\int a\,dt$ and $s=\int v\,dt$ .
Every integration adds a constant, use the given initial value (the velocity or displacement at a stated time) to find it before substituting.
Read the sign of your answer: a negative velocity means motion in the negative direction, and a negative acceleration means the particle is slowing down when it is moving forward.

How a teacher helps

When a student drops a mark here, it is usually a small, fixable habit, forgetting the constant of integration, or reading a negative velocity as a mistake rather than a direction. In a one-to-one lesson our teacher watches the exact line where the slip happens and corrects it on the spot, before it settles into a routine.

Because our teachers are experienced, you work with someone who explains why each step follows, not only what to write. Lessons are taught in English, while SPM papers are set in both Malay and English, so the notation reads the same to you either way.

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Frequently asked questions

How do I move between displacement, velocity and acceleration?

Differentiate to go down the chain and integrate to go back up. From displacement, $v=\frac{ds}{dt}$ and $a=\frac{dv}{dt}$ ; from acceleration, $v=\int a\,dt$ and $s=\int v\,dt$ .

Each integration needs a constant that you fix with a given value.

What does "momentarily at rest" mean?

It means the velocity is zero at that instant, $v=0$ , even though the acceleration may not be. Solve $v=0$ for $t$ ; a quadratic velocity can give two separate times at rest.

Why does my answer have a negative sign?

A negative velocity simply means the particle is moving in the negative direction, and a negative acceleration means the velocity is decreasing. The sign is information, not an error, keep it.

Do I always need the initial condition when I integrate?

Yes. Integration leaves an unknown constant, so without a stated value, the velocity or displacement at a particular time, you cannot pin down the exact function.

Substitute the given values first, then evaluate.

Source:SRC-DSKP-EN

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

Kinematics of Linear Motion, Worked Examples (easy)

What these examples cover

Worked examples

Key method points

How a teacher helps

Frequently asked questions

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