SingaporeAdditional MathematicsSyllabus dot point

How do we use differentiation to find velocity and acceleration from a displacement function?

Differentiate a displacement function to find velocity, and differentiate velocity to find acceleration, and find when a particle is at rest

A focused answer to the N(A)-Level Additional Mathematics outcome on using differentiation in kinematics. Differentiate displacement to get velocity and again for acceleration, and find when a particle is instantaneously at rest.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-06

Reviewed by: AI editorial process; not yet individually human-reviewed

Have a quick question? Jump to the Q&A page

Jump to a section

What this dot point is asking
The answer
Examples in context
Try this

What this dot point is asking

SEAB wants you to use differentiation to move from a displacement function to velocity, and from velocity to acceleration, for a particle moving in a straight line. You should also find when the particle is at rest (velocity zero) and when its acceleration is zero. This applies the differentiation rules from the calculus strand directly to motion, where each derivative has a clear physical meaning.

The answer

Velocity is the derivative of displacement

Because velocity is the rate of change of displacement, you find it by differentiating $s$ with respect to time:

v = \frac{ds}{dt}

So if the displacement is given as a function of $t$ , differentiate term by term using the power rule to get the velocity function.

Acceleration is the derivative of velocity

Acceleration is the rate of change of velocity, so differentiate again:

a = \frac{dv}{dt} = \frac{d^2s}{dt^2}

Acceleration is the second derivative of displacement. Two differentiations take you from displacement all the way to acceleration.

Finding when a particle is at rest

A particle is instantaneously at rest when its velocity is zero. So set the velocity function equal to zero and solve for $t$ :

v = 0

There may be more than one such time. These are the moments the particle momentarily stops, often when it changes direction.

Finding other instants

Similar questions ask for the time when the acceleration is zero (set $a = 0$ and solve), or for the velocity or acceleration at a particular time (substitute that value of $t$ into the relevant function). Always read whether the question wants velocity or acceleration.

Worked example

A particle has displacement $s = t^3 - 3t^2$ metres at time $t$ seconds. Find when it is at rest, and its acceleration at that later time.

Step 1: Differentiate for velocity

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

Step 2: Set velocity to zero

3t^2 - 6t = 0 \quad\Rightarrow\quad 3t(t - 2) = 0 \quad\Rightarrow\quad t = 0 \text{ or } t = 2

The particle is at rest at $t = 0\ \text{s}$ and $t = 2\ \text{s}$ .

Step 3: Differentiate again for acceleration

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

Step 4: Evaluate the acceleration at t = 2

a = 6(2) - 6 = 6\ \text{m s}^{-2}

At the later rest time $t = 2\ \text{s}$ , the acceleration is $6\ \text{m s}^{-2}$ in the positive direction.

Examples in context

Example 1. Turning points of motion. Setting velocity to zero in kinematics is the same algebra as finding stationary points of a curve in calculus: both solve "derivative equals zero". The particle's rest times are the stationary points of the displacement-time graph, which is why the two topics share a method.

Example 2. Maximum displacement. To find the furthest point a particle reaches, find when its velocity is zero (it stops moving outward there) and substitute that time into the displacement function. The same idea finds the highest point of a thrown object, where the upward velocity reaches zero.

Try this

Q1. Given $s = t^2 + 2t$ , find the velocity function. [1 mark]

Cue. $v = \dfrac{ds}{dt} = 2t + 2$ .

Q2. Given $v = 4t - 8$ , find the time when the particle is at rest. [2 marks]

Cue. $4t - 8 = 0$ , so $t = 2$ seconds.

Q3. Given $s = t^3 - 3t$ , find the acceleration at $t = 2$ . [3 marks]

Cue. $v = 3t^2 - 3$ , $a = 6t$ ; at $t = 2$ , $a = 12\ \text{m s}^{-2}$ .

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original4 marksA particle moves so that its displacement is

s = t^3 - 6t^2 + 9t

metres at time

t

seconds. Find expressions for its velocity and acceleration.

Show worked answer →

Velocity is the derivative of displacement: $v = \dfrac{ds}{dt} = 3t^2 - 12t + 9$ .

Acceleration is the derivative of velocity: $a = \dfrac{dv}{dt} = 6t - 12$ .

What markers reward: differentiating $s$ term by term to get $v$ , differentiating again for $a$ , and presenting both as functions of $t$ .

Original4 marksFor a particle with displacement

s = t^2 - 4t

metres, find the time at which it is instantaneously at rest.