Variable acceleration

The idea

When acceleration changes with time, SUVAT no longer works: use calculus. Differentiate displacement to get velocity, and velocity to get acceleration. Integrate to go back, using the starting values to find each constant.

$v = \frac{ds}{dt}$ and $a = \frac{dv}{dt}$ .
$s = \int v\,dt$ and $v = \int a\,dt$ .
Use a known value, such as $v$ at $t = 0$ , to find the constant.

Watch out Do not use SUVAT when the acceleration depends on $t$ .

See it worked

Worked example

A particle has $v = 3t^2 - 4t$ . Find its acceleration at $t = 2$ .

$a = \frac{dv}{dt} = 6t - 4$ .
At $t = 2$ : $12 - 4$ .
$a = 8\text{ m s}^{-2}$ .

Answer $8\text{ m s}^{-2}$

Your turn

Question 1Non-calculator3 marks

A particle has displacement $s = t^3 - 6t^2 + 9t$ metres. Find its velocity at $t = 2$ .

Question 2Non-calculator3 marks

For the same particle, find the times when it is momentarily at rest.

Question 3Non-calculator3 marks

A particle has acceleration $a = 6t\text{ m s}^{-2}$ and velocity $2\text{ m s}^{-1}$ at $t = 0$ . Find its velocity at $t = 2$ .

In past papers

17 official questions in 14 papers since 2018. The latest:

All A-level Maths past papers

A-level Maths Pro, Pearson Edexcel 9MA0

A-level Maths Pro

£129 once

Price rises to £199 on 1 January. Save £70 by joining before then.

At least 12 months of access

By paying you agree to our Terms and get access straight away, which ends your 14-day cancellation right.

No VAT is charged.