Learn7357Paper 2 · Pure and MechanicsKinematicsVariable acceleration

Kinematics · Subtopic 4 of 5

Variable Acceleration: Kinematics with Calculus

When acceleration changes with time, suvat is off the table. Differentiate to go from displacement to velocity to acceleration; integrate, with a constant each time, to come back.

Cambridge 9709 Paper 4Edexcel YMA01 WME02Edexcel 9MA0 Paper 3AQA 7357 Paper 2≈ 35 minRevision 6 · updated 2026-10-27

Section 1 of 14

Learning objectives

Use $v=\dfrac{ds}{dt}$ and $a=\dfrac{dv}{dt}$ when $s$ or $v$ is a function of $t$ .
Integrate $a$ or $v$ , using a starting condition to find the constant.
Find when a particle is at rest, and its greatest or least velocity.
Find the total distance travelled when the particle changes direction.

Pause and think

A particle has $v=6t-t^2$ . Can you use $v=u+at$ to find its velocity at $t=4$ ? Why or why not?

Section 2 of 14

Core idea

Start hereWhen suvat breaks

A rocket's acceleration increases as it burns fuel. Constant-acceleration formulas no longer apply. But velocity is the rate of change of displacement, and acceleration the rate of change of velocity: calculus handles any motion.

The big idea

$v=\frac{ds}{dt}$ , $a=\frac{dv}{dt}$ ; and $v=\int a\,dt$ , $s=\int v\,dt$ (with constants from initial conditions). For distance, split where $v$ changes sign.

1. Differentiating

s\ \xrightarrow{\ d/dt\ }\ v\ \xrightarrow{\ d/dt\ }\ a\qquad a\ \xrightarrow{\ \int dt\ }\ v\ \xrightarrow{\ \int dt\ }\ s

s = t³ − 6t² + 9t and its velocity

v=3t^2-12t+9=3(t-1)(t-3)

is zero at

t=1

and

t=3

: the particle stops and changes direction. Between them

v<0

and

s

decreases.

Worked example $s=t^3-6t^2+9t$

When is it at rest? Find the distance in the first 4 s.

Goal: Times; distance.
Strategy: $v=0$ ; positions at turning times.

$v=3t^2-12t+9=3(t-1)(t-3)$
Differentiate.
$t=1,3$
At rest.
$s(0)=0,\ s(1)=4,\ s(3)=0,\ s(4)=4$
Positions.
$4+4+4=12\text{ m}$
Out, back, out.

AnswerRest at

t=1,3

; distance

12

m (displacement

4

m).

✓ Check

v<0

between 1 and 3 (moving back) ✓.

2. Integrating

Worked example $a=6-2t$ from rest at the origin

Find $v$ , $s$ , max velocity, and when it's next at rest.

Goal: Values.
Strategy: Integrate twice; constants from $t=0$ .

$v=6t-t^2+c;\ v(0)=0\Rightarrow c=0$
Integrate.
$s=3t^2-\tfrac13t^3$
Integrate; $s(0)=0$ .
$a=0\Rightarrow t=3,\ v=9$
Max velocity.
$v=t(6-t)=0\Rightarrow t=6,\ s=108-72=36$
Next rest.

AnswerMax

9

m/s at

t=3

; at rest again at

t=6

36

m away.

✓ Check

\frac{dv}{dt}=6-2t

✓.

Common misconceptionForgetting the constant Every integration needs a constant; use the starting conditions.

Pause and think

$v=4t-t^2$ . Distance in the first 6 s?

Summary

Differentiate down; integrate up.
Split at stops for distance.

Variable acceleration: Acceleration that changes.

HistoryFluxions. Isaac Newton developed calculus in the 1660s largely to describe motion. He called a changing quantity a "fluent" and its rate of change a "fluxion", writing

\dot{x}

for the velocity of

x

. The dot notation survives in mechanics today:

\dot{s}=v

and

\ddot{s}=a

Fun factJerk, snap, crackle and pop. The derivative of acceleration is called jerk, and engineers really use it: lifts and roller coasters are designed to keep jerk small so riders are not thrown about. The next three derivatives have the jokey but genuine names snap, crackle and pop.

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