A-Level — mathsAdvanced2 min read

Variable Acceleration

Use calculus for variable acceleration problems at A-Level. Differentiate and integrate displacement, velocity, and acceleration.

Tutor AI Team2026-02-25T01:59:09.131Z

When acceleration is not constant, SUVAT equations don't apply. Instead, use calculus: differentiation and integration connect displacement, velocity, and acceleration.

The Relationships

$v = \frac{ds}{dt} \qquad a = \frac{dv}{dt} = \frac{d^2s}{dt^2}$

$s = \int v\,dt \qquad v = \int a\,dt$

Worked Example: Example 1

Problem

$s = 3t^2 - t^3$ . Find velocity and acceleration.

$v = 6t - 3t^2$ . $a = 6 - 6t$ .

At rest: $v = 0$ → $3t(2-t) = 0$ → $t = 0$ or $t = 2$ .

Solution

Worked Example: Example 2

Problem

$a = 4 - 2t$ , $v(0) = 3$ . Find $v$ and $s$ .

$v = 4t - t^2 + 3$ . $s = 2t^2 - \frac{t^3}{3} + 3t + C$ . If $s(0) = 0$ , then $s = 2t^2 - \frac{t^3}{3} + 3t$ .

Solution

Worked Example: Example 3

Problem

Max velocity: set $a = 0$ . $6 - 6t = 0$ → $t = 1$ . $v_{max} = 6(1) - 3(1) = 3$ m/s.

Solution

Practice Problems

1. $v = t^3 - 6t$ . Find when $a = 0$ and the displacement from $t = 0$ to $t = 3$ .
1. $a = 12 - 4t$ , $u = 0$ , $s(0) = 0$ . Find $s$ when $v = 0$ .

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Key Takeaways

✓
Differentiate $s$ to get $v$ , differentiate $v$ to get $a$ .
✓
Integrate $a$ to get $v$ , integrate $v$ to get $s$ .
✓
Use initial conditions to find constants of integration.

Tags:

#a-level#maths#mechanics#kinematics#calculus

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The Relationships

Worked Example: Example 1

Worked Example: Example 2

Worked Example: Example 3

Practice Problems

Key Takeaways

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