8.3MechanicsStretch

Variable Acceleration in 1D

When acceleration is not constant, suvat no longer applies. Instead, differentiation and integration link displacement, velocity and acceleration as functions of time.

26 min Further Kinematics

Edexcel A Level Maths: 8.3 Variable Acceleration in 1DCurated by Kepler, not produced by us. Watch it before the notes below.

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What you'll be able to do

Differentiate to go s → v → a
Integrate to go a → v → s
Use boundary conditions for constants
Find extreme values of motion

Calculus of motion

Velocity is the derivative of displacement, and acceleration is the derivative of velocity: $v = \dfrac{ds}{dt}$ , $a = \dfrac{dv}{dt}$ . Going the other way, integrate (and use conditions to find the constant).

v = \frac{ds}{dt}, \quad a = \frac{dv}{dt}

Differentiate down; integrate up.

Example 1If

s = t^3 - 4t

, find the velocity and acceleration.

v = \tfrac{ds}{dt} = 3t^2 - 4

a = \tfrac{dv}{dt} = 6t

Answer

v = 3t^2 - 4

a = 6t

Tip — Maximum/minimum velocity occurs where acceleration (dv/dt) is zero.

Formula recap

v = \frac{ds}{dt}, \ a = \frac{dv}{dt}

Differentiate.

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

Integrate (+ constant).

Common mistakes to avoid

Using suvat when acceleration varies.

suvat only works for constant acceleration; use calculus otherwise.

Forgetting the constant of integration.

Use a boundary condition to find it.

Key takeaways

v = ds/dt, a = dv/dt (differentiate down).
s = ∫v dt, v = ∫a dt (integrate up, + constant).
Extremes of velocity where a = 0.

Test yourself

Answer a few questions, then review anything you missed.

Prefer paper? Open the written self-check

Answer each question before opening its answer. This self-check does not record a quiz score.

Velocity is the derivative of:
- displacement
- acceleration
- force
- time
Show answer and explanation
Answer: displacement
v = ds/dt.
Acceleration is the derivative of:
- velocity
- displacement
- mass
- momentum
Show answer and explanation
Answer: velocity
a = dv/dt.
If $s = t^3 - 4t$ , then v =
- $3t^2 - 4$
- $6t$
- $t^2 - 4$
- $3t^2$
Show answer and explanation
Answer: $3t^2 - 4$
Differentiate s.
To go from acceleration to velocity you:
- integrate
- differentiate
- square
- use suvat
Show answer and explanation
Answer: integrate
v = ∫a dt.
Maximum velocity occurs where:
- $a = 0$
- $v = 0$
- $s = 0$
- $t = 0$
Show answer and explanation
Answer: $a = 0$
dv/dt = 0.

Next step

Keep practising

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