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 Video by Zeeshan Zamurred Further Kinematics

Edexcel A Level Maths: 8.3 Variable Acceleration in 1DWatch the full walkthrough 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: $v = \frac{d s}{d t}$ , $a = \frac{d v}{d t}$ . Going the other way, integrate (and use conditions to find the constant).

v = \frac{d s}{d t}, a = \frac{d v}{d t}

Differentiate down; integrate up.

v = \frac{d s}{d t} = 3 t^{2} - 4

a = \frac{d v}{d t} = 6 t

Answer

v = 3 t^{2} - 4

a = 6 t

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

Formula recap

v = \frac{d s}{d t}, a = \frac{d v}{d t}

Differentiate.

s = \int v d t, v = \int a d t

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.

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