Motion with changing acceleration, using calculus

When the equations stop working

A rocket car's engine pushes harder and harder: its acceleration grows every second. A car's driver lifts her foot slowly off the accelerator: its acceleration shrinks. In both, $a$ is not constant, so $v = u + at$ and its two partners give wrong answers. We need a tool that works for any motion: calculus.

Going down: differentiate

You already know the first step. If the position is given as a formula of time,

A worked example. A particle moves with $x = 2t^3 - 9t^2 + 12t$ (SI units).

1. Velocity. $v = 6t^2 - 18t + 12 = 6(t - 1)(t - 2)$.
2. When does it stop? $v = 0$ at $t = 1\ \text{s}$ and at $t = 2\ \text{s}$. Between them $v$ is negative: the particle moves back.
3. Acceleration. $a = 12t - 18$. At $t = 1\ \text{s}$, $a = -6\ \text{m/s}^2$; at $t = 2\ \text{s}$, $a = +6\ \text{m/s}^2$.

Try a cubic of your own.

Going up: integrate

Often we know the acceleration and want the velocity. Integration undoes differentiation:

On a graph, $\int a\,dt$ is the area under the acceleration-time graph, just as the displacement is the area under the velocity-time graph.

A worked example. A rocket car starts from rest with $a = 4t$ (SI units).

1. Velocity. $v = \int_0^t 4t\,dt = 2t^2$. At $t = 3\ \text{s}$, $v = 18\ \text{m/s}$.
2. Position. $x = \int_0^t 2t^2\,dt = \tfrac23 t^3$. At $t = 3\ \text{s}$, $x = 18\ \text{m}$.

Watch out. Using $v = u + at$ with the acceleration at the end, $a = 12\ \text{m/s}^2$, would give $36\ \text{m/s}$: twice too much, because the acceleration was smaller earlier on.

Check with the steady case. If $a$ is a constant, integrating gives $v = u + at$ and then $x = ut + \tfrac12 at^2$: the equations of motion are a special case of this method.

Practise with new numbers

Try it in Python

A computer does not need the formula for the integral. It steps through time in tiny steps: in each step the velocity grows by $a\,\Delta t$, and the position by $v\,\Delta t$. That is how games and weather forecasts move things.

Remember these

• $v = \dfrac{dx}{dt}$, $a = \dfrac{dv}{dt}$: differentiate to go down.
• $v = u + \int a\,dt$, $x = x_0 + \int v\,dt$: integrate to go up.
• $v = u + at$ and its partners only work when $a$ is constant.

Check yourself

1. A body moves with $x = t^3 - 12t$ (SI units). When is it at rest?

2. For $v = 8 - 2t$ (SI units), starting at $x = 0$, find the position at $t = 3\ \text{s}$.

3. A body starts from rest with $a = 3t^2$ (SI units). What is its velocity at $t = 2\ \text{s}$?

4. Why can you not use $s = ut + \tfrac12 at^2$ when $a = 4t$?