Variable acceleration

When the acceleration won't sit still.

Every one of the suvat equations assumes that $a$ never changes. Let it change and all five go out of the window — and calculus takes over. One particle, three graphs, and a ladder you can climb in both directions.

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Where it is

A particle moves along a straight line. Its displacement from $O$ after $t$ seconds is $s = t^3 - 6t^2 + 9t$ metres. It runs out to 4 m, comes back through $O$ at $t = 3$ , then sets off again.

Differentiate to go down

The gradient of the $s$ – $t$ graph is the velocity:

v = \frac{ds}{dt} = 3(t-1)(t-3)

Where the curve is flat, at $t = 1$ and $t = 3$ , the particle is momentarily at rest.

Once more

Do it again and the third graph appears — a straight line, but not a constant, which is why suvat fails here.

a = \frac{dv}{dt} = 6t - 12

It vanishes at $t = 2$ : not a moment of rest, but where the velocity stops falling and starts to climb, at $-3\text{ m s}^{-1}$ .

Integrate to go up

Climb back and areas replace gradients. The signed area under the $v$ – $t$ graph is the displacement: $+4$ , then $-4$ , then $+4$ .

\int_0^4 v\,dt = 4 = s(4)

Which is exactly where the top graph ends.

Not the same thing

Distance is not displacement. Add the sizes of the three pieces instead of letting them cancel and the particle has travelled 12 m, not 4.

Split the integral wherever $v = 0$ . Never read the distance off $s(4)$ alone.

Chapter 4

Two operations, one ladder.

Constants of integration from the initial conditions, maxima and minima of velocity, motion under a variable force, and twelve graded exam questions with full worked solutions are waiting.

Read the full chapter › Jump to the questions ›