Kinematics

The chain from displacement to acceleration

Kinematics is Section 3 calculus with different letters. Nothing new is being asked of you mathematically; the work is in reading the question.

So velocity is the rate at which displacement changes, and acceleration is the rate at which velocity changes.

Going the other way

Integration runs the chain backwards, and each step brings a constant that a condition in the question pins down:

$$a \;\xrightarrow{\;\int\;}\; v \;\xrightarrow{\;\int\;}\; s$$

The two conditions that appear most often are "starts from rest", meaning $v = 0$ when $t = 0$, and "starts at the origin", meaning $s = 0$ when $t = 0$.

At rest, and changing direction

Note what this does not mean. It does not mean $s = 0$ — the particle may be far from where it started. It does not mean $a = 0$ — a ball at the top of its flight is momentarily at rest while gravity is still pulling on it.

Solving $v = 0$ is how you find the moments the particle changes direction, which is what the next section depends on.

Distance is not displacement

Displacement is where the particle ends up relative to where it started: $s(t_2) - s(t_1)$. It can be negative, and it can be zero for a particle that has been moving the whole time.

Total distance travelled is how far it has actually gone, counting every stretch as positive.

They agree only while the particle keeps moving in one direction. As soon as it turns around, they part company.

For $v = 3t^2 - 12t + 9$ with $s = t^3 - 6t^2 + 9t$, the velocity is zero at $t = 1$ and $t = 3$. Then $s(0) = 0$, $s(1) = 4$, $s(3) = 0$ and $s(4) = 4$. The three legs are 4, 4 and 4 metres, so the particle travels 12 m in total — while its displacement after 4 seconds is only 4 m.

Velocity-time graphs

A straight line means constant acceleration; a horizontal line means constant velocity, so zero acceleration; the graph crossing the axis is the particle turning around.