9. Rectilinear Motion

Video coming soon3 worked examples

Theory

When a particle moves along a straight line, its position is described by a single displacement function $s(t)$ . Differentiating once gives velocity, and again gives acceleration:

v = \frac{ds}{dt}, \qquad a = \frac{dv}{dt} = \frac{d^2s}{dt^2}

Going the other way, integration recovers what differentiation removed — but each integration introduces a constant that must be found from the given conditions:

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

There is a third form of the acceleration that is invaluable when $a$ is given in terms of displacement rather than time. By the chain rule:

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

Most questions describe an event rather than giving a value, so learn the translations: “at rest” means $v = 0$ ; “returns to the starting point” means $s = 0$ ; “maximum velocity” means $a = 0$ .

The Golden Rule: distance and displacement are not the same thing. Displacement is simply $s$ at the end minus $s$ at the start, but if the particle changes direction — that is, if $v$ changes sign — the total distance must be found in separate stages and the magnitudes added.

⚠️ Common Examiner Traps

Distance vs displacement: always check whether $v = 0$ inside the time interval. If it does, the particle turned round and the two answers differ.
$a = 0$ is about velocity, not position: it gives maximum or minimum velocity. Maximum displacement occurs where $v = 0$ .
Losing the constants: integrating from acceleration to displacement needs two conditions, and they are often given at different times.
Choosing the wrong form of $a$ : if acceleration is given in terms of $s$ , use $a = v\frac{dv}{ds}$ . Trying to use $\frac{dv}{dt}$ leaves you with two variables and nowhere to go.

Worked examples

Example 1

A particle moves along a straight line so that its displacement after $t$ seconds is $s = t^3 - 6t^2 + 9t$ metres. Find the times at which it is at rest, and its acceleration at each of those times.

Step 1: Differentiate to obtain the velocity:

v = \frac{ds}{dt} = 3t^2 - 12t + 9

Step 2: “At rest” means $v = 0$ . Factorise and solve:

3(t^2 - 4t + 3) = 0 \implies 3(t-1)(t-3) = 0 \implies t = 1 \ \text{ or } \ t = 3

Step 3: Differentiate again for the acceleration:

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

Step 4: Evaluate at each time:

\begin{aligned} t = 1: &\quad a = 6 - 12 = -6\ \text{m/s}^2 \\ t = 3: &\quad a = 18 - 12 = 6\ \text{m/s}^2 \end{aligned}

The opposite signs tell us the particle is decelerating into the first stop and accelerating away from the second.

Example 2

For the same particle, $s = t^3 - 6t^2 + 9t$ , find its displacement and the total distance travelled during the first 4 seconds.

Step 1: Displacement is just the change in $s$ :

s(4) - s(0) = (64 - 96 + 36) - 0 = 4\ \text{m}

Step 2: For distance, first check whether the particle turned round. From the previous example $v = 0$ at $t = 1$ and $t = 3$ , and both lie inside the interval — so it changed direction twice.

Step 3: Evaluate $s$ at every turning point and at the ends:

\begin{aligned} s(0) &= 0 \\ s(1) &= 1 - 6 + 9 = 4 \\ s(3) &= 27 - 54 + 27 = 0 \\ s(4) &= 64 - 96 + 36 = 4 \end{aligned}

Step 4: Add the magnitude of each leg separately:

|4 - 0| + |0 - 4| + |4 - 0| = 4 + 4 + 4 = 12\ \text{m}

So the displacement is $4$ m but the distance travelled is $12$ m — the particle went forward, came back to the origin, then went forward again.

Example 3

A particle moves in a straight line with acceleration $a = 6s + 4$ , where $s$ is its displacement. When $s = 0$ its velocity is $2$ m/s. Find its velocity when $s = 2$ .

Step 1: The acceleration is given in terms of $s$ , not $t$ , so use the third form:

v\frac{dv}{ds} = 6s + 4

Step 2: Separate the variables and integrate both sides:

\int v\,dv = \int (6s + 4)\,ds \implies \frac{1}{2}v^2 = 3s^2 + 4s + C

Step 3: Apply the given condition, $v = 2$ when $s = 0$ :

\frac{1}{2}(4) = 0 + 0 + C \implies C = 2

Step 4: The relationship is therefore:

\frac{1}{2}v^2 = 3s^2 + 4s + 2

Step 5: Substitute $s = 2$ :

\frac{1}{2}v^2 = 12 + 8 + 2 = 22 \implies v^2 = 44 \implies v = 2\sqrt{11} \approx 6.6\ \text{m/s}