Kinematics

4.2 Kinematics of motion in a straight line

Scalars and vectors

A particle moving 5 m forward and 3 m back has travelled a distance of 8 m but has a displacement of 2 m. Average speed uses total distance, and average velocity uses displacement.

Motion graphs

The graph is a trapezium with parallel sides of 200 s and 300 s, and height 20:

$\text{distance} = \frac{1}{2}(200 + 300) \times 20 = 5000 \text{ m}$

The constant acceleration formulae

These are valid only when the acceleration is constant and the motion is in a straight line.

List the quantities you know and want, then pick the formula that contains them.

Motion under gravity

An object moving freely in a vertical line has acceleration $g = 10$ m s⁻² downwards. Choose a positive direction and keep to it.

Taking up as positive, $a = -10$. At the top, $v = 0$:

$0 = 20^2 + 2(-10)s \quad\Rightarrow\quad s = 20 \text{ m}$

It returns when $s = 0$: $0 = 20t - 5t^2$, so $t = 4$ s.

$15 = 20t - 5t^2$, so $t^2 - 4t + 3 = 0$ and $t = 1$ s or $t = 3$ s. There are two answers, one going up and one coming down.

Two particles

When particles meet or one catches the other, write each one's displacement using the same clock, then set up the condition that links them.

$15t = \tfrac{1}{2}(2)t^2$, so $t^2 = 15t$ and $t = 15$ s.

Variable acceleration

When the acceleration changes, the formulae above do not apply. Use calculus.

At rest: $3(t - 1)(t - 3) = 0$, so $t = 1$ or $t = 3$. Its acceleration is $a = 6t - 12$, which is $-6$ m s⁻² at $t = 1$ and $6$ m s⁻² at $t = 3$.

Its displacement is $s = t^3 - 6t^2 + 9t$, with no constant, since $s = 0$ at $t = 0$.

$s(1) = 4$ and $s(3) = 0$. It moves 4 m forward, then 4 m back, so the distance is 8 m even though the displacement is 0.

Stages of motion

Many questions join two or more stages: a car accelerating, then cruising, then braking. The final velocity of one stage is the initial velocity of the next. A velocity–time sketch makes these questions easier, even when not asked for.