Kinematics

2.1 Equations of motion

Definitions

A car going once round a circular track has travelled a large distance but has zero displacement, and its average velocity is zero.

Motion graphs

For a curved displacement–time graph, the velocity at an instant is the gradient of the tangent at that point. Area below the time axis on a velocity–time graph counts as negative displacement.

Deriving the equations

These equations apply only when the acceleration is constant and the motion is in a straight line. Let $u$ be the initial velocity, $v$ the final velocity, $a$ the acceleration, $t$ the time and $s$ the displacement.

From the definition of acceleration, $a = (v-u)/t$, so:

$v = u + at$

The area under the velocity–time line is a trapezium, so $s = \tfrac{1}{2}(u+v)t$. Substituting for $v$ and for $t$ in turn gives:

$s = ut + \tfrac{1}{2}at^2 \qquad v^2 = u^2 + 2as$

Measuring the acceleration of free fall

Drop a steel ball through a measured height $h$, starting a timer when it is released and stopping it when it reaches a trapdoor or light gate at the bottom. From rest, $h = \tfrac{1}{2}gt^2$.

Repeat for several heights and plot $h$ against $t^2$. This is a straight line through the origin with gradient $\tfrac{1}{2}g$, so $g = 2 \times \text{gradient}$. Using a graph averages out random errors, and an intercept reveals a systematic error such as a delay in releasing the ball.

Projectile motion

When an object moves with constant velocity in one direction and constant acceleration at right angles to it, treat the two directions separately. They share only the time.

For a projectile with no air resistance:

• Horizontally: no force, so the velocity is constant and $s_x = u_x t$.
• Vertically: acceleration $g$ downwards, so use the equations of motion.

The path is a parabola. At the top, the vertical velocity is zero but the horizontal velocity is unchanged, so the speed is not zero there.