Kinematics in One Dimension

≈ 25 min

Kinematics describes motion without asking what causes it. Along a single straight line we track position $x(t)$ , and define

v = \frac{dx}{dt}, \qquad a = \frac{dv}{dt} = \frac{d^2x}{dt^2}

Velocity is the rate of change of position; acceleration is the rate of change of velocity. Both are signed: the sign records direction along the chosen axis, so you must choose a positive direction before you substitute a single number.

When $a$ is constant, integrating twice gives the three equations of motion:

v = v_0 + at \qquad \Delta x = v_0 t + \tfrac{1}{2}at^2 \qquad v^2 = v_0^2 + 2a\,\Delta x

The third is the first two with $t$ eliminated — useful whenever time is neither given nor wanted.

Worked example. A minibus taxi pulls away from a rank from rest with a constant acceleration of $2.5\ \text{m/s}^2$ for $8.0\ \text{s}$ .

Speed after $8.0$ s:

v = v_0 + at = 0 + (2.5)(8.0) = 20\ \text{m/s}

Distance covered:

\Delta x = v_0 t + \tfrac{1}{2}at^2 = 0 + \tfrac{1}{2}(2.5)(8.0)^2 = 80\ \text{m}

A cross-check with the third equation: $v^2 = 0 + 2(2.5)(80) = 400$ , so $v = 20\ \text{m/s}$ — consistent.

A bakkie accelerates uniformly from rest to $20\ \text{m/s}$ in $8.0\ \text{s}$ . What is its acceleration, in $\text{m/s}^2$ ?

On a velocity–time graph, the area between the graph and the time axis represents which quantity?

The displacement of the object

The acceleration of the object

The average speed of the object

The rate of change of acceleration

A stone is dropped from rest off a bridge over the Vaal River and reaches the water $2.0\ \text{s}$ later. Taking $g = 9.8\ \text{m/s}^2$ and ignoring air resistance, how far did it fall, in metres?

A minibus travelling at $25\ \text{m/s}$ brakes with a constant acceleration of $-5.0\ \text{m/s}^2$ until it stops. How many metres does it travel while braking?