Motion along a straight lineAQA A-Level Physics: Revision notes

Displacement $s$ is the straight-line distance from the start in a stated direction (a vector); distance is the total length of path (a scalar).

• Speed = distance ÷ time (scalar); velocity = displacement ÷ time (vector): $v = \Delta s / \Delta t$
• Acceleration is the rate of change of velocity (vector): $a = \Delta v / \Delta t$

The average velocity is the total displacement divided by the total time. The instantaneous velocity is the velocity at one moment, found as the gradient of the displacement–time graph at that moment, i.e. $\Delta s/\Delta t$ for an extremely short interval.

Worked example. A runner completes a 400 m lap in 50 s and returns to the start: average speed = 8.0 m s⁻¹ but average velocity = 0, because the displacement is zero.

• Displacement–time: the gradient is the velocity. A straight line means constant velocity; a curve means changing velocity; a horizontal line means at rest.
• Velocity–time: the gradient is the acceleration and the area between the graph and the time axis is the displacement.
• Acceleration–time: the area under the graph is the change in velocity.

For non-uniform acceleration the graph is a curve; use a tangent to find the gradient at a point and count squares to find an area.

Bouncing ball (upward positive): the velocity–time graph is a straight line with gradient −9.81 m s⁻² while the ball is in the air, because the acceleration is constant. At each bounce the velocity changes suddenly from negative to positive with a smaller magnitude as energy is lost; the displacement–time graph is a series of curved arcs.

For constant acceleration along a straight line, with initial velocity $u$, final velocity $v$, time $t$ and displacement $s$:

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

Choose the equation that does not contain the quantity you neither know nor need. Take one direction as positive and give signs to every vector.

Worked example. A cyclist accelerates from rest at 1.2 m s⁻² for 10 s: $v = 0 + 1.2 \times 10 = 12$ m s⁻¹ and $s = \tfrac{1}{2} \times 1.2 \times 10^2 = 60$ m.

Near the Earth's surface, all objects in free fall (with air resistance negligible) have the same downward acceleration, $g = 9.81$ m s⁻². For an object dropped from rest, $s = \tfrac{1}{2}gt^2$. For an object thrown upwards at $u$, the maximum height is found from $v^2 = u^2 + 2as$ with $v = 0$ and $a = -g$.

Worked example. Thrown up at 15 m s⁻¹: $0 = 15^2 - 2 \times 9.81 \times h$, so $h = 11.5$ m.

A steel ball is held by an electromagnet at height $h$ above a trapdoor. When the current is switched off a timer starts; it stops when the ball opens the trapdoor, giving the fall time $t$. Since $u = 0$:

$h = \tfrac{1}{2}gt^2 \quad\Rightarrow\quad t^2 = \frac{2}{g}h$

Plot $t^2$ against $h$: the gradient is $2/g$, so $g = 2/\text{gradient}$.

• Random errors (scatter in t): repeat each drop and take the mean; use a wide range of heights.
• Systematic errors: a constant delay (e.g. the magnet releasing late) gives a non-zero intercept; an incorrectly measured h also shifts the line.
• Measure h with a ruler to the bottom of the ball and fix the ruler's zero at the trapdoor.