Uniformly accelerated motion and motion graphsEdexcel A-Level Physics: Revision notes

Displacement s is the distance moved in a stated direction from a starting point (a vector). Velocity v is the rate of change of displacement, and acceleration a is the rate of change of velocity: a = (v − u)/t when the acceleration is constant. Velocity is in m s⁻¹ and acceleration in m s⁻².

In a straight line, direction is shown by sign: choose one direction as positive and keep to it. A decelerating object moving in the positive direction has a negative acceleration.

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

• v = u + at
• s = ½(u + v)t
• s = ut + ½at²
• v² = u² + 2as

Each equation leaves out one of the five variables, so choose the equation that omits the quantity you neither know nor need. The second equation comes from average velocity = ½(u + v); the others follow by combining it with v = u + at.

List what you know (s, u, v, a, t), identify the unknown, then pick the equation.

Worked example. A car travelling at 20 m s⁻¹ brakes uniformly and stops in a distance of 40 m. Find the deceleration and the stopping time.

Known: u = 20 m s⁻¹, v = 0, s = 40 m. Use v² = u² + 2as: 0 = 400 + 2a(40), so a = −5.0 m s⁻². Then v = u + at gives 0 = 20 − 5.0t, so t = 4.0 s.

For vertical motion under gravity, a = g = 9.81 m s⁻² downwards, so a ball thrown upwards has a = −9.81 m s⁻² if upwards is positive. At the highest point v = 0.

On a displacement–time graph the gradient is the velocity. A straight sloping line means constant velocity, a horizontal line means the object is at rest, and a curve means the velocity is changing.

The velocity at an instant is the gradient of the tangent to the curve at that point. The average velocity over an interval is the gradient of the straight line (chord) joining the two points. A curve getting steeper shows an increasing speed; one flattening shows a decreasing speed. A uniformly accelerating object gives a parabola.

On a velocity–time graph the gradient is the acceleration and the area between the graph and the time axis is the displacement. Areas above the axis are positive and below the axis negative, so the net area is the displacement and the sum of the absolute areas is the distance travelled.

For uniform acceleration the graph is a straight line and areas are triangles, rectangles or a trapezium. For non-uniform acceleration the graph is curved: find the acceleration from the gradient of a tangent, and find the displacement by counting squares or splitting the area into strips.

On an acceleration–time graph the area under the line is the change in velocity.

A steel ball is held by an electromagnet and released when the current is switched off, which starts an electronic timer. The timer stops when the ball breaks a light gate or hits a trapdoor switch. Measure the fall height h with a metre rule for several heights and the time t for each.

Since u = 0, h = ½gt². A graph of h against t² is a straight line through the origin with gradient g/2, so g = 2 × gradient.

Improve accuracy by: repeating each height and averaging; using a large range of heights; measuring h from the bottom of the ball to the trapdoor and avoiding parallax; and using a thin sheet of paper or a non-magnetic release to avoid a delay due to residual magnetism. Electronic timing removes human reaction-time error.