Motion graphsEdexcel International A Level Maths: Revision notes

On a displacement–time graph the gradient is the velocity. A straight line means constant velocity, a horizontal line means the object is at rest, and a curve means the velocity is changing (the gradient of a tangent gives the instantaneous velocity). A negative gradient means motion back towards the start. Example: a displacement rising from 0 to 20 m in 5 s has gradient $20/5=4$ m s⁻¹; falling from 20 m to 0 in 6 s gives $-20/6=-3.33$ m s⁻¹.

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, areas below are negative. For total distance, add the sizes of all the areas. A horizontal line is constant velocity (zero acceleration). Example: velocity rising uniformly from 0 to 12 m s⁻¹ in 4 s, then constant for 6 s, then falling uniformly to 0 in 6 s has distance $\frac12(4)(12)+(6)(12)+\frac12(6)(12)=132$ m.

A speed–time graph has gradient equal to the rate of change of speed, and its area is always the distance travelled (speed is never negative). On an acceleration–time graph the area is the change in velocity. Example: a lift with acceleration 1.5 m s⁻² for 4 s gains $1.5\times4=6$ m s⁻¹; then 3 s at $-2$ m s⁻² loses 6 m s⁻¹, so it stops.

Distance is the total length of path travelled; displacement is the straight-line change in position, with direction. Average speed $=\dfrac{\text{total distance}}{\text{total time}}$. A cyclist who rides 20 m away and 20 m back has travelled 40 m but has a displacement of 0. In the sprinter example, average speed is $132/16=8.25$ m s⁻¹, not $(0+12)/2$.

A journey with a uniform acceleration, a constant-speed phase and a uniform deceleration gives a trapezium. With acceleration $a_1$, maximum speed $V$ and deceleration $a_2$, the times are $V/a_1$ and $V/a_2$, the remaining time is at constant speed, and the total distance is the trapezium area. Setting this equal to the given distance may give a quadratic; reject any root that makes a time negative. Example: $a_1=0.5$, $a_2=1$, $T=160$ s, $D=2600$ m gives $3V^2-320V+5200=0$, so $V=20$ m s⁻¹ (the root $86.7$ is rejected).