Displacement, velocity and acceleration, the constant-acceleration (suvat) equations, motion under gravity, displacement-time and velocity-time graphs, and using calculus when acceleration varies.

What this dot point is asking

OCR wants you to define displacement, velocity and acceleration, use the constant-acceleration (suvat) equations including for vertical motion under gravity, interpret displacement-time and velocity-time graphs (gradients and areas), and use calculus (differentiation and integration) to handle motion where the acceleration is not constant.

The answer

Displacement, velocity and acceleration

Displacement is position relative to a fixed point (a vector). Velocity is the rate of change of displacement, and acceleration is the rate of change of velocity. In a straight line we treat them as signed quantities, with one direction chosen as positive.

The suvat equations

When acceleration is constant, the five quantities $s$ (displacement), $u$ (initial velocity), $v$ (final velocity), $a$ (acceleration) and $t$ (time) are linked by the standard equations.

Motion under gravity

A body moving freely under gravity has constant acceleration $g = 9.8$ m s$^{-2}$ downwards. Choose a positive direction and keep signs consistent: if up is positive then $a = -9.8$, and at the highest point $v = 0$.

Motion graphs

On a displacement-time graph the gradient is velocity. On a velocity-time graph the gradient is acceleration and the area under the graph is the displacement. Reading gradients and areas turns a graph into the motion.

Examples in context

Choosing the right suvat equation

The skill is matching the equation to the four quantities involved. List $s, u, v, a, t$, mark what you know and what you want, and pick the equation missing the one variable you neither know nor need.

Variable acceleration and calculus

When acceleration varies with time the suvat equations do not apply; instead, differentiate to go from displacement to velocity to acceleration, and integrate to go back. To find total distance (not displacement) when the direction changes, split the motion at the times when $v = 0$ and add the magnitudes.

Try this

Q1. A car travels from rest at $a = 2$ m s$^{-2}$ for $5$ s. Find its final speed and the distance covered. [3 marks]

• Cue. $v = 0 + 2(5) = 10$ m s$^{-1}$; $s = \tfrac{1}{2}(2)(5^2) = 25$ m.

Q2. A particle has displacement $s = 2t^3 - 9t^2$. Find its acceleration at $t = 2$. [3 marks]

• Cue. $v = 6t^2 - 18t$, $a = 12t - 18$, so at $t = 2$, $a = 6$ m s$^{-2}$.