Quantities and units in mechanics, displacement, velocity and acceleration, motion graphs, and the constant-acceleration (suvat) equations including vertical motion under gravity.

What this dot point is asking

WJEC wants you to use the correct quantities and units in mechanics (SI units, scalars and vectors), to define displacement, velocity and acceleration, to read and sketch motion graphs, and to apply the constant-acceleration equations (the suvat equations) including vertical motion under gravity. This is the foundation of the mechanics section and feeds the forces work and the A2 calculus kinematics.

The answer

Quantities and units

Mechanics is built on SI units: length in metres, time in seconds, mass in kilograms. Velocity is in $\text{m s}^{-1}$, acceleration in $\text{m s}^{-2}$, and force in newtons.

Motion graphs

Splitting a velocity-time graph into triangles and rectangles is the quickest way to find displacement when the motion has several phases.

The constant-acceleration equations

For motion in a straight line with constant acceleration:

Here $u$ is initial velocity, $v$ final velocity, $a$ acceleration, $s$ displacement and $t$ time. Choose the equation that contains your three knowns and the one unknown, so you avoid simultaneous equations.

Vertical motion under gravity

A body moving vertically under gravity has acceleration $g \approx 9.8\,\text{m s}^{-2}$ directed downward. Choose a positive direction and apply the suvat equations with $a = g$ (or $-g$ if up is positive). At the highest point of an upward throw the velocity is momentarily zero.

Examples in context

Example 1. A dropped object. A stone is dropped from rest down a well and hits the water after $2\,\text{s}$. With $u = 0$ and $a = 9.8$, the depth is $s = \tfrac{1}{2}(9.8)(2)^2 = 19.6\,\text{m}$, and the impact speed is $v = 9.8 \times 2 = 19.6\,\text{m s}^{-1}$. One suvat equation gives the depth, another the speed.

Example 2. Reading a velocity-time graph. A cyclist accelerates uniformly from rest to $12\,\text{m s}^{-1}$ in $6\,\text{s}$, holds it for $10\,\text{s}$, then stops in $4\,\text{s}$. The total displacement is the area: $\tfrac{1}{2}(6)(12) + (10)(12) + \tfrac{1}{2}(4)(12) = 36 + 120 + 24 = 180\,\text{m}$. The area under the graph captures all three phases at once.

Try this

Q1. A car travels at constant $25\,\text{m s}^{-1}$ for $8\,\text{s}$. Find the distance covered. [2 marks]

• Cue. Constant velocity, so $s = vt = 25 \times 8 = 200\,\text{m}$.

Q2. A ball is dropped from rest. Find its speed after $3\,\text{s}$ (take $g = 9.8\,\text{m s}^{-2}$). [2 marks]

• Cue. $v = u + at = 0 + 9.8 \times 3 = 29.4\,\text{m s}^{-1}$.

Q3. A particle decelerates uniformly from $30\,\text{m s}^{-1}$ to rest in $120\,\text{m}$. Find the deceleration. [3 marks]

• Cue. $v^2 = u^2 + 2as$: $0 = 900 + 2a(120)$, so $a = -3.75\,\text{m s}^{-2}$ (a deceleration of $3.75\,\text{m s}^{-2}$).