Equations of uniformly accelerated motionEdexcel International A Level Physics: Revision notes

Uniformly accelerated motion means motion in a straight line with constant acceleration, so the velocity changes by the same amount every second. The five variables are displacement $s$, initial velocity $u$, final velocity $v$, acceleration $a$ and time $t$.

Acceleration is the rate of change of velocity, $a = \frac{v - u}{t}$, in m s⁻². Velocity, displacement and acceleration are vectors, so choose a positive direction and give each quantity a sign.

For constant acceleration:

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

The second equation comes from the area under a velocity–time graph: the average velocity is $(u + v)/2$ because the velocity changes at a constant rate. Combining it with $v = u + at$ gives the other two. Each equation leaves out one variable, which is how you choose between them.

List the three known quantities and the one you want. The equation to use is the one that does not contain the variable you have neither been given nor asked for.

• No $s$: $v = u + at$
• No $a$: $s = \frac{(u + v)t}{2}$
• No $v$: $s = ut + \frac{1}{2}at^2$
• No $t$: $v^2 = u^2 + 2as$

For a body starting from rest, $u = 0$. For a body that stops, $v = 0$. Taking the direction of the initial velocity as positive, a braking vehicle has a negative $a$.

A car brakes uniformly from $24\ \text{m s}^{-1}$ to rest in $6.0\ \text{s}$.

Acceleration: $a = \frac{0 - 24}{6.0} = -4.0\ \text{m s}^{-2}$.

Total distance: $s = \frac{(24 + 0) \times 6.0}{2} = 72\ \text{m}$.

Distance in the first $3.0\ \text{s}$: $s = 24 \times 3.0 - \frac{1}{2} \times 4.0 \times 3.0^2 = 54\ \text{m}$.

Check: speed at 3.0 s is $24 - 4.0 \times 3.0 = 12\ \text{m s}^{-1}$, and $12^2 = 24^2 - 2 \times 4.0 \times 54 = 144$, which agrees.

For an object moving freely under gravity with air resistance negligible, the acceleration is $g = 9.81\ \text{m s}^{-2}$ downwards, whether it moves up, down or is at its highest point.

For an object thrown upwards at $15\ \text{m s}^{-1}$ (upwards positive, $a = -9.81\ \text{m s}^{-2}$):

• At the top $v = 0$, so $s = \frac{u^2}{2g} = \frac{225}{19.62} = 11.5\ \text{m}$
• Time to return to the start: $s = 0$ gives $t = \frac{2u}{g} = 3.06\ \text{s}$

The return speed equals the launch speed (but the velocity is reversed).