15.4 The constant-acceleration formulae — Practise with help

1. $v - u = at$, so $t = \dfrac{v - u}{a}$.
2. $20 + 24 = 44$.
3. $5t = 30$, so $t = 6$.

If question 1 was hard, revise rearranging formulae before you continue. Every formula on this page has to be rearranged at some point.

You know $u = 4$, $a = 6$ and $t = 5$. Use the formula that has those three letters and the one you want.

For the velocity, use $v = u + at$. For the displacement, use $s = ut + \frac{1}{2}at^2$.

"From rest" means the initial velocity is $0$. So $u = 0$, $a = 4$ and $t = 7$.

With $u = 0$, the formula $s = ut + \frac{1}{2}at^2$ simplifies to $s = \frac{1}{2}at^2$.

The final velocity is $0$, because the car stops. So $u = 20$, $v = 0$ and $t = 8$.

Use $v = u + at$ for the acceleration. For the distance, use $s = \dfrac{u + v}{2}t$.

The acceleration is negative, as expected for braking.

You know $u = 8$, $v = 28$ and $s = 100$. The time is not wanted first, so use the formula without $t$.

Use $v^2 = u^2 + 2as$ for the acceleration. Then use $v = u + at$ for the time.

With upwards positive, $u = 15$ and $a = -9.8$. Use $t = 2$.

Use $v = u + at$ for the velocity, then $s = ut + \frac{1}{2}at^2$ for the height.

The velocity is negative, so the ball is already falling at $2$ s.

At the greatest height the ball is instantaneously at rest, so $v = 0$.

Use $v^2 = u^2 + 2as$ with $v = 0$ and $a = -9.8$ to find $s$. Then use $v = u + at$ to find $t$.

The car stops, so $v = 0$. With forwards positive, $a = -5$.

Use $v^2 = u^2 + 2as$ for the distance, then $v = u + at$ for the time.