Equations of motion Examples

Example 1: dropped object

A ball is dropped from rest from a height of $20\mathrm{m}$. Ignore air resistance. Take downward as positive, so $u=0$, $a=9.81\mathrm{m}{\mathrm{s}}^{-2}$, $s=20\mathrm{m}$.

Use the no-time equation:

$$v^2=u^2+2as=0+2(9.81)(20)=392.4$$ $$v\approx 19.8\mathrm{m}{\mathrm{s}}^{-1}$$

The impact speed is about $19.8\mathrm{m}{\mathrm{s}}^{-1}$ downward.

Example 2: braking car

A car moving at $15\mathrm{m}{\mathrm{s}}^{-1}$ brakes uniformly to rest over $30\mathrm{m}$. Let forward be positive: $u=15$, $v=0$, $s=30$.

$$v^2=u^2+2as$$ $$0=15^2+2a(30)$$ $$a=-3.75\mathrm{m}{\mathrm{s}}^{-2}$$

The negative sign means acceleration points opposite the chosen positive direction, so the car slows down.

See Equations of motion Equations and definitions for the formula set and Equations of motion Common pitfalls for sign checks.

Example 3: acceleration from rest

A trolley starts from rest and accelerates at $2\mathrm{m}{\mathrm{s}}^{-2}$ for $4\mathrm{s}$. Then $u=0$, $a=2$, $t=4$.

$$s=ut+\frac{1}{2}a{t}^2=0+\frac{1}{2}(2)(4^2)=16\mathrm{m}$$

Its final velocity is $v=u+at=8\mathrm{m}{\mathrm{s}}^{-1}$.

Mini sanity check

If the object starts from rest, $u=0$, then $v=at$ and $s=frac12a{t}^2$. That is the classic “dropped or launched from rest” simplification. If your algebra still contains $u$ after setting the object initially at rest, you probably forgot to substitute it early.