Sometimes the time is neither given nor wanted: a car's braking distance, a jet's runway length. You could find $t$ first from the constant-acceleration equations, but one more equation skips that step.
Key idea
$v^2 = v_0^2 + 2a\Delta x$ links the speeds and the displacement without the time. It also needs constant acceleration.
Why it works
Start from the area under the straight $v$–$t$ line: a trapezoid, $\Delta x = \tfrac12(v_0 + v)\,t$. The first equation gives $t = \dfrac{v - v_0}{a}$. Substitute: $$\Delta x = \dfrac{(v + v_0)(v - v_0)}{2a} = \dfrac{v^2 - v_0^2}{2a}.$$ Multiply both sides by $2a$ and the time is gone: $v^2 = v_0^2 + 2a\Delta x$.
Worked examples
Worked example: Stopping distance
A car moving at $25\ \text{m/s}$ brakes at $5\ \text{m/s}^2$. How far does it go while stopping?
- Take the motion as positive: $v_0 = 25$, $v = 0$, $a = -5\ \text{m/s}^2$.
- $0 = 25^2 + 2(-5)\Delta x$, so $\Delta x = \dfrac{625}{10} = 62.5\ \text{m}$.
Answer: $62.5\ \text{m}$
Worked example: Speed after a stretch
A skier at $4\ \text{m/s}$ speeds up at $2\ \text{m/s}^2$ over $30\ \text{m}$ of slope. Find the final speed.
- $$v^2 = 4^2 + 2(2)(30) = 16 + 120 = 136.$$
- $v = \sqrt{136} \approx 11.7\ \text{m/s}$.
Answer: about $11.7\ \text{m/s}$
Worked example: Required acceleration
A jet must reach $70\ \text{m/s}$ from rest on a $1{,}400\ \text{m}$ runway. What acceleration does it need?
- $70^2 = 0 + 2a(1400)$.
- $a = \dfrac{4900}{2800} = 1.75\ \text{m/s}^2$.
Answer: $1.75\ \text{m/s}^2$
| Equation | Leaves out |
|---|---|
| $v = v_0 + at$ | $\Delta x$ |
| $\Delta x = v_0 t + \tfrac12 a t^2$ | $v$ |
| $v^2 = v_0^2 + 2a\Delta x$ | $t$ |
| $\Delta x = \tfrac12(v_0 + v)\,t$ | $a$ |
Because $\Delta x = \dfrac{v_0^2}{2|a|}$ when stopping, doubling the speed makes the braking distance four times as long, not twice. Traffic engineers build that square into stopping-distance tables.
Common mistake
Forgetting the 2 in $2a\Delta x$, or stopping at $v^2$ and forgetting the square root. Also, the square root only gives a size: $v^2 = 36$ means $v = +6$ or $-6\ \text{m/s}$, and you choose the sign from the direction of motion. Check: units of $2a\Delta x$ are $\text{m}^2/\text{s}^2$, the same as $v^2$.