v² = u² + 2as
A third equation with no time in it.
How far before it stops?
A driver sees an obstacle and brakes. We know the car's velocity when braking begins and how strongly the brakes slow it, but we do not know for how long it brakes.
Which equation do you think could still give the stopping distance?
What this lesson covers
The idea
Eliminating t from the first two equations gives v² = u² + 2as, relating initial and final velocities, acceleration and displacement for constant acceleration.
How far before it stops?
A driver sees an obstacle and brakes. We know the car's velocity when braking begins and how strongly the brakes slow it, but we do not know for how long it brakes.
Which equation do you think could still give the stopping distance?
- One that has no time in it
- One that needs the time
- None, the time is needed
Braking distance
Set the velocity and the braking acceleration.
Time is eliminated
From v = u + at we get t = (v − u)/a. Putting this into s = ut + ½at² removes t and leaves v² = u² + 2as. For a car brought to a stop, v = 0, so 0 = u² + 2as and s = u²/8 when a = −4 m s⁻². The car at 15 m s⁻¹ needed 28.1 m, and the car at 30 m s⁻¹ needed 112.5 m.
Eliminating t from the first two equations gives v² = u² + 2as. It relates the initial and final velocities, the acceleration and the displacement for constant acceleration.
Notes
v² = u² + 2as: obtained by eliminating t from the first two equations; no time needed.
Check yourself
A car moving at 15 m s⁻¹ brakes with a = −4 m s⁻² until it stops (v = 0). How far does it travel? Give your answer to one decimal place.
Answer: 28.1 m
s = 15² ÷ 8 = 225 ÷ 8 = 28.125 m, about 28.1 m.
Which quantities are related by v² = u² + 2as?
How is v² = u² + 2as obtained from the first two equations?
A car at 15 m s⁻¹ stops in 28.1 m with a given braking acceleration. With the same braking, a car at 30 m s⁻¹ needs…
- Initial velocity, final velocity, acceleration and time. Those are the quantities of v = u + at. In v² = u² + 2as the displacement replaces the time.
- Initial velocity, final velocity, acceleration and displacement — correct. Yes!
- Displacement, time, initial velocity and acceleration. Those are the quantities of s = ut + ½at². In v² = u² + 2as the final velocity replaces the time.
- By adding the two equations. Adding does not remove t. The book puts t = (v − u)/a into the second equation.
- By making the acceleration zero. If a were zero there would be no acceleration left in the equation.
- By eliminating t — correct. Yes!
- about four times the distance (112.5 m) — correct. Yes!
- about twice the distance. Doubling u makes u² four times larger, so the stopping distance is four times larger.
- the same distance. The stopping distance depends on the velocity at which the brakes are applied.