We were handed the three constant acceleration equations and told to memorise them. Two of them I can see the sense of: v = u + at is just the definition of acceleration, and s = ut + ½at² comes from the average velocity.
But v² = u² + 2as looks unrelated to either, and I cannot see why there is no t in it. Is it derived from the other two or is it separate?
Accepted Answer
It is the other two with t eliminated, and that is the whole reason it exists.
From v = u + at, t = (v − u)/a. Put that into s = ut + ½at²:
s = u(v − u)/a + ½a(v − u)²/a²
s = (uv − u²)/a + (v − u)²/(2a)
s = [2uv − 2u² + v² − 2uv + u²] / (2a)
s = (v² − u²)/(2a)
Rearranged, v² = u² + 2as. The t is missing because you deliberately removed it, which makes this the equation to reach for when the question never mentions time.
It is the work energy theorem wearing a hat. Multiply through by m/2 and you get ½mv² = ½mu² + mas, and mas is just F·s.
If you want it in one line rather than four, use the chain rule to write acceleration in terms of position instead of time:
a = dv/dt = (dv/ds)(ds/dt) = v dv/ds
So a ds = v dv, and integrating both sides with a constant gives as = (v² − u²)/2. This version also shows you where it stops being true: the moment a is not constant, the left side is ∫a ds and you cannot pull a out.
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