## motion, constant acceleration

A car slows down from 23 m/s to rest in a distance of 85m. what was its acceleration, assumed constant?

a=Δv/Δt x=1/2at^2

i don't know where to start
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 what is the velocity-displacement formula?
 v^2=2ax

## motion, constant acceleration

in that equation is v the change in velocity?
 v^2=2ax
 okay thanks i guess my teacher messed up
 I'm sorry, I just gave you some bad advice...ignore my first response.
 I don't know what I was thinking, but yeah you use (final velocity)^2 = (initial velocity)^2 + 2*a*x and just substitute the stuff you know and solve for a.
 Does anyone know how to derive V^2=V0^2+2as Ratch
 Take the velocity-time equation: $$v = v_{0} + a \, t$$ and the position time equation: $$x = v_{0} \, t + \frac{1}{2} \, a \, t^{2}$$ and eliminate time t.

 Quote by Ratch Does anyone know how to derive V^2=V0^2+2as Ratch
Work energy theorem
KEv2-KEv0=mas