# 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

in that equation is v the change in velocity?

Last edited:
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.

Does anyone know how to derive V^2=V0^2+2as

Ratch
Work energy theorem
KEv2-KEv0=mas