PHYSICS
posted by Ami .
A particle confined to motion along the x axis moves with constant acceleration from x = 2.0 m to x = 8.0 m during a 2.5s time interval. The velocity of the particle at x = 8.0 m is 2.8 m/s. What is the acceleration during this time interval?

v = Vo + a t
x = Xo + Vo t + .5 a t^2
at 8 m
2.8 = Vo + 2.5 a
8 = 2 + 2.5 Vo + .5 a (6.25)
so
Vo = 2.8 2.5 a
8 = 2 + 2.5(2.82.5a) + 3.125 a
solve for a
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