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 2.5-s 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?

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  1. 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)

    Vo = 2.8 -2.5 a
    8 = 2 + 2.5(2.8-2.5a) + 3.125 a
    solve for a

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