Physics

The makers of a certain automobile advertise that it will accelerate from 15 to 50 mi/hr in high in 13 sec. Compute (a) the acceleration in ft/ sec2, and (b) the distance the car travels in this time, assuming the acceleration to be constant.

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  1. V1 = 15mi/h * 5280Ft/mi * 1h/3600s = 22Ft/s.
    V2 = 50/15 * 22 = 73.3 Ft/s.

    a. a = (V2-V1)/t=(73.3-22)/13 = 3.95 Ft/s.

    b. d = V1*t + 0.5*a*t^2.
    t = 13 s.
    Solve for d.

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