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An object dropped from restfrom the top of a tall building falls y=16t^2 feet in the first t seconds.

Find the speed of the object at t=3 seconds and confirmalgebraically. The answer is 96 feet/sec but how do i figure it out? cans omeone show me steps? thanks

  • math -

    The given function is
    y(t) = 16 t²
    The speed is the derivative of y with respect to t, i.e.
    y'(t)=dy/dt=16*(2t)=32t
    So the speed at t=3 seconds,
    y'(t)=y'(3)=32*3=96 ft/s.

    To check algebraically, we take the distance travelled at 2.99 and 3.01 seconds, and divide the difference in distance by (3.01-2.99)=0.02 seconds.
    y(2.99)=143.0416 ft
    y(3.01)=144.9616 ft
    approximate speed
    = (y(3.01)-y(2.99))/(3.01-2.99)
    = (144.9616-143.0416)/(0.02)
    = 96 ft/sec. checks with previous calculations

    Note: if you have not done calculus before, please let me know.

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