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2 January, 22:38

A model rocket is fired vertically upward from rest. its acceleration for the first three seconds is a (t) = 72t, at which time the fuel is exhausted and it becomes a freely "falling" body. sixteen seconds later, the rocket's parachute opens, and the (downward) velocity slows linearly to - 28 ft/s in 5 s. the rocket then "floats" to the ground at that rate. (a) determine the position function s and the velocity function v (for all times t

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  1. 3 January, 00:48
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    Velocity function = V (T) = Integral of 72t

    = ∫72xdx

    = 72∫xdx

    Now solving for ∫xdx

    Apply the power rule

    ∫x^ndx = xn + 1 / n + 1 with n=1:

    = x^2 / 2

    Plug that in the solved integrals:

    = 72∫xdx

    = 36 x^2

    So ...

    72t^2/2 = 36 (t^2) = this is the velocity function

    Position = S (T)

    = Integral of 36 (t^2)

    = ∫36x^2dx

    = 36∫x^2dx

    Now solving for ∫x^2dx

    Apply the power rule

    ∫x^ndx = xn + 1 / n + 1 with n = 2:

    = x^3 / 3

    Plug that in the solved integrals:

    = 36∫x^2dx

    = 12x^3

    So the position of S (t) = 12 (t^3)
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