A small rocket shot from the ground experiences an acceleration of 3 - 2t ft/s2.

Question

A small rocket shot from the ground experiences an acceleration of 3 - 2t ft/s2. Find a formula for the height h of the rocket, in feet, after t seconds, if its initial velocity is 40 ft/s. h(t) = What is the maximum height reached by the rocket? h = ft When is the rocket at this point? t = sec

Explanation / Answer

a) a(t) = 3 - 2t => v(t) = 3t - t^2 + 40 => h(t) = (3/2)t^2 - (1/3)t^3 + 40t b) h'(t) = v(t) = 3t - t^2 + 40 => t^2 - 3t - 40 = 0 => (t - 8)(t + 5) = 0 => t = 8 or t = -5 h(8) = (3/2)(8^2) - (1/3)(8^3) + 40(8) = 245.33 c) t = 8

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