Question 989333: A body subject to air resistance is allowed to fall from the top of a 36 m tower. Given that the distance s metres, through which it falls in t seconds, is given by
s = 5 t^2 − t^3 / 3, 0 ≤ t ≤ 3
what is the acceleration after 3 seconds, in m/s2. Please give an exact, possibly fractional, answer (no decimals).
Do I differentiate twice?
So, (10t - 2t^2) / 3
Then (10 - 4t) / 3
I'm stuck!
THANK YOU
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! A body subject to air resistance is allowed to fall from the top of a 36 m tower. Given that the distance s metres, through which it falls in t seconds, is given by
s = 5 t^2 - t^3 / 3, 0 < t < 3
what is the acceleration after 3 seconds, in m/s2. Please give an exact, possibly fractional, answer (no decimals).
Do I differentiate twice?
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Acceleration is the 2nd derivative of displacement.
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So, (10t - 2t^2) / 3 ***** = (10t - 3t^2)/3
Then (10 - 4t) / 3 ***** = (10 - 6t)/3
At t = 3: s" = (10 - 18)/3 = -8/3
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Seems odd that the acceleration is negative.
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Maybe it's:
s = 5t^2 - t^3/3
s' = 10t - t^2
s" = 10 - 2t
s"(3) = 4 m/sec/sec
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I have doubts about the given equation.
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