The minimum acceleration attained on the interval #0<=t<=4# by the particle whose velocity is given by #v(t)=t^3-4t^2-3t+2# is?

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I think I should take the derivative of this velocity function to get a function for acceleration. Then find zeros of it. But after that, I don't know what to do.

Thank you!

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Answer 1 · 2018-04-11T03:53:05

The minimum acceleration is #-25/3# m/s^2.

You are right about the first couple of steps. Differentiate to get the acceleration function which will be

#a(t) = 3t^2 - 8t - 3#

Now you need to get the minimum of the accerlation function, so differentiate again.

#a'(t) = 6t - 8#

#0 = 6t - 8#

#a = 2(3t - 4)#

#t = 4/3#

This is a minimum because, #a'(t) < 0# to the left of #t# and #a'(t) > 0# to the right of #t#.

Thus, the minimum acceleration occurs at #t = 4/3# and this accerlation would be #a(4/3) = 3(4/3)^2 - 8(4/3) - 3 = 16/3 - 32/3 - 3 = -25/3#

Hopefully this helps!