Question

The maximum acceleration attained on the interval 0≤t≤3 by the particle whose velocity is given by v(t) = t^3 - 3(t^2) + 12t + 4 is
(A) 9
(B) 12
(C) 14
(D) 21
(E) 40

Answer 1

Velocity is given by v(t). We want to determine the time at which the acceleration reaches its maximum. Remember that acceleration is the derivative of velocity.

Thus, find this derivative, obtaining an expression for acceleration: a(t).

Now treat a(t) as a new function. Maximize it. Forget about velocity (it's no longer relevant.) Find the derivative of a(t). Set this derivative = to 0. solve for t. That t value is the time at which the acc'n reaches its max. Substitute that t value back into the eq'n for a(t). The result is your maximum acc'n.

Answer 2

so v'(t) = 3t^2 - 6t + 12
then v"(t) = 6t - 6 = 6(t-1)
and when (t-1)=0 --> t = 1
then substitute back in...
v'(0) = 12
v'(1) = 9
v'(3) = 21
so 21 is the answer? @mathmale