Question

The velocity of a particle moving along the x-axis for....

Answer 1

\[t \ge0\]is given by\[v_{x}=24-3t^{3}\](a) What is the particle's acceleration when it first achieves a velocity of zero?
(b) What is the particle's acceleration when it achieves its maximum displacement in the +x-direction?

Answer 2

(a) Find when the velocity is zero...

0 = 24 - 3t^3
3t^3 = 24
t^3 = 8
t = 2

then plug 2 into acceleration equation, which you find by taking derivative of velocity...

a(t) = -9t^2
a(2) = -9(2)^2
a(2) = -36

(b) To find the maximum displacement you set the slope of the tangent line = 0...
aka x'(t) = v(t) = 0.

Seems like we just did this and got a = -36

Answer 3

Thank you Piglet for showing me how to do this.