Expert: Socrates - 5/14/2006

Question
If I can please see steps to this as well....

If s(t)=.5t^4-5t^3+12t^2, find velocity when acceleration is zero.

Do I take the antiderivative of the acceleration?

Answer
You don't take the antiderivative of acceleration , because you are given the position function to begin with.

The first derivative of position gives velocity

v(t) = 2t^3 - 15t^2 + 24 t

The derivative of velocity gives acceleration

a(t) = 6t^2 - 30t + 24


We want velocity when acceleration is 0. So we need to find out when acceleration is 0. Set the expression for acceleration equal to 0 and get

6t^2 - 30t + 24 = 0

t^2 - 5t + 4 = 0

(t-1)(t-4) = 0

so acceleration is 0 when t=1 or t=4

To find the velocity when acceleration is 0 , substitute these values for t into the expression for velocity.

v(1) = 2 - 15 + 24 = 11

v(4) = (2)(64)- (15)(16) + (24)(4) = -16


the two velocities are 11 and -16