Question

calc help please (:

The position, in meters, of a body at time t sec is s(t)=t^3-6t^2+9t. Find the body's acceleration each time the velocity is zero.

Answer 1

velocity is the derivative of position. so take the derivative of the position function and set the result equal to 0 to get the t-values.

acceleration is the derivative of velocity, so take the derivative of the velocity function you calculated in the previous step, and plug in the t-values you got.

Answer 2

so when i take the derivative and get \[v(t)=3t^2-12t+9\] and then change v(t) to 0, and factor out the 3 common to all terms, i'll get \[0=3(t^2-4t+3)\]
ending with t=3 and t=1. When plugging into the acceleration equation, end with -6m/s^2
and 6m/s^2 right?

Answer 3

yup good