calculus help velocity

Question

the position of an object at time T is given s(t)=2t^3-13^2+24t+100. a) find the velocity and the acceleration of the particle at t=1 b) find the times at which the particle comes to a rest

Answer 1

s(t) = 2t³ - 13t² + 24t + 100

a) If s(t) = position, then velocity and acceleration are:

v(t) = ds(t)/dt = 6t² - 26t + 24

a(t) = dv(t)/dt = ____?

b) When the object comes to rest, v(t) = 0

v(t) = 6t² - 26t + 24

0 = 6t² - 26t + 24

0 = 2·(3t² - 13t + 12)

Use factoring or the quadratic formula to solve for t. There are two answers.