Acceleration ???

Question

Q) An object travels along a straight line so that its distance s at time t is given by s = 3t^3 - 9t^2 + 7t - 2  for  t ≥ 0

At what time is its acceleration zero?

 

Do you just differentiate twice and make the function equal zero and then solve? Would differentiating once and doing this solve for velocity?

Answer 1

s = 3t^3 - 9t^2 + 7t - 2

Velocity of the object, v =ds/dt

v =d(3t^3 - 9t^2 + 7t - 2)/dt

Using the formula:(d/dt)(t^n) =nt^(n-1) and (d/dt)(c) =0

v =9t^2-18t+7

Acceleration of the object, a=dv/dt

a =d(9t^2-18t+7)/dt

  =18t-18

Given condition:acceleration =0

a=0

18t-18=0

t =1 s