In need of help with calc

Question

A particle moves along a straight line with equation of motion s=t^6-5t^5. Find the value of t (other than 0) at which the acceleration is equal to zero.

my work:

s=t^6-5t^5

v=6t^5-25t^4

a=30t^4-100t^3

0=30t^4-100t^3

10(3t^4-10t^3)=0 I am not sure if this is even right but I don't know where to go from here.

Answer 1

This looks good so far. But you haven't taken out the greatest common factor which will allow you to solve it.

0=30t⁴-100t³

0=10t³(3t-10)

Set both factors equal to zero and solve.

10t³=0

t=0 (They don't want this one)

3t-10=0

3t=10

t=10/3