Very confuse on this?

Question

A particle moves along a straight line with equation of motion s = t5 − 5 t4 Find the value of t (other than 0 ) at which the acceleration is equal to zero.

Answer 1

s = t^5 − 5 t^4

If s is position, ds / dt will be velocity...

s' = 5t^4 - 20t^3

...and the next derivative will be acceleration...

s'' = 20t^3 - 60t^2

You want to know where the acceleration is zero

0 = 20t^3 - 60t^2
0 = 20t - 60
60 = 20t
3 = t

The acceleration is zero at time t = 3.

Answer 2

is the formula s=t^5-5t^4 ?
If so, all you have to do is find the second derivative

The derivative of the distance (s) formula is its velocity (v) formula. The derivative of this velocity formula equals the acceleration (a) formula.

So the first derivative of t^5 - 5t^4 is 5t^4 - 20t^3 (i'm assuming you know the power rule to derive functions)

and the second derivative is 20t^3 - 60t^2.
Set this equation equal to zero, as this is the formula for acceleration and you want to find out what "t" is then "a" is zero.
Therefore:
20t^3 - 60t^2 = 0
20t^2(t - 3) = 0
t - 3 = 0
t = 3 t = 0(from dividing out the 20t^2)
so, acceleration is zero when time is 3. Hope this helped.