s(t)=t-3/t^2 when is the velocity 0? (t is equal to time)?

Question

i found the velocity with no problem. but i forget how to find the time when velocity is 0. can someone help?

Answer 1

s'(t) =(t-3)(-2t^{-3}) + t^{-2}
=( -2t+6 +t)/ t^{-3}
=(6-t) / t^{-3}
so the velocity is 0 when the top =0
i.e 6-t-0
t=6
s

Answer 2

ok, so you find the derivative for the velocity, which would be [(t)(t^2) - (t-3)(2t)]/t^4 = (t^2 - 2t - 3)/t^3. Now you set that equal to 0

0 = (t^2 - 2t - 3)/t^3
t^3 = 0
t = 0
OR
0 = t^2 - 2t - 3
0 = (t -3)(t + 1)
t = -1, 3

so t = 0, 3 s

Answer 3

if you found the velocity then you know that the S is distance and the velocity function is the first derivative.
Take the first derivative ( dS/dt = 1-6t) and set it equal to 0. Solve for t.

Answer 4

0=t-3/t^2
3=t^3
t=3^(1/3)

Answer 5

s'(t) =(t-3)(-2t^{-3}) + t^{-2}
=( -2t+6 +t)/ t^{-3}
=(6-t) / t^{-3}
6-t-0
t=6
s