how do you find the maximum velocity of
s(t)=-93.75t^3+93.75t^2+45t
step by step to the answer would help a ton...
Assuming that you know how to use the power rule...
$\displaystyle s(t)=-93.75t^3+93.75t^2+45t $
The derivative of position, s(t), gives you velocity
$\displaystyle v(t)=-281.25t^2+187.5t+45 $
The derivative of velocity is...
$\displaystyle a(t)=-562.5t+187.5 $
set it equal to 0
$\displaystyle a(t)=-562.5t+187.5=0$
I hope you can solve t...and get $\displaystyle t=\frac {1}{3} $
Use the first derivative test to see if it is a maximum (it is)
....
$\displaystyle a(t)=-562.5t+187.5=0
$
Solve for t..
$\displaystyle -562.5t=-187.5 $
$\displaystyle t=\frac{-187.5}{-562.5}=\frac {1}{3} $
I surely hope you knew how to do that.
Also, that is WHERE the maximum velocity occurred at on v(t).
To find the maximum velocity, you would need to plug in 1/3 into V(t).
To determine whether it is a max or a min, you need to do a first derivative test in this case. I cannot teach you how to do it, that is something your teacher should do.