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...

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- Nov 26th 2008, 04:22 PMunknown2221HELP!!!! NEED HELP FINDING MAXIMUM VELOCITY...
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... - Nov 26th 2008, 04:25 PMLinnus
$\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 $

Now, do you know how to find the maximum of a function? - Nov 26th 2008, 04:27 PMunknown2221
- Nov 26th 2008, 05:01 PMLinnus
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) - Nov 26th 2008, 05:07 PMunknown2221
- Nov 26th 2008, 05:19 PMLinnus
....

$\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. - Nov 26th 2008, 05:26 PMunknown2221
so you just plug in v(t) like this to find max velocity

-281.25(1/3)^2+187.5t(1/3)^2+45 = ? - Nov 26th 2008, 05:30 PMLinnus
yes...

- Nov 26th 2008, 05:32 PMLinnus
yes...except...where did you get the "t(1/3)^2" in

-281.25(1/3)^2+187.5t(1/3)^2+45 = ?

because that wasn't in the v(t) equation...