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

The derivative of position, s(t), gives you velocity

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...

The derivative of position, s(t), gives you velocity

The derivative of velocity is...

set it equal to 0

I hope you can solve t...and get

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
....

Solve for t..

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...