HELP!!!! NEED HELP FINDING MAXIMUM VELOCITY...

• Nov 26th 2008, 04:22 PM
unknown2221
HELP!!!! 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 PM
Linnus
$\displaystyle s(t)=-93.75t^3+93.75t^2+45t$
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 PM
unknown2221
Quote:

Originally Posted by Linnus
$\displaystyle s(t)=-93.75t^3+93.75t^2+45t$
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?

can you do it step by step to the answer so i can see how to do it right

thanks
• Nov 26th 2008, 05:01 PM
Linnus
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 PM
unknown2221
Quote:

Originally Posted by Linnus
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)

so after -562.5t+187.5=0
its 187.5 X -3 = 562.5t

can you show me how the max and min is found leading to the max velocity please
• Nov 26th 2008, 05:19 PM
Linnus
....
$\displaystyle a(t)=-562.5t+187.5=0$

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 PM
unknown2221
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 PM
Linnus
yes...
• Nov 26th 2008, 05:32 PM
Linnus
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...