Thread: HELP!!!! NEED HELP FINDING MAXIMUM VELOCITY...

1. 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...

2. $s(t)=-93.75t^3+93.75t^2+45t$
The derivative of position, s(t), gives you velocity

$v(t)=-281.25t^2+187.5t+45$

Now, do you know how to find the maximum of a function?

3. Originally Posted by Linnus
$s(t)=-93.75t^3+93.75t^2+45t$
The derivative of position, s(t), gives you velocity

$v(t)=-281.25t^2+187.5t+45$

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

4. Assuming that you know how to use the power rule...

$s(t)=-93.75t^3+93.75t^2+45t$
The derivative of position, s(t), gives you velocity

$v(t)=-281.25t^2+187.5t+45$

The derivative of velocity is...
$a(t)=-562.5t+187.5$
set it equal to 0
$a(t)=-562.5t+187.5=0$
I hope you can solve t...and get $t=\frac {1}{3}$

Use the first derivative test to see if it is a maximum (it is)

5. Originally Posted by Linnus
Assuming that you know how to use the power rule...

$s(t)=-93.75t^3+93.75t^2+45t$
The derivative of position, s(t), gives you velocity

$v(t)=-281.25t^2+187.5t+45$

The derivative of velocity is...
$a(t)=-562.5t+187.5$
set it equal to 0
$a(t)=-562.5t+187.5=0$
I hope you can solve t...and get $t=\frac {1}{3}$

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

6. ....
$a(t)=-562.5t+187.5=0
$

Solve for t..
$-562.5t=-187.5$

$t=\frac{-187.5}{-562.5}=\frac {1}{3}$

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.

7. so you just plug in v(t) like this to find max velocity

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

8. yes...

9. 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...

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maximum velocity equation calculus

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