# Thread: Calc I: velocity problem

1. ## Calc I: velocity problem

Suppose an arrow is shot upward on the moon with a velocity of 39 m/s, then its height in meters after t seconds is given by h(t)=39t-0.83t^2. Find the average velocity over the given time intervals.

[8, 9]:
[8, 8.5]:
[8, 8.1]:
[8, 8.01]:
[8, 8.001]:

I understand that you need to figure out slope and simplify but my answer never comes out correctly.

2. Originally Posted by 151
Suppose an arrow is shot upward on the moon with a velocity of 39 m/s, then its height in meters after t seconds is given by h(t)=39t-0.83t^2. Find the average velocity over the given time intervals.

[8, 9]:
[8, 8.5]:
[8, 8.1]:
[8, 8.01]:
[8, 8.001]:

I understand that you need to figure out slope and simplify but my answer never comes out correctly.
really? are you sure you are using the right equation?

note, the average velocity over the time interval $[a,b]$ is given by

$\text{Avg Vel. } = \frac {h(b) - h(a)}{b - a}$

3. yeah I'm using that equation but I changed it to look like this:
[y(8+h)-y(8)]/[(8+h)-8]

4. maybe my algebra skills are just horrible but for the average velocity equation, I simplifiied it and got this:
[-.83h^2+25.72h+273]/h

Is that even close?

5. Originally Posted by 151
yeah I'm using that equation but I changed it to look like this:
[y(8+h)-y(8)]/[(8+h)-8]
why go through all that trouble?

Originally Posted by 151
maybe my algebra skills are just horrible but for the average velocity equation, I simplifiied it and got this:
[-.83h^2+25.72h+273]/h

Is that even close?
and no, it is not correct, but it is close.

...are you sure there should be 273 there?

6. The reason I altered to velocity equation is because that's what the example problem in the book did. If you use the original equation, how do you plug in the intervals?

For the first interval, would it look like this?
(39-.83(9)^2)-(39-.83(8)^2)/(9-8)

7. Originally Posted by 151
The reason I altered to velocity equation is because that's what the example problem in the book did. If you use the original equation, how do you plug in the intervals?

For the first interval, would it look like this?
(39-.83(9)^2)-(39-.83(8)^2)/(9-8)
yes, but you can continue doing it the first way, since you already went through all the trouble

you have: $\frac {25.72h - 0.83h^2}h$

your h values are 1, 0.5, 0.1, 0.01, 0.001

8. Okay, thanks I got it now.

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# suppose an arrow is shot upward on the moon with a velocity of

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