- #1

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find the average velocity for the time period beginning when t=2 and lasting

1. .5 sec

2. .1 sec

3. .05 sec

4. .01 sec

- Thread starter afcwestwarrior
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- #1

- 457

- 0

find the average velocity for the time period beginning when t=2 and lasting

1. .5 sec

2. .1 sec

3. .05 sec

4. .01 sec

- #2

berkeman

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Post your work so far.

- #3

Gib Z

Homework Helper

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So ima bend the rules abit and give him abit of help.

You have a function for displacement. Do some magic and make that into one for velocity. Find the area underneath the periods of time concerned, and squash it into a rectangle with the width of the period of time, the height will be the average value.

Sorry I cheated berkeman, Ill see you guys in a while. Been fun hanging with you all.

- #4

HallsofIvy

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How high is the ball at t= 2 s? How high is it at t= 2.5s? How high is it when t= 2.1 s? How high is it when t= 2.05 s? How high is it when t= 2.01 s?

Of course the average velocity is the difference in heights divide by the time interval.

(If this problem had asked for average

- #5

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- #6

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- #7

HallsofIvy

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Do you understand what that formula says? It doesn't do you any good to memorize formulas if you don't understand what they are saying. "s(T)" in the formula is the position at time T. Here, you are throwing a ball up so s(T) is the HEIGHT at time T. You said "height in feet after t seconds is given by y=40t-16tsq." by which I think you mean 40t+ 16t^2 (in ASCII) or

40t- 16t

Now: y(2)= 40(2)- 16(2)

y(2.5)= 40(2.5)- 16(2.5)

y(2.1)= 40(2.1)- 16(2.1)

y(2.05)= 40(2.05)- 16(2.05)

y(2.01)= 40(2.01)- 16(2.01)

Okay, you "plugged 2 into the height formula" and got 16

Finally, the ".5", ".1", ".05", and ".01"

- #8

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i got 0, then i got 13.44 and so on

- #9

HallsofIvy

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Are you under the impression that this willi got 0, then i got 13.44 and so on

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