Average velocity question

  • Thread starter vysero
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  • #1
vysero
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Here is my problem:

A stone is tossed in the air from ground level with an initial velocity of 20m/s. Its height at time t is h(t) = 20t-4.9t^2 m. Compute the stones avg velocity over the time interval [1.5,2.5].

I understand how to compute average velocity its Vavg=change in distance/change in time. However, I am running into a problem trying to figure a change in distance from the equation they gave me.

I tried searching in the library for a similar question but could not find one.
 

Answers and Replies

  • #2
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What is the height at time t = 2.5 s?
What is the height at time t = 1.5 s?
How much time elapsed between time t = 1.5 s and time 2.5 s?
 
  • #3
vysero
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Oh I get it so I plug those values into the formula to get the height and then I compute the average velocity? Cool thanks!
 
  • #4
HallsofIvy
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Important point- if you throw a rock from ground level so that it goes to a very great height, then comes down and hits the ground again, obviously it must have had a large velocity at certain time. But since it started at ground level and got back to ground level, its net distance traveled is 0 and its average velocity over that time is 0.
 

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