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An average velocity problem

  1. Jan 30, 2007 #1
    if a ball is thrown up into the air with a velocity of 40ft/s it's height in feet after t seconds is given by y=40t-16tsq.
    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. jcsd
  3. Jan 30, 2007 #2


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    Staff: Mentor

    Post your work so far.
  4. Jan 30, 2007 #3

    Gib Z

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    Homework Helper

    This is going to be my last post for a while i think, ima be busy for a while, So cya guys den!

    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.
  5. Jan 30, 2007 #4


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    Since it asks for average velocity, you don't need to use calculus.
    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 speed the answer might have been different!)
  6. Jan 30, 2007 #5
    i know the formula s(T)-S(T)/time elapsed, but i plug in the numbers and i can't figure out s, and s is 4.9m per second, but that's only for m/s problems, i have the answers for this problem, but i don't understand this problem
  7. Jan 30, 2007 #6
    here's what i did i plugged 2 into the height formula and got 16 and i used that for seconds, then i did this 16(.5)sq.-16(2)sq./1.5 and i don't get the 1st answer which is 32 feet per second, i get
  8. Jan 31, 2007 #7


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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- 16t2. Starting at t= 2 seconds and "lasting .5 second" means from t= 2 to t= 2.5. "lasting .1 seconds" means from t= 2 to t= 2.1", etc.
    Now: y(2)= 40(2)- 16(2)2
    y(2.5)= 40(2.5)- 16(2.5)2= what?
    y(2.1)= 40(2.1)- 16(2.1)2= what?
    y(2.05)= 40(2.05)- 16(2.05)2= what?
    y(2.01)= 40(2.01)- 16(2.01)2= what?

    Okay, you "plugged 2 into the height formula" and got 16 feet. That's height, not time- you can't "use that for seconds!" Is the "16" in "16(.5)^2- 16(2)^2/1.5 that 16? Where did you get THAT formula? Since you are told that y= 40t- 16t^2 your "s(T)-s(t)/time elapsed" should be y(T)-y(t)/time elapsed= (40T-16T^2-(40t-16t^2))/time elapsed- although it is easier to do what I suggested above- calculate the postions at each of those times and subtract y(2) from each rather that doing y(2) over and over again.

    Finally, the ".5", ".1", ".05", and ".01" are "time elapsed", not specific times (the problem said "lasting"). The specific ending times of each interval, starting at t= 2 and "lasting" those periods, are 2.5, 2.1, 2.05, and 2.01
  9. Feb 1, 2007 #8
    i got 0, then i got 13.44 and so on
  10. Feb 2, 2007 #9


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    Are you under the impression that this will mean anything to someone reading it? You got "0, then 13.44 and so on" for WHAT? What calculation are you doing? Please show exactly what you are doing.
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