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Average Velocity between time to time

  1. Jan 21, 2009 #1
    1. The problem statement, all variables and given/known data
    I have to find the average velocity of an equation from time 2 to time 4. This equation here is of the position.
    P = -5*t^3+5*t^2 -4*t.
    2. The attempt at a solution

    Since this is the position i have to take derivative of that.
    -15t^2-10t-4

    Then i substitute 2 for t and i get -44 then i substitute 4 and get -204

    ( -44 + -204) / 2 = -124 average velocity. Is this correct? The website where i submit my answers to says it's wrong.
    Thanks in advance.
     
  2. jcsd
  3. Jan 21, 2009 #2

    PhanthomJay

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    Your equation for average velocity is valid for constant acceleration only. In general, average velocity must be calculated using [tex]\Delta s/\Delta t [/tex]
     
  4. Jan 21, 2009 #3
    It looks like you took your derivative wrong. I think the second term should be positive, ie [tex]-15t^2+10t-4[/tex]
     
  5. Jan 21, 2009 #4

    i'm sorry that was a typo
    but yes though the position formula is = 5*t2 -4*t -5*t3.

    thats 10t-4-15t^2
    10x2 - 4 - 15x2^2
    20 - 4 - 15x4
    16 - 60
    -44

    and

    10x4 - 4 - 15x4^2
    40 - 4 - 240
    36 - 240
    -204

    (-44 + -204)/2
    -124

    so what is the average velocity between 2 and 4 seconds?
    and i punch in -124 and nothing :(
     
  6. Jan 21, 2009 #5
    i found it by guessing starting at -100 and going negative one by one.. it said ok at -113

    how can that be?
     
    Last edited: Jan 21, 2009
  7. Jan 21, 2009 #6

    PhanthomJay

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    Try re-reading my first response.:wink:
     
  8. Jan 21, 2009 #7
    If you have access to calculus, just use mean value theorem.
    If not, listen to phantom jay.
     
  9. Jan 21, 2009 #8
    Jay is right. Since acceleration is changing in time you just need to divide the total change in position by the total time, ie

    [tex]\frac{\Delta S}{\Delta T} = \frac{-256-(-28)}{2} = -113[/tex]
     
    Last edited: Jan 21, 2009
  10. Jan 21, 2009 #9
    hhmm so what goes where? I'm just starting physics. Its actually my 4th day in the class and we have like 3 formulas that are all for constant acceleration. The funny thing is what is on our homework is much more advanced vs what we're learning in class.
     
  11. Jan 21, 2009 #10
    Maybe the letters are confusing you.

    In physics, [tex]\Delta S[/tex] means Total change in position. And [tex]\Delta T[/tex] means Total time elapsed. So to get the -256 number you just plug in 2 for t in your position equation (the one you were given). To get the -28 number, you plug in 4 for t.

    Physically, these two numbers are your position in coordinate space at t=2 and t=4. You want the difference between these two values (as indicated by [tex]\Delta[/tex]) so you subtract them. You find the total time elapsed by subtracting the values you plugged in for t (4 - 2 = 2).

    Think about it this way:

    If you take 1 hour to walk 5 miles, you walked an average of 5 miles per hour. It doesn't matter how many times you stopped or started...whether you ran really fast sometimes and then slower other times...overall you took 1 hour to go 5 miles. So by dividing 5 miles by 1 hour you get the average velocity of 5 miles per hour.

    You're doing the same thing here except that instead of hours you're using seconds. Instead of "5 miles" you are given an equation which plots your position. But it's the same concept exactly.

    Cheers.
     
  12. Jan 21, 2009 #11
    Ahh that makes sense. i get it now :)
    Thank you all for your responses. I appreciate the help a ton.

    This forum is definitely going in the favorites. :)
     
  13. Jan 22, 2009 #12

    PhanthomJay

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    Well, I see that things haven't changed since I last took Calc based Physics 201 some 40 years ago. The teaching concentration by far is on constant acceleration problems, but then they throw the other harder stuff at you , and you're on your own. But sometimes, it's the best way to learn.
     
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