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Average velocity question

  1. Oct 19, 2009 #1
    I need to calculate the average velocity of

    d= Et²+Ft+G over the interval [t1;t2]

    I after resolving i get to this function

    avg v = (E(t2)² + Ft2 - E(t1)² + Ft1) / (t2-t1)


    my question is it possible to go any further than this?
     
  2. jcsd
  3. Oct 19, 2009 #2

    Mark44

    Staff: Mentor

    Your average velocity has an error. The Ft1 term in the numerator should be -Ft1. After you fix that, you can simplify your expression.

    avg v = (E(t2)² + Ft2 - E(t1)² - Ft1) / (t2-t1)
    = [E(t22 - t12) +F(t2 - t1)]/(t2 - t1)

    Now factor (t2 - t1) out of each term in the numerator, and you can cancel with the same factor in the denominator.
     
  4. Oct 19, 2009 #3
    yea i for got to distribute the - sign

    would this be it?

    [t2 - t1 (E(t2-t1) + F)]/ t2 - t1 and than the solution would be

    (E(t2-t1) + F)

    correct?
     
  5. Oct 19, 2009 #4

    Mark44

    Staff: Mentor

    No. a2 - b2 = (a - b)(a + b).
     
  6. Oct 19, 2009 #5
    than like this

    [E (t2-t1)(t2+t1) + F (t2-t1)] / t2-t1 =

    than i can cancel the top (t2-t1) with the bottom giving

    E (t2-t1) + F
     
  7. Oct 19, 2009 #6

    Mark44

    Staff: Mentor

    No again.

    Also, you should write your expression this way: [E (t2-t1)(t2+t1) + F (t2-t1)] / (t2-t1)
     
  8. Oct 19, 2009 #7
    crossed out the wrong one, i meant

    [E (t2+t1) + F] correct?
     
    Last edited: Oct 19, 2009
  9. Oct 19, 2009 #8

    Mark44

    Staff: Mentor

    Yes, that's better.
     
  10. Oct 19, 2009 #9
    and than for instantaneous i would take the derivative which is

    2et² + f correct?
     
  11. Oct 19, 2009 #10

    Mark44

    Staff: Mentor

    Yes, but I would use the same letters as in the original problem.
    v = 2Et2 + F
     
  12. Oct 19, 2009 #11
    thanks for your help, now just have to figure out my other thread
     
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