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Helping friend

  1. Mar 21, 2005 #1
    I am trying to help a friend through this problem

    train problem - 2 trains going oposite direction leaving from same station - train A going 120 km/hr (velocity)
    - train B going 20 km / hr / hr (acceler) Time when are they 500 km apar

    What i did was

    Trying to solve

    [tex] 500= \int_{0}^{t} (20x^2+120x) [/tex]

    however when I put that into the solver I am getting negitive answers
    with guess and check with the integral I cam out to around 2.549
     
  2. jcsd
  3. Mar 21, 2005 #2
    So then I guess you could do

    [tex] 500= \frac{20t^3}{3}+60t^2}-(\frac{20x^3}{3}+60x^2}) [/tex]

    [tex] 500=\frac{20t^3}{3}+60t^2} [/tex]

    but hmm when you solve that you still get negitive answers
    What am I doing wrong
     
  4. Mar 21, 2005 #3
    lol I just inserted the
    [tex] \frac{20t^3}{3}+60t^2 [/tex] into y1
    then 500 into y2
    find where they intersect
    it should be 2.54841

    [edit--- this is wrong]
     
    Last edited: Mar 21, 2005
  5. Mar 21, 2005 #4
    Can someone please show me why I can't go back and do basic math??
     
  6. Mar 21, 2005 #5
    I know you can solve it with .5*20*x^2+120x=500 but the thing I don't get is why you reduce the dimesnions of the original equations when putting it under the integral
     
  7. Mar 21, 2005 #6
    Set up a differential equation

    [tex] x^{\prime \prime} (t) = 20 \frac{\mbox{km}}{\mbox{h}^2}[/tex]

    with ICs [tex] x^\prime (0) = 120 \frac{\mbox{km}}{\mbox{h}}, \ x(0) = 0[/tex] and solve for [itex]x(t)[/tex]. Once you've done that, just plug in [itex] x(t_0) = 500 \mbox{km}[/itex] and solve for [itex]t_0[/itex].
     
    Last edited: Mar 21, 2005
  8. Mar 21, 2005 #7
    The reason that your integral didn't work is that it is set up wrong. Why would you integrate distance (although your integrand isn't quite distance either, but it's close enough for me! :smile:) to get distance?

    If you want it illustrated more clearly that your method doesn't work, then just tell me:What are the units of your [tex]\left(20 \frac{\mbox{km}}{\mbox{h}^2}\right)t^3[/tex] term? Do they match the units of [itex]500 \mbox{km}[/itex]?
     
    Last edited: Mar 21, 2005
  9. Mar 21, 2005 #8

    Doc Al

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    Staff: Mentor

    Don't know why you are integrating (instead of just using kinematics equations), but this is the integral you want:
    [tex] 500= \int_{0}^{t} (20t +120) dt [/tex]

    Time is in hours, of course.
     
    Last edited: Mar 21, 2005
  10. Mar 21, 2005 #9
    Time to nitpick:

    To clean up notation, you should actually use

    [tex] \int_0^t (20s + 120) ds[/tex]

    or some other random variable instead of [itex]s[/itex]
    ~
     
  11. Mar 21, 2005 #10
    Thanks... yeah I got that now... I had added an extra dimension for no reason. And I agree the kinematic equations work much easyier but he wasn't allowed to do it. We got what we think is the right answer now of [tex] \sqrt{86}-6[/tex]
     
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