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Velocity time graph

  1. Feb 3, 2013 #1
    1. The problem statement, all variables and given/known data
    http://imageshack.us/photo/my-images/96/caracarbgraph2.png/[PLAIN]


    [b]2. Relevant equations[/b]
    This image is 2 cars, and Car 1 is starting at a higher velocity rate than Car 2, which is constant @ t = 0. Now then, the car is ahead and then decreases speed over time, I am trying to figure out when car 2 well eventually catch up to car 1.


    [b]3. The attempt at a solution[/b]
    I am not sure if there is any equation to this, but it looks like they go the same speed at 5 seconds, the slope of car 1 is 6 m/s^2 (70 m/s - 10 m/s = 60 m/s/10s = 6m/s^2) However I'm not even sure if that actually helps.. since car 2 has a slope of 0.
     
  2. jcsd
  3. Feb 3, 2013 #2

    rollingstein

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    http://img96.imageshack.us/img96/7369/caracarbgraph2.png [Broken]
     
    Last edited by a moderator: May 6, 2017
  4. Feb 3, 2013 #3

    ehild

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    You need to know what was the initial distance between the cars.

    ehild
     
  5. Feb 3, 2013 #4

    rollingstein

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    [STRIKE]40 t = 70 t + 1/2 * (-14) t^2[/STRIKE]

    My bad.

    40 t = 70 t + 1/2 * (-6) t^2
     
    Last edited: Feb 3, 2013
  6. Feb 3, 2013 #5

    ehild

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    That would be true when they start from the same place (if you used the correct acceleration).

    ehild
     
  7. Feb 3, 2013 #6

    rollingstein

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    In the absence of other info. I assumed that, yes.
     
  8. Feb 3, 2013 #7
    Why -14?
     
  9. Feb 3, 2013 #8

    rollingstein

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    What's acceleration?
     
  10. Feb 3, 2013 #9

    ehild

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    That is not -14. Check.

    ehild
     
  11. Feb 3, 2013 #10

    rollingstein

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    Right. My blunder. -6.
     
  12. Feb 3, 2013 #11

    rollingstein

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    40 t = 70 t + 1/2 * (-6) t^2
     
  13. Feb 3, 2013 #12
    It's a quadratic right? How did you know to use that formula?
    so A = -3, B = 30, and C= 0
    = -5 (+/-) SQRT(1) = 0, and 10

    So 10 seconds is the answer.
     
  14. Feb 3, 2013 #13

    rollingstein

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    Which?
     
  15. Feb 3, 2013 #14

    ehild

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    What kind of motions are involved?

    ehild
     
  16. Feb 3, 2013 #15
    Displacement, acceleration, initial velocity, and time?
     
  17. Feb 3, 2013 #16

    ehild

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    Car A moves with constant acceleration. You should know the formula for displacement in terms of time.

    ehild
     
  18. Feb 3, 2013 #17
    x =v0t + ½*at2

    But it was setup that x was 40t because the distance was needed between the 2 cars correct?
     
  19. Feb 4, 2013 #18

    ehild

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    If the set-up was that both cars started from the same position x=0, then the position of car A is xA=70t-3t2 as function of the time t and the position of car B is xB=40t. Car A leaves car B at the start, but slows down and car B catches it at the end. At that instant, both cars are at the same place again, xA=XB. 40t=70t-3 t2

    ehild
     
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