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Kinematics: when will the two cars meet?

  1. Nov 4, 2013 #1
    1. The problem statement, all variables and given/known data
    Two cars start from rest and travel in opposite directions.
    Car A travels at 6m/s^2 to a maximum of 30m/s. Car B travels 5m/s^2 to maximum of 35m/s. 200m between them. When will they meet?

    2. Relevant equations
    d=vt
    d=vit + 1/2 at2


    3. The attempt at a solution
    at 6 seconds:
    car A:
    d=vit + 1/2 at2 +30m
    =105m

    Car B:
    d=vit + 1/2 at2 = 90m

    5m left in between them. How do I find when they meet?
     
  2. jcsd
  3. Nov 4, 2013 #2
    Clearly the cars will meet between 5 and 7 seconds. Since t = 5 s, car A moves uniformly, car B keeps accelerating. It may be convenient to compute the distance at t = 5 s.
     
  4. Nov 4, 2013 #3
    At 5s:
    Car 1
    d= 75m

    Car 2
    d = 62.5m

    Yes, I know they'll be meeting in between 6 and 7 seconds but how do I find the exact time?
     
  5. Nov 4, 2013 #4
    How much distance is there to cover yet? What is the combined speed of converging between 5 and 7 seconds?
     
  6. Nov 4, 2013 #5

    NascentOxygen

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    If in t more seconds of travel car A moves a distance "d" in the same time that car B travels a distance "5-d" they will meet at time t at that position d.
     
  7. Nov 4, 2013 #6
    62.5m left. what do you mean by combined speed of converging between 5 and 7 seconds?
    Why are we looking for the combined speed?
     
  8. Nov 4, 2013 #7
    No it doesn't work that way because their speed is not constant.
     
  9. Nov 4, 2013 #8
    If you know the combined speed, which is uniformly accelerated, then finding the time to cover 62.5 m is very simple.
     
  10. Nov 4, 2013 #9
    Ok, I still don't understand why I need to combine the speed. One car is going at 30m/s after 6 seconds and the other is not yet going at 35m/s because it takes 7 seconds for it to reach maximum velocity..
     
  11. Nov 4, 2013 #10

    NascentOxygen

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    It works even when their speed is not constant.
     
  12. Nov 5, 2013 #11

    haruspex

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    It's the definition of 'meeting'. They are at the same place at the same time.
     
  13. Nov 7, 2013 #12
    But why is it 5-d? car A is accelerating at 6m/s^2 and car B is accelerating at 5m/s^2. At t= 1, they are not at 30m/s nor 35m/s respectively. It takes car A 5 seconds to reach 30m/s and then the speed stays constant. Car B takes 7 sec to reach maximum velocity at 35m/s.

    Please help!
     
  14. Nov 7, 2013 #13

    haruspex

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    Because in the OP you calculated that after 6 seconds they would still be 5m apart. So if the point where they will meet is another d metres from A then it's 5-d from B.
    Why is t=1 interesting? Take it from how things stand after 6 seconds:
    5 metres apart
    A at constant 30m/s
    B currently at 30m/s and still accelerating
    You can either do it by relative motion:
    What is their relative velocity?
    What is their relative acceleration?
    How long will it take to close the gap, assuming B's acceleration doesn't change?
    or by supposing they meet t seconds later, A having traveled d and B having traveled 5-d, and write out the equations for that.
     
  15. Nov 7, 2013 #14
    oK, I don't really understand relative velocity, so I am going to suppose they meet t seconds later.

    For Car B:
    D=Vit +1/2 at^2 so D= 5-d, vi= 0 and a= 5m/s^2. I isolated t and equated to Car A.

    t=d/30 = sqrt((10-2d)/5) ? Solve for d, then solve for t?
    is that the correct way to approach it?
     
  16. Nov 7, 2013 #15

    haruspex

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    Think again about vi.
    What corresponding equation do you have for car A?
     
  17. Nov 7, 2013 #16
    T= D/30 for car A
     
  18. Nov 7, 2013 #17

    haruspex

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    OK. What about that vi for car B?
     
  19. Nov 9, 2013 #18
    vi= 30m/s? Ok, I think I got the answer. But it's a bit complicated to do it this way. Can somebody explain the relative velocity way? Thank you
     
    Last edited: Nov 10, 2013
  20. Nov 10, 2013 #19
    At times greater than 5 seconds, car 1's speed is 30 m/s, and its distance is 75 +30(t-5) = 30t - 75
    At times greater than 5 seconds (but less than 7 seconds), car 2's distance is 2.5 t2. If they meet between 5 and 7 seconds, the sum of these two distances must total 200.
     
  21. Nov 12, 2013 #20
    how do I do it with relative velocity?
     
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