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Homework Help: Kinematics 1 dimensional motion acceleration question

  1. Aug 15, 2010 #1
    A race driver has made a pit stop to refuel. After refueling, he starts from rest and leaves the pit area with an acceleration whose magnitude is 6.8 m/s2; after 4.1 s he enters the main speedway. At the same instant, another car on the speedway and traveling at a constant velocity of 66.0 m/s overtakes and passes the entering car. The entering car maintains its acceleration. How much time is required for the entering car to catch up with the other car?

    b.2 Equations used
    and other equations

    3. The attempt at a solution

    distance = v2=u2+2as
    s= v2 - u2/2a
    s= 263.1m

    The answer is t = 11.2s

    I am really stuck guys. I understand the final velocity of the entering car must be greater to catch up to the car travelling constant acceleration, but the displacement is the same. But I am finding it difficult to proceed without knowing the final velocity of entering car or the displacement.

    Can someone offer a worked example for me - with different values?
  2. jcsd
  3. Aug 15, 2010 #2
    Try breaking up the problem into two parts:
    1- accelerating up to the race track edge, right when the passing vehicle is aside the car.
    2- accelerating up to the vehicle ahead, to when the car catches up.

    Hint: What distance has the accelerating car traveled when it's at the track edge? Can this be used to find the velocity of the accelerating car at that moment? You are right about the displacement of the two cars being the same when they've caught up, but it is from this point, not the start of the pit-stop car's acceleration.
  4. Aug 16, 2010 #3
    Thanks for the suggestion, I might have to wait for my lecturer to get back to me, I am just burnt out on this problem. I just don't know how to work out the displacement when I don't know the accelerating car's final velocity when overtaking the car, and cannot seem to understand..

    My assumption that entering car's final velocity isn't equal to 66m/s is the point where I lose the problem..
  5. Aug 16, 2010 #4
    Don't give up. Try this:
    d = ut + 1/2at^2
    where u = 0m/s, a = 6.8m/s^2, t = 4.1 s
    Calculate d
    Then try using:
    v^2 = u^2 + 2ad
    Then go from there.

    Get back to me with your solution.
    Last edited: Aug 16, 2010
  6. Aug 16, 2010 #5
    1. d = 57.15m
    2. v = 27.87m/s

    Thanks, I understand where I went wrong with D now. I am still just very daft with how to find out velocity of the car when it catches up with the other car. I know it must be some assumption that I'm missing. thanks so much for your help!
  7. Aug 16, 2010 #6
    Excellent. You're almost there! Keep going!
    Now you can equate the displacements of each car (like you said in your first post, "the displacement is the same") from that point and solve for the time it takes them to reach the distance at which they catch up.
    Hint: Da = Vat and Db = Ubt + 1/2at^2
    where Da = displacement of car 'a', Va = velocity of car 'a', and similar for car 'b'
    and Da = Db so...

    I should be more clear, car 'a' is the one passing at this moment. Car 'b' is the pit-stop car. Hope I didn't confuse you.
    Last edited: Aug 16, 2010
  8. Aug 16, 2010 #7
    D = VT
    = 57.15/66.0
    = 0.9s

    D= ut + 1/2 at2
    57.15 = 0t + 1/2 (6.8) (t2)
    t2 = 114.3/6.8 = 16.8
    t = 4.1s

    oops that looks familiar

    D= ut + 1/2 at2
    57.15 = 27.87t + 1/2 at2

    is this the one I should use... but that looks weird too

    the answer said t = 11sec

    So sorry :/ ....
  9. Aug 16, 2010 #8
    Don't be sorry, you're so close. You're doing a good job. I'll give you a little nudge.

    Da = Vat = 66t
    Db = 27.87t + 1/2(6.8)t^2 = 27.87t + 3.4t^2
    Da = Db, plug in for each what we have above
    66t = 27.87t + 3.4t^2
    Do you know how to solve this for t?

    Remember that the first time the question gave was for the pit-stop car to travel to the race track (t = 4.1s was for the 'first' part of the question), not to catch up. Now that we know the speed of each car at the moment they are side by side, and the acceleration of car 'b', we can calcluate the time t it takes the cars to meet up from the moment the one passes the other.
  10. Aug 16, 2010 #9

    I thought I did but remembered I cant make it = 2t2, as I can't add them. Ah yes, I see what I did with Da, silly me.

    You don't have to explain about finding t, you've already spent so much time on me! if you could send me a link or something that'd be wonderful :)
  11. Aug 16, 2010 #10
    I'm enjoying helping you. And you're so close already. Let's continue, shall we? You're just a hop, skip, and a jump away from the answer.

    First thing, subtract 27.87t from both sides of the equation. What do you get?
  12. Aug 16, 2010 #11
    66t/28.87 = 2.368

    will ts cancel as they are both associated so,

    66t-27.87t = 3.4t^2
    38.13 = 3.4t^2
    t^2 = 11.21 !!!!

  13. Aug 16, 2010 #12
    and then hang on.... might be too soon

    t^2 = 11.21
    t = 3.35
  14. Aug 16, 2010 #13
    I think it's just a typo, since you understood to cancel the common t, but it should read:

    38.13 = 3.4t
    11.2 s = t

    You're welcome. Your interest and tenacity got you the right answer. Kudos.

    Oh, and btw. Did you notice how you didn't even need the final velocity of the pit-stop car to ascertain the time asked for in the question?

    Happy studying.
    Last edited: Aug 16, 2010
  15. Aug 16, 2010 #14
    Thanks again - you really, really helped my confidence :)
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