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Another Catch Problem I'm lost at

  1. Sep 13, 2006 #1
    Another Catch Problem I'm lost at!!!

    My physics teacher is spitting these catch problems at me like a maniac. Here's the question (it was on my test and I was partially clueless):

    A girl is trying to running as fast as she can at a constant velocity of 4.5 m/s . She is trying to catch her boyfriend's car which is at rest behind the red light. Just when the girl is 10m behind the car, the light changes and the car accelerates at 1 m/s^2. How far will the girl have to run in order to catch the car?

    Also 'she can not catch the car' was one of the choices in the multiple answers.

    In order for the girl to catch the car, they need to be at the same position in the same time. I sketched a d/t graph for this. The graph simply summarizes my thoughts on this problem. Can someone please walk me through the calculations needed to solve this problem.

    Also, if someone out there has some tips about these kinds of problems please tell them to me. I'm really struggling at these.
     

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  3. Sep 13, 2006 #2

    chroot

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    Write an equation for the position of the runner, in terms of time.

    Write an equation for the position of the car, in terms of time.

    Set them equal to each other (since the two positions must be equal for a "catch" to occur).

    Solve for time -- the time at which the positions of car and runner are equal.

    Use that value of time in either the car or the runner equation to find the position at that time.

    - Warren
     
  4. Sep 13, 2006 #3
    Oh my god thank you chroot!

    I did what you said and ended up with t = 9s for the time at which they are at the same position. Then I solved for the position at which they meet and it was 40.5m. So the girl ran 50.5m

    Now can someone confirm that's the right answer?! Just so I can relax and jump up and down in my room! :D
     
  5. Sep 13, 2006 #4

    chroot

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    I think perhaps you wrote you equations for the car and runner incorrectly, because your answer is different from mine.

    Can you please write down the equations you used?

    - Warren
     
  6. Sep 13, 2006 #5
    I used the following two equations to calculate time:

    (sorry for the crappy mathematical expressions, I don't know a fast way to type equations)

    d = ((Vf+Vi)/2)t for the runner, so:

    d = (4.5) t

    then,

    d = Vi(t) + (1/2)(a)(t)^2 for car. Since Vi = 0 and a = 1 then:

    d = (1/2)t^2

    Therefore:

    (1/2)t^2 = 4.5 t

    thus

    t = 9s

    That's what I did...I am famous for making stupid math errors so if I did something wrong just point it out I'm used to it :)
     
  7. Sep 13, 2006 #6

    chroot

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    So, the equations you found are:

    car(t) = 0.5 t^2
    runner(t) = 4.5t

    Your equations are not quite right, since you forgot to include the car's 10m head start.

    - Warren
     
  8. Sep 13, 2006 #7
    Yes, I found where the problem is. Thank you anyways :)

    I did forget to calculate the 10 m head start...I'm so happy now... :D
     
  9. Sep 13, 2006 #8

    chroot

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    So what was your final answer? (Just to be sure.)

    - Warren
     
  10. Sep 13, 2006 #9
    I found two answers for t...one 4 and the other is 5.

    We only care about 4, so then after the calculations we end up with the fact that the girl will catch the bus after running for 18 m. At that time the bus has traveled 8 m. The difference is 10. :)

    And I know this answer is right because I just recieved the same answer from many of my classmates.
     
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