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Kinematics Equation Problem!

  1. Sep 17, 2013 #1
    1. The problem statement, all variables and given/known data

    Sam is running at 3.8 m/s and is 75m behind John who is running at a constant velocity of 4.2m/s. If Sam accelerates at 0.15m/s^2, how long will it take him to catch John?

    2. Relevant equations

    d = (vf+vi/2)t
    vf = vi + at
    d = vit + 1/2at^2
    vf^2 = vi^2 + 2ad
    d = vft - 1/2at^2


    3. The attempt at a solution

    I have asked this before, but have now come back with a little more understanding than before.
    I've done the following so far:

    Xs (displacement of Sam) = viΔt + 1/2aΔt^2
    = (3.8m/s)Δt + (0.075m/s^2)Δt^2

    Xj (displacement of John) = 75 + 4.2m/sΔt

    [-> from: (xf-xi) = VavΔt
    xf = xi + VavΔt ]

    ---

    ... Ok so now I do (3.8m/s)Δt + (0.075m/s^2)Δt^2 = 75 + 4.2m/sΔt
    to find the time when they are in the same position/displacement, correct?

    But I'm so confused as to what I would do after that?

    EDIT; Also, I got 1005.33s for the time, and I don't think that's right?
     
    Last edited: Sep 17, 2013
  2. jcsd
  3. Sep 17, 2013 #2
    What you got is a quadratic equation where delta t is unknown. Solve it.
     
  4. Sep 17, 2013 #3
    ax^2 + bx + c = 0?

    This may be a stupid question but how would I know which variable goes for which coefficient?
    (I need to brush up on my math...)
     
  5. Sep 17, 2013 #4
    ##x## is the unknown. In your case, the unknown is ##\Delta t##. You should be able to match the other symbols.
     
  6. Sep 17, 2013 #5
    Okay I think I get it but then what was this for: (3.8m/s)Δt + (0.075m/s^2)Δt^2 = 75 + 4.2m/sΔt ?
    I thought I was trying to find time with that?
     
  7. Sep 17, 2013 #6
    I do not understand your questions. What does ##\Delta t## mean to you?
     
  8. Sep 17, 2013 #7
    time, is it not?
     
  9. Sep 17, 2013 #8
    It is time, and it is what you are after. What is not clear to you?
     
  10. Sep 17, 2013 #9
    Well, my teacher said I had to find the position/displacement of both Sam and John, which I did:
    (3.8m/s)Δt + (0.075m/s^2)Δt^2 for Sam and
    75 + 4.2m/sΔt for John

    and if I do: (3.8m/s)Δt + (0.075m/s^2)Δt^2 = 75 + 4.2m/sΔt
    won't that give me time? So why would I have to do the quadratics equation?
    Or am I doing something wrong here...
     
  11. Sep 17, 2013 #10
    Because the equation you got is a quadratic equation. It will give you time - when you solve it.
     
  12. Sep 17, 2013 #11
    OH, now I understand what my teacher was saying!
    I got 32s, is that right?
     
  13. Sep 17, 2013 #12
    You can check your answer yourself: you have the formula for displacement for Sam, and the formula for John. Do you get the same displacements at 32 s?
     
  14. Sep 17, 2013 #13
    I got 34.4 seconds
     
  15. Sep 17, 2013 #14
    (3.8m/s)(32s) + (0.075m/s^2)(32s)^2 = 198.4 m
    75 + 4.2m/s(32s) = 209.4 m

    so I am a bit off...
    this is what i did:
    ---
    (3.8m/s)Δt/t + (0.075m/s^2)Δt^2 = 75 + 4.2m/sΔt/t
    3.8m/s - 3.8m/s + (0.075m/s^2)Δt^2 = 75 + 4.2m/s - 3.8m/s
    (0.075m/s^2)Δt^2 / 0.075m/s^2 = 75 + 4.2m/s - 3.8m/s / 0.075m/s^2
    Δt^2 = 75 + 4.2m/s - 3.8m/s / 0.075m/s^2
    (square root)Δt = (square root)1005.33 s

    Δt = 32 s
     
    Last edited: Sep 17, 2013
  16. Sep 18, 2013 #15
    I do not know what you are doing, but you are not solving the equation correctly.

    First you need to bring it to the standard form: ## a \Delta t^2 + b \Delta t + c = 0 ##.

    Then use the (supposedly) well-known formula for its roots.
     
  17. Sep 18, 2013 #16
    Okay I finally got it. Thank you so much and sorry for all the troubles!
     
  18. Sep 18, 2013 #17
    No trouble at all. I hope this has been useful for you :)
     
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