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Homework Help: Kinematic (suvat) equations?

  1. Dec 16, 2007 #1
    okay, this is simple and i feel ridiculous for asking, but i'm reviewing for my final and i do not remember how to do this type of problem..

    Runner A runs toward a flag pole at a constant speed of 5.0 m/s. Runner B begins to run toward the flagpole at 3.0 m/s^2 when Runner A is 15 m ahead. When will Runner B catch Runner A?

    I know I'll be setting two distance equations equal to each other, but beyond that I don't know what to do.
     
  2. jcsd
  3. Dec 16, 2007 #2

    Hootenanny

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    Have you come across any kinematic (suvat) equations?
     
  4. Dec 16, 2007 #3
    yeah, from kinematics:

    D: distance travelled (m)
    A: acceleration m/(s^2)
    t: time (s)
    D0: initial distance

    D = 0.5*A*t^2 + V*t + D0

    if you start your stop watch (t=0) when runner B starts to run, then runner A is already 15 m away. therefore:

    for runner A:
    DA = 5*t + 15

    for runner B:
    DB = 0.5*3*t^2

    where
    DA: distance travelled by runner A
    DB: distance traveleed by runner B

    set DA = DB, you get a quadratic.
    solve the quadratic and the right answer is 5.24 s

    So it takes runner B be 5.24 s to catch up to runner runner A after he/she starts to run.

    If you start your stopwatch when runner A starts to run, then your time is 5.24 + 15/5 = 8.24 seconds

    Just double check my math please. hope this helps.
     
  5. Dec 16, 2007 #4
    Runner B needs to cover some extra distance so are you sure about the equation for runner B?
     
  6. Dec 16, 2007 #5
    I think he accounted for it with Runner A ( +15)
     
  7. Dec 17, 2007 #6
    Actually yes, I am sorry, I didn't see that.
     
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