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At what time will runner B catch up with runner A?

  1. Sep 7, 2004 #1
    Runner A, who runs with an average speed of 3.0 m/s, starts out at 3:00 P.M. Runner B, who runs with an average speed of 4.0 m/s, starts after A from the same place exactly 5 min later.
    a.) At what time will runner B catch up with runner A?
    b.) If the runners stop when B catches A, how far do they run?

    So far I have 4.0 m/s*300 s = 1.2 Km and 3.0 m/s x 300s= 0.9 km, but I don't even know if this is the right aproach, can anyone help me from here?
     
  2. jcsd
  3. Sep 7, 2004 #2

    JasonRox

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    They are two linear equations(with one solution I believe), find the point of intersection.

    I will not tell you the answer, but merely try to open your eyes.

    You know the velocities of A and B.

    What formula do you use to find the velocity?

    Try to combine the formulas to come up with an answer.

    Hint: You need to use algebra. The distance travelled must be equal for both because that is when B catches up to A.

    Show me some more work, and I'll be back to give more advice.

    Note: I can tell you more right now, like most people, but doing it on your feels so much better.
     
  4. Sep 7, 2004 #3
    Trying

    so....
    If I let the time A runs be "t", then the time B runs is "t- 5".
    ...4.0 m/s= t-5
    and 3.0= t

    So... 4.0 m/s= 3.0-5
    I'm I on the right track?
     
  5. Sep 7, 2004 #4

    JasonRox

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    Huh?

    To take that route I believe you will need to convert the minutes into seconds since your velocity is 4.0m/s.

    Look at what you wrote. You said 4=3-5. We know that's incorrect already.

    The formula for velocity is [tex]v=d/t[/tex].

    Because there is two different velocities, there is two different equations to work with.

    With the two equations that you get, combine them.

    Note: Remember, d will be the same for both.
     
  6. Sep 7, 2004 #5
    Still trying

    so...for B v=1.2 km/ 3.0

    and for A v= .9 km/3.0, IF RIGHT, WHAT UNITS IS 3.0 IN, HR?
     
  7. Sep 7, 2004 #6

    JasonRox

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    Wrong. I'm having a hard time trying to understand where you got this.

    I'll give you the equation for B, since you kind of got it in your second post. A is very similiar.

    [tex]4.0m/s = \frac{d}{t_1-300s}[/tex] , 300s = 5min (converted into seconds).

    You had t-5, but that was incorrect. Find equation A. Remember d is the same for both.

    Combine both equations by isolating a term(d or t_1).

    Note: I must go bang my head trying to fully comprehend SR.

    Hopefully someone takes over if you are still stuck.
     
  8. Sep 7, 2004 #7

    JasonRox

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    Also, read the first 2 chapters of your textbook. It will talk about dimensionalities(metres, seconds) and most likely average velocities.

    That would help a lot.

    There is also a forum for homework help, and they have helpful people there. Believe it or not, they have regulars for helping out.

    Please avoid homework related problems in this forum. I made the mistake myself, so it's not big deal, so I'm letting you know.
     
  9. Sep 7, 2004 #8
    ?

    so distance equal to 2.1?
     
  10. Sep 8, 2004 #9

    pervect

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    Let's start with runner a.

    If runner a moves at a velocity of 3 m/s, and he runs for 1 second, how many meters does he move?

    If he runs for 2 seconds, how many meters does he move?

    More advanced:
    Can you write an equation for the position of runner a as a function of time?
     
  11. Sep 8, 2004 #10

    JasonRox

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    That's what I was saying. I even gave the equation itself.

    I would love to show you how it's done, but it is better to work it out yourself. Once you get that breakthrough, everything else becomes a little easier.

    Note: The answer might be 2.1m because I never checked.
     
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