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Average speed problem What am I missing?

  1. Aug 31, 2009 #1
    1. The problem statement, all variables and given/known data
    A person walks from point a to point b at a speed of 5 m/s. Then she turns around and walks from point b to point a at a speed of 3 m/s. What is her average speed?


    2. Relevant equations
    Average speed is equal to d/ change in t.


    3. The attempt at a solution
    So I know the distance is 2d. I also know the acceleration is constant, though I don't know if I need it for the problem. So I tried to do 5m/s + 3m/s /2 to get the answer. That led to the wrong path. Since we don't have time, I'm assuming that maybe I use a kinematic equation. However, we don't know what acceleration is, other than it is constant.
     
  2. jcsd
  3. Aug 31, 2009 #2

    rl.bhat

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    Here the person is walking with uniform velocity.
    From a to b the time taken is t1 = d/5 m/s
    From b to a the time taken is t2 = d/3 m/s
    Find total time and total distance. Then the average velocity = total distance / total time.
     
  4. Aug 31, 2009 #3
    I see that total distance is 2, or 2d. I am having trouble seeing how to get the time.
     
  5. Aug 31, 2009 #4

    rl.bhat

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    Total time t = t1 + t1 = d/(5 m/s) + d/(3 m/s)
     
  6. Aug 31, 2009 #5
    Tried working it out, but I'm not seeing it here. I see that by manipulation you get t=d/v at which point you plug in the speed that I am given. Can you explain a bit more. I'm quite confused here.
     
  7. Aug 31, 2009 #6

    rl.bhat

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    Average velocity = 2d/t = 2d/( d/5 + d/3)
     
  8. Aug 31, 2009 #7
    Oooooooo!!! It all makes sense! Got it! Thanks so much! Really appreciate it. The trick was seeing that the d's would cancel! Thanks!
     
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