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Average Speed vs. Velocity

  1. Aug 24, 2012 #1
    1. The problem statement, all variables and given/known data
    The cheetah can run as fast as 113 km/h, the falcon can fly as fast as 161 km/h, and the sailfish can swim as fast as 105 km/h. The three of them run a relay with each covering a distance L at maximum speed. What is the average speed of this relay team for the entire relay?


    2. Relevant equations
    Average Speed = [itex]\frac{Total distance traveled}{Total Time}[/itex]

    Average Velocity = [itex]\frac{Total displacement}{Total Time}[/itex]


    3. The attempt at a solution
    Each covers distance L, so a total of 3L, and the time it would take is their maximum velocity/L; 113/L + 161/L + 105/L. Is there some information I'm not accounting for?

    I spoke with my professor and he had this to offer:
    The average speed would be equal to one-third the sum of the three
    speeds as you calculated if the three speeds were each maintained for
    the same length of time. Instead, they are each maintained for the
    same distance, which will give a slightly different result. Given a
    distance and speed, you can calculate the time.
     
  2. jcsd
  3. Aug 24, 2012 #2

    Doc Al

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    Staff: Mentor

    OK.
    Double check that.
     
  4. Aug 24, 2012 #3
    This is incorrect, which is easily seen from the resultant units. What is the correct equation?
     
  5. Aug 24, 2012 #4
    I'm bad about not using units. I really need to stop that.


    Ok, so the total time would be [itex]\frac{3L}{(113 + 161 + 105)}[/itex] .... [itex]\frac{km}{km/hr}[/itex] = hr
     
  6. Aug 24, 2012 #5

    Doc Al

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    Staff: Mentor

    Almost, but not quite. (But you have the right units now. Careful how you add those fractions.)
     
  7. Aug 24, 2012 #6
    The total time is the sum of the times of all the participants. What are those times?
     
  8. Aug 24, 2012 #7
    [itex]\frac{L}{113}[/itex]+[itex]\frac{L}{161}[/itex]+[itex]\frac{L}{105}[/itex] = [itex]\frac{L}{379}[/itex] ...But isn't the total distance covered 3L? This shouldn't be so difficult for me, I don't know why I'm having so much trouble.
     
  9. Aug 24, 2012 #8

    Doc Al

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    That addition is incorrect. (1/3 + 1/3 + 1/3 ≠ 1/9 !!) Convert each fraction to a decimal, then add.
    That's true.
     
  10. Aug 24, 2012 #9
    Well, how our teacher taught us, it's ALL the distance over ALL the time. Try following that.
     
  11. Aug 24, 2012 #10

    Doc Al

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    That's what he's been trying to do. See the first post.
     
  12. Aug 25, 2012 #11
    Unit L = 1

    [itex]\frac{1}{113}[/itex]+[itex]\frac{1}{161}[/itex]+[itex]\frac{1}{105}[/itex] = 0.0246

    And the total distance is 3L = 3, so average speed would be 3/.0246 = 122km/h....correct?

    That seems like a reasonable answer given the velocities of each.
     
    Last edited: Aug 25, 2012
  13. Aug 25, 2012 #12

    Doc Al

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    Looks good!
     
  14. Aug 26, 2012 #13
    Thank you!
     
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