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Average Speed vs. Velocity

  • Thread starter cowmoo32
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  • #1
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Homework Statement


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?


Homework Equations


Average Speed = [itex]\frac{Total distance traveled}{Total Time}[/itex]

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


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.
 

Answers and Replies

  • #2
Doc Al
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Each covers distance L, so a total of 3L,
OK.
and the time it would take is their maximum velocity/L;
Double check that.
 
  • #3
6,054
390
the time it would take is their maximum velocity/L; 113/L + 161/L + 105/L
This is incorrect, which is easily seen from the resultant units. What is the correct equation?
 
  • #4
122
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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
 
  • #5
Doc Al
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44,871
1,119
Ok, so the total time would be [itex]\frac{3L}{(113 + 161 + 105)}[/itex]
Almost, but not quite. (But you have the right units now. Careful how you add those fractions.)
 
  • #6
6,054
390
The total time is the sum of the times of all the participants. What are those times?
 
  • #7
122
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Almost, but not quite. (But you have the right units now. Careful how you add those fractions.)
[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.
 
  • #8
Doc Al
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44,871
1,119
[itex]\frac{L}{113}[/itex]+[itex]\frac{L}{161}[/itex]+[itex]\frac{L}{105}[/itex] = [itex]\frac{L}{379}[/itex]
That addition is incorrect. (1/3 + 1/3 + 1/3 ≠ 1/9 !!) Convert each fraction to a decimal, then add.
...But isn't the total distance covered 3L?
That's true.
 
  • #9
109
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Well, how our teacher taught us, it's ALL the distance over ALL the time. Try following that.
 
  • #10
Doc Al
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Well, how our teacher taught us, it's ALL the distance over ALL the time. Try following that.
That's what he's been trying to do. See the first post.
 
  • #11
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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.
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:
  • #12
Doc Al
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Looks good!
 
  • #13
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Thank you!
 

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