Average Velocity for Three People's Trips

AI Thread Summary
The discussion revolves around calculating the average velocity of three individuals traveling at different speeds: Person A at 10.2 m/s, Person B at 10.4 m/s, and Person C at 3.51 m/s. The initial calculation of simply averaging the speeds resulted in 8.0 m/s, which differs from the textbook's answer of 4.89 m/s. Participants suggested that the average velocity should be calculated using total displacement and total time, rather than just averaging the speeds. There was also mention of using the root mean square method, though it did not yield the textbook answer either. Ultimately, the correct approach involves considering the time or distance each individual traveled to accurately determine the average velocity.
Balsam
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Person A travels 10.2m/s[forward]. Person B travels 10.4m/s[forward]. Person C travels 3.51m/s[forward]. Find the average velocity for the trio.I added all 3 velocities(just the magnitudes) and divided by 3, but my answer was different from the textbook's. I got 8.0m/s[forward], but the answer in the book is 4.89m/s[forward]. I don't know what I did wrong.
 
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Balsam said:
Person A travels 10.2m/s[forward]. Person B travels 10.4m/s[forward]. Person C travels 3.51m/s[forward]. Find the average velocity for the trio.I added all 3 velocities(just the magnitudes) and divided by 3, but my answer was different from the textbook's. I got 8.0m/s[forward], but the answer in the book is 4.89m/s[forward]. I don't know what I did wrong.
Please remember to use the Homework Help Template when posting here in the schoolwork forums. Thank you.

Now, please show your detailed calculations so we can look for any errors. :smile:
 
berkeman said:
Please remember to use the Homework Help Template when posting here in the schoolwork forums. Thank you.

Now, please show your detailed calculations so we can look for any errors. :smile:

10.2+10.4+3.51=24.11--> 24.11/3= ~8.0m. <-- That is what I did
 
Balsam said:
10.2+10.4+3.51=24.11--> 24.11/3= ~8.0m. <-- That is what I did
Your work looks correct. Are you sure you copied the problem statement correctly? Are you sure they all were moving in the same direction (forward)? Are they all facing the same direction as they move?
 
berkeman said:
Your work looks correct. Are you sure you copied the problem statement correctly? Are you sure they all were moving in the same direction (forward)? Are they all facing the same direction as they move?
Yes, I'm sure. All of the individual velocity values are given in the textbook, I copied them correctly.
 
Balsam said:
Yes, I'm sure. All of the individual velocity values are given in the textbook, I copied them correctly.
Is there a figure that goes with the question? Can you upload a picture of the question?
 
Balsam said:
Person A travels 10.2m/s[forward]. Person B travels 10.4m/s[forward]. Person C travels 3.51m/s[forward]. Find the average velocity for the trio.
Are these traveling for identical times OR for identical distances?

Example: 3 friends in a car take turns at driving ...

EDIT: either way, still can't get the book's answer though
 
NascentOxygen said:
Are these traveling for identical times, or over identical distances?

Example: 3 friends in a car take turns at driving ...

EDIT: either way, still can't get the book's answer though
Try root mean square. Ans comes close.
That could be the method.
 
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NascentOxygen said:
Are these traveling for identical times OR for identical distances?

Example: 3 friends in a car take turns at driving ...

EDIT: either way, still can't get the book's answer though
I figured it out-- you have to add all of their displacements to get the total displacement, then add all of their times to get the total time, and then divide those 2 numbers to get their average velocity.
 
  • #10
Balsam said:
I figured it out-- you have to add all of their displacements to get the total displacement, then add all of their times to get the total time, and then divide those 2 numbers to get their average velocity.
Does that get you to the book's answer?
 
  • #11
256bits said:
Try root mean square. Ans comes close.
That could be the method.
Interesting...but on what basis could you justify RMS?
 
  • #12
NascentOxygen said:
Does that get you to the book's answer?
Yeah, it does.
 
  • #13
Balsam said:
Yeah, it does.
Using your method I don't get the textbook's answer.
 
  • #14
NascentOxygen said:
Interesting...but on what basis could you justify RMS?
RMS doesn't give the book answer either I believe, but I was considering a velocity distribution ( 3 points ?? ).
Is that valid for velocities in one direction - I can's say.
The answer seems to be weighted towards the slower individual though.
 
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