Average velocity? Average speed?

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Homework Help Overview

The problem involves a cow running in two segments: first 108 m east and then 93 m north, with questions regarding the average speed and average velocity over the entire trip. The context is rooted in kinematics, specifically focusing on concepts of distance, displacement, and vector quantities.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the calculations for average speed and average velocity, with some expressing confusion about the average velocity calculation. There are references to using Pythagorean theorem for determining displacement, and questions about the interpretation of directions and distances involved.

Discussion Status

There is ongoing exploration of the problem, with participants questioning their assumptions and calculations. Some guidance has been offered regarding the use of vectors and the importance of displacement in calculating average velocity. Multiple interpretations of the problem setup are being considered.

Contextual Notes

Participants note discrepancies in understanding the directions involved in the problem, with some confusion about whether the cow's movements were correctly interpreted. There is also mention of the need to consider the vector nature of average velocity.

CMATT
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Homework Statement


A cow runs in a straight line, moving 108 m east in 16.4 seconds. It then turns and runs 93 m north in 7.64 seconds. a) What was the cow's average speed over the whole trip?
b) What was the cow's average velocity over the whole trip?

Homework Equations



average speed = total distance traveled / time

average velocity = displacement / time

The Attempt at a Solution


a) average speed = (108 m + 93 m) / (16.4 s + 7.64 s) = 8.36 m/s

b) average velocity = (108 m - 93 m) / (16.4 s + 7.64 s) =

I keep getting the wrong answer for average velocity ... what am i doing wrong? It says on my homework the right answer is 5.93 m/s northeast

Thanks!
 
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CMATT said:

Homework Statement


A cow runs in a straight line, moving 108 m east in 16.4 seconds. It then turns and runs 93 m north in 7.64 seconds. a) What was the cow's average speed over the whole trip?
b) What was the cow's average velocity over the whole trip?

Homework Equations



average speed = total distance traveled / time

average velocity = displacement / time

The Attempt at a Solution


a) average speed = (108 m + 93 m) / (16.4 s + 7.64 s) = 8.36 m/s

b) average velocity = (108 m - 93 m) / (16.4 s + 7.64 s) =

I keep getting the wrong answer for average velocity ... what am i doing wrong? It says on my homework the right answer is 5.93 m/s northeast

Thanks!
The magnitude of the displacement is the straight-line distance from the beginning point to the ending point. Consult Mr. Pythagoras .
 
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The problem said west and then north. You seem to have thought it was west and then east.
 
HallsofIvy said:
The problem said west and then north. You seem to have thought it was west and then east.
So do I add them together then?
 
Have you at least drawn a picture? Did you think about SammyS's remark about "Pythagoras"?
 
HallsofIvy said:
Have you at least drawn a picture? Did you think about SammyS's remark about "Pythagoras"?
Yes, I drew a graph with N/S/E/W directions and lines showing where the cow is relative to the direction. She is traveling east at first, and then changes direction to go north. Both positive directions...but even when I add them and divide I'm still getting it wrong. As for the pythagoras, I'm not sure what to plug into where
 
CMATT said:
Yes, I drew a graph with N/S/E/W directions and lines showing where the cow is relative to the direction. She is traveling east at first, and then changes direction to go north. Both positive directions...but even when I add them and divide I'm still getting it wrong. As for the pythagoras, I'm not sure what to plug into where

If you go 108m east then 93m north, how far are you from where you started?
 
CMATT said:

Homework Statement


A cow runs in a straight line, moving 108 m east in 16.4 seconds. It then turns and runs 93 m north in 7.64 seconds. a) What was the cow's average speed over the whole trip?
b) What was the cow's average velocity over the whole trip?

Homework Equations



average speed = total distance traveled / time

average velocity = displacement / time

The Attempt at a Solution


a) average speed = (108 m + 93 m) / (16.4 s + 7.64 s) = 8.36 m/s

b) average velocity = (108 m - 93 m) / (16.4 s + 7.64 s) =

I keep getting the wrong answer for average velocity ... what am i doing wrong? It says on my homework the right answer is 5.93 m/s northeast

Thanks!
Keep in mind that an average velocity is a *vector* because displacement is a vector. Find the total displacement (a vector) and divide it by the total time interval. By the way, their answer is not completely correct; northeast means 45 degrees north of east but the correct answer is not exactly 45 degrees north of east.
 
PeroK said:
If you go 108m east then 93m north, how far are you from where you started?
201 m ?
 
  • #10
CMATT said:
201 m ?

Okay, an interesting answer! What about this.

You have a large football field that is 108m long and 93m wide. You put a flag in one corner, then you walk 108m east and 93m north to the opposite corner and put another flag there. How far are the flags apart?
 

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