# Homework Help: Average velocity? Average speed?

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1. Dec 11, 2015

### CMATT

1. The problem statement, all variables and given/known data
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?

2. Relevant equations

average speed = total distance traveled / time

average velocity = displacement / time

3. 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 hw the right answer is 5.93 m/s northeast

Thanks!

2. Dec 11, 2015

### SammyS

Staff Emeritus
The magnitude of the displacement is the straight-line distance from the beginning point to the ending point. Consult Mr. Pythagoras .

3. Dec 12, 2015

### HallsofIvy

The problem said west and then north. You seem to have thought it was west and then east.

4. Dec 12, 2015

### CMATT

So do I add them together then?

5. Dec 12, 2015

### HallsofIvy

Have you at least drawn a picture? Did you think about SammyS's remark about "Pythagoras"?

6. Dec 12, 2015

### CMATT

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

7. Dec 12, 2015

### PeroK

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

8. Dec 12, 2015

### nrqed

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.

9. Dec 12, 2015

### CMATT

201 m ?

10. Dec 12, 2015

### PeroK

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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