Avg velocity and speed question

[niteshadw]

I'm having a difficulty determining the average velocity of a problem:
A person walks first at a constant speed of 4.80 m/s along a straight line from point A to point B and then back along the line from B to A at a constant speed of 3.40 m/s.

I figured out average speed to be:

avg speed = distance traveled/time of travel

[4.8+3.4]/2 = 4.1
(4.8)(3.4) = 16.32
16.32/4.1 = 3.98 m/s

but I can't figure out the average velocity:

avg velocity = displacement/time

I know direction matters in velocity but I can't figure it out even if I make up numbers for the distance and use the velocities provided.

Does anyone know how to figure out the avg velocity. Help would be much appreciated, thank you.

 

[moose]

"[4.8+3.4]/2 = 4.1 "

If the two distances are the same (A to B, and B to A), then wouldn't you think that the distance is the same ;)
You could substitute any number you want in there. For example 10 meters and then divide by how long it took each way to get that far. You had it that the distances were different, when in fact they are identical.

 

[ek]

Is it not zero?

?

 

[Andre]

Every problem should be made as simple as possible, but not simpler. Average speed can only be averaged out if the lapsed time of two speeds is the same. Consider this. I travel 60 miles in one hour with 60 mph and back with 30 miler per hour. Is my average speed 45 mph? No because it took one hour for the first part and two hours for the second. So I traveled 120 miles in three hours and that makes 40 mph, not 45.

Could that help?

 

[HallsofIvy]

[4.8+3.4]/2 = 4.1

Okay, you took the arithmetic average of the two speeds. Why? This is not "time"

(4.8)(3.4) = 16.32

And now you multiplied the speeds! Again: why? This is not "distance"

16.32/4.1 = 3.98 m/s

And finally divide one of the distance by the other! But you yourself said average speed was "time divided by distance". What is remarkable is that it gives the correct answer!

Here's how I would do the problem: since the actual distance from A to B is not given, let's just call it "x". How much time did it take to go from A to B at 4.8 m/s? That's a distance of x m at 4.8 m/s: \frac{x}{4.8} seconds. Now how much time will it take to go the x m from B back to A at 3.4 m/s? That's \frac{x}{3.4} seconds. Roundtrip time \frac{x}{4.8}+\frac{x}{3.4}= \frac{3.4x+ 4.8x}{(4.8)(3.4)} seconds. Of course, the total distance is 2x m so
speed= distanc/time= 2x(\frac{(4.8)(3.4)}{3.4x+ 4.8x})= \frac{(4.8)(3.4)}{3.4+ 4.8}, exactly what you had. Okay, I'm impressed- but 'fess up did you really come up with that formula or were you just lucky?!
As far as the vector velocity is concerned, that's easy: You went from A to B and then back to A again. What is your actual "displacement"- that is, how far are you from your starting point at the end? (See ek's response!)

 

[niteshadw]

Okay, I'm impressed- but 'fess up did you really come up with that formula or were you just lucky?!

I don't know if I'm lucky or not but that's the arithmetic I've used to determine my answer, I just did not bother to write it out as neatly as you have, it should work for any other numbers, but I think it should. The problem is modeled after similar ones from a book; which I do not have so I don't really know what's the procedure to solve them.

Thanks for the help, I did not account for the sign but instead added the two distances - I kept thinking that the traveled distance is just added then divided by the time. Thank you all again!