Finding Average Velocity of Entire Trip


by Bashyboy
Tags: average, entire, trip, velocity
Bashyboy
Bashyboy is offline
#1
Mar8-12, 07:31 PM
P: 877
I have been solving these sort of problems, and they follow this basic paradigm: first, they will tell me I traveled a distance D at a velocity of A, then they traveled the same distance D at a velocity of B. Why can't I simply just simply perform the following computation: (A+B)/2. I have been searching the internet for some unequivocal explanation to this sort of problem, but have found no such explanations that suffice in aiding my comprehension in solving these kinds of problem. So, it would be much appreciated if someone would could give me a good explanation of this.

Thank you, in advance.

Edit: I forgot to mention that the velocity A does not equal the velocity B
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Gordianus
Gordianus is offline
#2
Mar8-12, 08:17 PM
P: 217
Average speed is computed as traveled distance divided into elapse time.
No doubt in your example the total distance is 2D. However the elapsed time is (D/A+D/B).
The first term is the time employed to travel distance D at speed A and the second is the same for B.
Now, if you compute the average speed, won't find (A+B)/2
Bashyboy
Bashyboy is offline
#3
Mar9-12, 11:29 AM
P: 877
Oh, so the reason why I cannot simply just add the two velocities, and divide by two, is because the time it takes to travel distance D is different with each velocity, meaning the elasped time is different than when I just add them and divide by two?

Gordianus
Gordianus is offline
#4
Mar9-12, 11:54 AM
P: 217

Finding Average Velocity of Entire Trip


You're right.
cmb
cmb is offline
#5
Mar9-12, 12:34 PM
P: 623
FWIW, velocity is a vector quantity, so this particular question will also be dependent on the geometric relationship between the start and finishing points.

Whereas the above may be interpreted to be discussing speed rather than velocity.
Bashyboy
Bashyboy is offline
#6
Mar9-12, 02:41 PM
P: 877
cmb, I actually just sort of made this problem on the spot. I posed the question with the pretense that each velocity were in the same direction, but I certainly see what you are saying.


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