Average velocity question?

Question

In my textbook there is a formula for average velocity and its messing me up. It says "average velocity= displacement/time elapsed" and right under it " average velocity= distance final- distance initial/ time final-time initial". Isn't time elapsed, and time final subtract time initial, completely different. There is also a question that states A ball rolls 10.0m south in a Time of 6.00s, hits a wall and rolls back a distance of 15.0m north in a time of 10.0s what is the average velocity. To find out the answer wouldn't it be 15.0(north) + (-10.0south)/10.0 - 6.00.

Answer 1

My definition of average velocity is total distance divided by total time.



And that is the same as your two definitions. The differences you see are just terminology. Just remember "total distance divided by total time".



Re your sample, remember, total distance, and that is 5 meters, divided by total time, 16 seconds or (5/16)m/s. Time elapsed is 16 seconds.



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Answer 2

"Isn't time elapsed, and time final subtract time initial, completely different"

Nope. If you start at a time of t = 2 and end at t=6, that means the elapsed time is 4, which is 6-2 (final-initial time).



"There is also a question that states A ball rolls 10.0m south in a Time of 6.00s, hits a wall and rolls back a distance of 15.0m north in a time of 10.0s what is the average velocity. To find out the answer wouldn't it be 15.0(north) + (-10.0south)/10.0 - 6.00."



Assuming you meant (15 - 10) / (10-6), yep (don't forget PEMDAS). This only works for average velocity, and it'll be a completely different story for average speed though :P

Answer 3

1. Elapsed time (time elapsed) is exactly the difference between the end time (time final) minus the start time (time initial).



2. In the second question, consider the difference between speed and velocity. Velocity is a vector quantity, while speed is not. The speed and velocity for the first part is that the ball rolls 10 m in 6 seconds, or 1.66 m/sec (speed) and -1.66 m/sec (velocity assuming that moving north would be positive and south negative. Similarly, the second part is 1.5 m/sec (speed) and +1.5 m/sec (velocity, assuming north is positive again).



3. The average velocity is 5 m in 16 seconds, or 0.3125 m/sec. The average speed would be 1.66 x 6 + 1.5 x 10 / 16, which is to weight the speed average across the time. 1.5625 m/sec.



Your last statement (wouldn't it be ...) has part of the idea -- net displacement (where it started - where it ended), but also subtracts the times (which shouldn't be subtracted to find the average velocity.