Question 1092449: A horse canters away from its trainer in a straight line, moving 120 m away in 15.0 s. It then turns abruptly and gallops halfway back in 4.8 s.
(a) Calculate its average speed.
m/s
(b) Calculate its average velocity for the entire trip, using "away from the trainer" as the positive direction.
m/s
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! average speed is distance traveled divided by time it takes.
average velocity is change in location traveled divided by the time it takes.
here's a decent explanation from the web.
http://www.dummies.com/education/science/physics/distinguishing-average-speed-and-average-velocity/
this applies to your problem in the following manner.
A horse canters away from its trainer in a straight line, moving 120 m away in 15.0 s. It then turns abruptly and gallops halfway back in 4.8 s.
(a) Calculate its average speed in meters per second.
it traveled 120 meters in 15 seconds and then traveled the same 120 meters in 4.8 seconds.
the total distance traveled was 240 meters.
the total time it took was 19.8 seconds.
the average speed was total distance divided by time = 12.12121212... meters per second = 12.12 meters per second rounded to two decimal places.
the fractional equivalent would be 400/33 meters second which is equal to (12 + 4/33) meters per second.
(b) Calculate its average velocity for the entire trip, using "away from the trainer" as the positive direction.
the average velocity is the change in location divided by the time it took to make that change.
that would be 0 meters per second because the horse took 19.8 seconds and wound up in the same location that it started from.
the horse ran 120 meters in a positive direction away from the trainer and then ran 120 meters in a negative direction towards the trainer.
the distance traveled was 240 meters but the change in location was 0.
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