A horse trots away from its trainer in a straight line, moving 38 m away in 9.0 s. It then turns abruptly and gallops halfway back in 1.8 s. Calculate
- its average speed and
- its average velocity for the entire trip, using "away from the trainer" as the positive direction.
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This is Giancoli Answers with Mr. Dychko. This horse travels 38 meters away in 9 seconds and then turns around and gallops back half the distance which is 38 divided by 2 or 19 meters in 1.8 seconds and we are gonna calculate its average speed first of all by taking the total distance that it travels divided by the total time so that's 38 meters away plus the 19 meters back all divided by 9 seconds away plus the 1.8 seconds back which gives us 5.3 meters per second is the average speed. The average velocity is calculated differently; it is the displacement divided by the change in time. So the horse's final position x 2 is here 19 meters away from the trainer positive because we are told to take the position away from the trainer as positive and minus the initial position which is at the trainer which is position 0 and divided by the time it takes to get to this position 2 which is 9 seconds plus 1.8 seconds minus the time at position 0 which is time 0 and we get positive 1.8 meters per second is the average velocity.
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