A horse canters away from its trainer in a straight line, moving 110 m away in 16.0 s. It then turns abruptly and gallops halfway back in 5.4 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
You seem to have an identity problem, Cameron Lawrence/Jon Tres.
Please use the same name for your posts.
I apologize for that.
Needing help!
To calculate the average speed, we need to use the formula:
average speed = total distance ÷ total time
(a) To find the total distance, we need to add the distance the horse moved away from the trainer and the distance it moved back halfway:
total distance = distance away + distance back
= 110 m + (110 m ÷ 2)
= 110 m + 55 m
= 165 m
The total time is the sum of the time taken to move away and the time taken to move halfway back:
total time = time away + time back
= 16.0 s + 5.4 s
= 21.4 s
Now we can use the formula for average speed:
average speed = total distance ÷ total time
= 165 m ÷ 21.4 s
≈ 7.71 m/s
Therefore, the horse's average speed is approximately 7.71 m/s.
(b) To calculate the average velocity, we need to consider both magnitude and direction. In this case, "away from the trainer" is taken as the positive direction. Since the horse ends up back at its starting position, the displacement is zero, regardless of the distance covered.
The average velocity is then zero, as there is no change in position over the entire trip.
Therefore, the horse's average velocity for the entire trip is 0 m/s.