A horse trots away from its trainer in a straight line, moving 38m away in 9.0s. It then turns abruptly and gallops halfway back in 1.8 s. Calculate (a) its average speed and (b) its average velocity for the entire trip, using "away from the trainer" as the positive direction.

a) The average speed of the horse is 5.3 m/s.

b) The average velocity of the horse is 1.8 m/s

Explanation:

Hi there!

a) The average speed (a. s) is calculated as the traveled distance (d) divided the time in which that distance was covered (t):

a. s = d/t

The distance covered by the horse is 38 m away from the trainer and then halfway (38/2 m) back:

d = 38 m + 38/2 m = 57 m

The time it took the horse to cover that distance was 9.0 s plus 1.8 s on the way back:

t = 9.0 s + 1.8 s = 10.8 s

Then the avereage speed will be:

a. s = 57 m / 10.8 s

a. s = 5.3 m/s

The average speed of the horse is 5.3 m/s.

b) The average velocity (a. v) is calculated as the displacement (Δx) divided the time (t) in which that displacement took place.

a. v = Δx/t

The displacement is calculated as the distance between the final position and the initial position:

Δx = final position - initial position

The horse moved 38 m away from the trainer (in the positive direction) and then returned 19 m (the horse moved 19 m in the negative direction). The final position of the horse is 19 m away from the trainer (the origin of the frame of reference). Then, the displacement will be:

Δx = final position - initial position

Δx = 19 m - 0 m (notice that the horse started from the origin, i. e., the trainer).

Δx = 19 m

And the average velocity will be:

a. v = Δx/t

a. v = 19 m / 10.8 s

a. v = 1.8 m/s

The average velocity of the horse is 1.8 m/s