A horse canters away from its trainer in a straight line, moving 130 m away in 13.0 s. It then turns abruptly and gallops halfway back in 5.4 s.
a) Calculate its average speed.
(b) Calculate its average velocity for the entire trip, using "away from the trainer" as the positive direction.
avg speed=distance/time=(130+130)/18.4 sec
avg velocity=0 entire trip
To calculate the average speed of the horse, we need to find the total distance traveled and divide it by the total time taken. Let's break down the problem step by step.
a) To calculate the average speed, we need to find the total distance traveled. The horse moves 130 m away from its trainer and then gallops halfway back, which is a distance of 130/2 = 65 m in the opposite direction. So, the total distance traveled is 130 m + 65 m = 195 m.
Next, we calculate the total time taken. The horse moves away from its trainer for 13.0 seconds and then returns halfway back in 5.4 seconds, which gives us a total time of 13.0 s + 5.4 s = 18.4 s.
Finally, we can calculate the average speed by dividing the total distance traveled by the total time taken:
Average Speed = Total Distance Traveled / Total Time Taken
Average Speed = 195 m / 18.4 s
Average Speed ≈ 10.60 m/s
Therefore, the average speed of the horse is approximately 10.60 m/s.
b) To calculate the average velocity, we need to consider both the direction and magnitude of displacement. In this question, "away from the trainer" is considered the positive direction.
Since the horse moves away from its trainer and then returns halfway back, it ends up at a displacement of 0 m (since it ultimately returns to the starting point). Therefore, the average velocity is 0 m/s.
The average velocity is zero because velocity is a vector quantity and is calculated as the displacement divided by the time taken. Since the horse ends up at the same position as it started, the displacement is zero, and therefore the average velocity is also zero.