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 (a) its average speed and (b) its average velocity for the entire trip, using “away from the trainer” as the positive direction.
The Correct Answer and Explanation is:
Let’s break this problem down into two parts: the horse’s trot away from the trainer and then its gallop halfway back.
Part 1: The Horse Trot
- Distance: 38 m
- Time: 9.0 s
Part 2: The Horse Gallops
- The horse moves halfway back, so the distance traveled is half of 38 m:
Distance=382=19 m\text{Distance} = \frac{38}{2} = 19 \, \text{m}Distance=238=19m - Time: 1.8 s
Now we can calculate the average speed and average velocity.
(a) Average Speed
Formula for average speed: Average Speed=Total DistanceTotal Time\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}}Average Speed=Total TimeTotal Distance
Total distance = 38 m (away) + 19 m (back) = 57 m
Total time = 9.0 s (trot) + 1.8 s (gallop) = 10.8 s
So, the average speed is: Average Speed=57 m10.8 s=5.28 m/s\text{Average Speed} = \frac{57 \, \text{m}}{10.8 \, \text{s}} = 5.28 \, \text{m/s}Average Speed=10.8s57m=5.28m/s
(b) Average Velocity
Formula for average velocity: Average Velocity=Total DisplacementTotal Time\text{Average Velocity} = \frac{\text{Total Displacement}}{\text{Total Time}}Average Velocity=Total TimeTotal Displacement
Displacement is the difference between the final and initial positions, taking into account the direction:
- Initial position: 0 m (starting point)
- Final position: 38 m away – 19 m back = 19 m (since “away” is positive direction)
So, the displacement is 19 m.
Total time = 10.8 s.
Thus, the average velocity is: Average Velocity=19 m10.8 s=1.76 m/s\text{Average Velocity} = \frac{19 \, \text{m}}{10.8 \, \text{s}} = 1.76 \, \text{m/s}Average Velocity=10.8s19m=1.76m/s
Summary
- Average Speed: 5.28 m/s
- Average Velocity: 1.76 m/s
