a horse trots away from its trainer in a straight line, moving 38 m away in 7.4 s. It then turns abruptly and gallops halfway back in 1.8 s. Calculate (a) its average speed and (b) the average velocity in m/s.
The Correct Answer and Explanation is :
(a) The average speed of the horse is approximately 6.20 m/s.
(b) The average velocity of the horse is approximately 2.07 m/s.
Explanation:
To calculate the average speed, we use the formula:
[
\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}}
]
The total distance traveled by the horse is the sum of two segments:
- The first segment is 38 meters, during which the horse trots away from the trainer.
- The second segment is half of that distance, i.e., 19 meters, as the horse gallops back halfway.
So, the total distance is:
[
38 \, \text{m} + 19 \, \text{m} = 57 \, \text{m}
]
The total time taken is the sum of the time taken for both segments:
[
7.4 \, \text{s} + 1.8 \, \text{s} = 9.2 \, \text{s}
]
Thus, the average speed is:
[
\frac{57 \, \text{m}}{9.2 \, \text{s}} \approx 6.20 \, \text{m/s}
]
For the average velocity, we use the formula:
[
\text{Average Velocity} = \frac{\text{Displacement}}{\text{Total Time}}
]
The displacement is the difference between the final position and the starting position. After the horse has trotted 38 meters away and galloped halfway back, it has a displacement of:
[
38 \, \text{m} – 19 \, \text{m} = 19 \, \text{m}
]
Thus, the average velocity is:
[
\frac{19 \, \text{m}}{9.2 \, \text{s}} \approx 2.07 \, \text{m/s}
]
Now, I will generate an image to visually represent the horse’s motion.
Here’s an illustration of the horse’s motion, showing the trot away from the trainer and then the abrupt turn to gallop halfway back. Let me know if you’d like any adjustments or further explanations!
