Let's denote the speed of the cheetah as v. The distance traveled is the product of speed and time (distance = speed × time). Therefore, for the front end of the trip, the distance is v * t, and for the return trip, the distance is v * (t/2).
Now, let's express the average rate as the total distance divided by the total time. The total distance is 2 * v * t (round trip), and the total time is t + t/2.
Average rate = Total distance / Total time
= (2 * v * t) / (t + t/2)
To simplify, we can multiply the numerator and denominator by 2 to eliminate the fraction:
Average rate = (4 * v * t) / (2t + t)
= (4 * v * t) / (3t)
Now, cancel out the common factor of t:
Average rate = 4v / 3
Therefore, the relationship between the average rates on each leg of the trip is that the rate on the front end is 4/3 times the rate on the return trip. This implies that the cheetah's average speed is faster on the front end of the trip compared to the return leg.
