Suppose the cheetah sprinted at maximum speed for 8 minutes and then slowed to 40 mph for the next 8 minutes. a. how far would the cheetah have traveled in the first 8 minutes? show how you arrived at your answer. b. how far would the cheetah have traveled in the next 8 minutes? show how you arrived at your answer. c. how much farther did the cheetah travel in the first 8 minutes than in the second 8 minutes? d. the cheetah traveled 1.75 times faster for the first 8 minutes than it did for the second 8 minutes. was the distance traveled during the first 8 minutes 1.75 times greater than the distance traveled during the second 8 minutes? show the calculation to justify your answer. e. if the cheetah made a round-trip and took half the amount of time on the return trip as on the front end of the trip, what would be the relationship between the average rates on each leg of the trip? using a complete sentence, explain how you arrived at this conclusion.

A] the maximum speed traveled by cheetah is 70 mph, given that the cheetah traveled at maximum speed for 8 minutes, then the distance traveled will be given by: distance=speed×timespeed=70 mphtime=8/60 hoursdistance=28/3 =9 1/3 miles.b] distance traveled for the next 8 minutes will be: speed=40 mphtime=8/60 hourshence plugging in our values in the formula we get: distance=8/60×40=16/3= 5 1/3 milesc] to get how much farther the cheetah traveled in the first 8 minutes than in the second 8 minutes we get the difference of the distance covered in the two times.(distance traveled in the first 8 traveled in the next 8 minutes)=9 1/3-5 1/3=4 milestherefore the cheetah traveled 4 miles farther in the first 8 min compared to the next 8 min.d] given that the cheetah traveled 1.75 times faster for the first 8 min as compared to the next 8 minutes, to compare if the distance traveled in the first 8 min is 1.75 times the next 8 minute, we get the ratio of the distance as follows: (distance covered in the first 8 minutes)/(distance covered in the next 8 minutes)=[28/3]÷[16/3]=7/4=1.75therefore we conclude that the the distance covered in the first period is 1.75 times that of the second period.e. the time taken for the first trip was 8 minutes, if the cheetah made a round-trip, and took half the amount of time on the return trip as on the front end trip then the time taken to travel will be: 16/2=8 min=8/60 hoursthe speed traveled back would be: average speed=(9 1/3+5 1/3)/(8/60)=110 miles/per hoursthe average speed for the first leg will be: average speed=(70+40)/2=55 mphtherefore we conclude from above that the the average speed of the second leg was twice that of the first leg.