Step 2: Calculating distance using varied speeds 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 traveled 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 have 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? Use a complete sentence, explain how you arrived at this conclusion.

Part A

The instructions don't say the max speed, but we can find the max speed is 70 mph (since 1.75*40 = 70).

Convert that to miles per minute

70 mph = 70/60 = (7/6) miles per minute

In 8 minutes, the cheetah travels about (7/6)*8 = 9.333 miles at max speed.

Approximately 9.333 miles

Part B

40 mph = (40/60) miles per minute

40 mph = (2/3) miles per minute

distance = rate*time

distance = (2/3)*8

distance = 5.333 miles approximately

Approximately 5.333 miles

Part C

Subtract the results from parts A and B

9.333 - 5.333 = 4

The decimal portions cancel out completely since both have the '3's go on forever.

  4 miles

Part D

d1 = distance traveled the first 8 minutes

d1 = (rate)*(time)

d1 = (1.75r)*(8)

d1 = 14r

where r is the speed in miles per minute traveled during the next 8 minutes

d2 = distance traveled the second 8 minute period

d2 = (rate)*(time)

d2 = r*8

d2 = 8r

Now divide the distances d1 over d2

d1/d2 = (14r)/(8r)

d1/d2 = 14/8

d1/d2 = 1.75

Yes the distance traveled during the first 8 minutes is 1.75 times greater compared to the distance traveled during the second 8 minutes.

Part E

I'm assuming you meant to say "half" instead of "have".

The first leg of the round trip is 8+8 = 16 minutes.

The second leg, or return journey, of the trip is 16/2 = 8 minutes.

The total distance is d1+d2 = 14r+8r = 22r where r is r = 2/3 calculated back in part B

This means the total distance traveled is 22r = 22*(2/3) = 44/3 miles.

Let's calculate the average speed on the first leg of the trip

d = r*t

44/3 = r*16

16r = 44/3

r = (44/3)*(1/16)

r = (44*1)/(3*16)

r = 44/48

r = 11/12

r = 0.917 miles per minute approximately

Now let's calculate the return speed

d = r*t

44/3 = r*8

8r = 44/3

r = (44/3)*(1/8)

r = 44/24

r = 11/6

r = 1.833 miles per minute approximately

Let's compare the results 11/12 and 11/6

We can see that 11/6 is twice as large compared to 11/12

Note that (11/12)*2 = 22/12 = 11/6

So if the cheetah goes from point A to point B in some amount of time, and goes from B to A in half the amount of time, then it must have traveled twice as fast to get back home.

A simple example could be as follows. Let's say a car travels 120 miles and does so for 4 hours. Its speed is 120/4 = 30 mph. Now let's say it comes back home and takes 2 hours (which is half of 4) so its speed is now 120/2 = 60 mph. The jump from 30 to 60 is times 2, showing the return speed is twice as fast to effectively counterbalance the fact we halved the time value.

The cheetah's average speed on the return trip is twice as fast compared to the average speed on the first leg of the trip.

just needed some points but try socratic