What function gives the time (t) it takes the rider to finish the race in terms of (s)?

Question

In a bike race, a rider covers a 5-mi flat stretch of road at a speed of (s) mi/h. She then doubles her speed down a hill 1 mi long. Finally, she reduces her downhill speed by 12 mi/h as she rides the last 3 mi of the race. What function gives the time (t) it takes the rider to finish the race in terms of (s)?

Answer 1

Remember that distance = (rate)(time). For you, it will be useful to use distance/rate = time.

 

 

 

For the first stretch of the race, the biker travels 5 miles at s mph. That's a time of 5/s.

 

For the downhill portion, she travels 1 mile at 2s mph. That's a time of 1/(2s).

 

For the last portion, she travels 3 miles at 2s - 12 mph. That's a time of 3/(2s - 12).

 

 

 

So, the total time is:

 

t(s) = 5/s + 1/(2s) + 3/(2s - 12)

 

 

 

You can simplify that if you wish:

 

t(s) = 10/(2s) + 1/(2s) + 3/(2s - 12)

t(s) = (10 + 1)/(2s) + 3/(2s - 12)

t(s) = 11(2s - 12)/(2s(2s - 12)) + (6s)/(2s(2s - 12))

t(s) = (22s - 132 + 6s)/(2s(2s - 12))

t(s) = (28s - 132)/(2s(2s - 12))

t(s) = (14s - 66)/(2s(s - 6))

t(s) = (7s - 33)/(s(s - 6))

t(s) = (7s - 33)/(s² - 6s)