Question

In a bike race, a rider covers a 5-mi flat stretch of road at a speed of s mph. She then doubles her speed down a hill 1 mi long. Finally, she reduces her downhill speed by 12 mph 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.

Distance = rate • time
d=rt

1) Solve d=rt for t.


2) Write an expression for time, t, for each part of the race using the equation from 1).
Hint: Start with t=5/s for the first stretch.


First stretch: t=

Second stretch: t=

Third stretch: t=


3) Add the expressions from 2) and simplify.
Hint: LCD is (2s-12)(2s)(s).

Answer 1

First stretch: t=5/s
Second stretch: 1/(2s)
Third stretch: 3/(2s-12)\[\frac{5}{s}+\frac{1}{2s}+\frac{3}{2s-12}=\frac{7 s-33}{s(s-6) } \]