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 bu 12 mi/h as she rides the last 3 mi of the race. What function gives the time t it takes the rider finish the race in terms of s?
time = distance / speed
time at s mi/h = 5/s
time at 2s = 1/2s
time at 2s-12 = 3/(2s-12)
total time = 5/s + 1/2s + 3/(2s-12)
= (4s-11) / s(s-6)
9/(s)(2s)(2s-12)
I agree with #1
To find the function that gives the time it takes for the rider to finish the race in terms of s, we can break down the race into three parts: the flat stretch, the downhill, and the final 3 miles.
1. Flat stretch:
The rider covers a 5 mi flat stretch of road at a speed of s mi/h. The time it takes to cover this distance is given by:
time_flat = distance / speed = 5 mi / s mi/h = 5/s hours
2. Downhill:
After the flat stretch, the rider doubles her speed down a 1 mi long hill. So her speed becomes 2s mi/h. The time it takes to cover this distance is given by:
time_downhill = distance / speed = 1 mi / (2s) mi/h = 1/(2s) hours
3. Final 3 miles:
Finally, the rider reduces her downhill speed by 12 mi/h, so her speed becomes (2s - 12) mi/h. The time it takes to cover this distance is given by:
time_final = distance / speed = 3 mi / (2s - 12) mi/h = 3/(2s - 12) hours
Now, we can add up the times for each part to get the total time t it takes the rider to finish the race:
t = time_flat + time_downhill + time_final
t = 5/s + 1/(2s) + 3/(2s - 12)
So, the function that gives the time t it takes the rider to finish the race in terms of s is:
t(s) = 5/s + 1/(2s) + 3/(2s - 12)