SOLUTION: in a biathlon run over a 16 km (each way) course, the first half is the outgoing leg which is running and the returns section is covered by bicycle. Sam takes 1 hour 15 min. on the

Algebra ->  Finance -> SOLUTION: in a biathlon run over a 16 km (each way) course, the first half is the outgoing leg which is running and the returns section is covered by bicycle. Sam takes 1 hour 15 min. on the      Log On


   

Question 970270: in a biathlon run over a 16 km (each way) course, the first half is the outgoing leg which is running and the returns section is covered by bicycle. Sam takes 1 hour 15 min. on the run leg. What speed must Sam cycle the returns leg to bring about a 50% improvement on his average speed for the entire race ?
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
In a biathlon run over a 16 km (each way) course, the first half is the outgoing leg which is running and the returns section is covered by bicycle.
Sam takes 1 hour 15 min. on the run leg.
What speed must Sam cycle the returns leg to bring about a 50% improvement on his average speed for the entire race ?
:
Find the average speed on the run leg, 1 hr 15min = 1.25 hrs
16/1.25 = 12.8 km/hr
Find the 50% improvement in speed
1.5 * 12.8 = 19.2 km/hr
:
let s = speed required on the return trip to average 19.2 km/hr
Write a time equation; time = dist/speed
:
run time + bike time = total time
16%2F12.8 + 16%2Fs = 32%2F19.2
Reduce fractions
5%2F4 + 16%2Fs = 5%2F3
multiply by 12s, cancel the denominators and we have
3s(5) + 12(16) = 4s(5)
15s + 192 = 20s
192 = 20s - 15s
192 = 5s
s = 192/5
s = 38.4 km/hr, ride his bike on the return leg to average 19.2 km/hr
: