Can Lance Armstrong Beat a Healthy Young Male in a Hill-Climbing Bike Race?

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SUMMARY

Lance Armstrong can generate a sustained power output of 500 Watts, while a healthy young male (HYHM) produces 300 Watts. In a hill-climbing race with a 6% incline over five miles, Armstrong's calculated velocity is 12.7496 m/s, resulting in a total time of 631.135 seconds. The HYHM's velocity is 10.1982 m/s, leading to a time of 789.033 seconds. Armstrong wins the race by a margin of 2 minutes and 37.898 seconds.

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Homework Statement


1. It has been estimated that Tour-de-France champion Lance Armstrong could generate a sustained 500 Watts of power over a 20-minute period, while a healthy young human male (HYHM) can generate about 300 Watts of power for 20-minutes. Lance and HYHM are going to race (on bicycles) up a hill with a 6% upgrade, that is five miles long, and the elevation at the top of the hill is 5000 feet. Both rider/bicycle combinations weigh 170 lbs, with frontal area 0.36m2 and coefficient of drag 0.88 (values being typical of bicyclists in crouched racing positions). The coefficient of rolling resistance for both bicycles is 0.01. (1) Who gets to the top first? (2) How much longer does it take the loser to make it to the top?

(1) The winner is ______________________________
(2) The winner's margin of victory is:
________________hours __________________minute _________________seconds


Homework Equations


P= FxV
Resistance of Air = ρ/2(coefficient of drag*Frontal area*v^2
Rolling Resistance = coefficient of rolling resistance*W
Grade Resistance = Weight*Grade
p=1.0567 kg/m^3



The Attempt at a Solution


I converted all the units to meters and Newtons. I added all resistances together to find F but since velocity is not known I left it as F=0.16738v^2+12.0092
I then applied P=Fxv equation by plugging in the power for both persons and the force equation above to find the velocity of each.
500=0.16738v^3+12.0092v
300=0.16738v^3+12.0092v

I got
v(lance)=12.7496 m/s
v(HYHM)=10.1982 m/s

With the distance of 5 miles = 8046.72m, I divided the distance by each velocity and got
t(lance)=631.135 seconds
t(HYHM)=789.033 seconds

After subtracting both, the margin of victory is

t= 2 minutes and 37.898 seconds.

Did I do this correctly? Thank you for your responses.
 
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Where do you consider the incline? 12N is certainly too low to account for that.
It is probably beyond the scope of this problem, but you could consider that the air at this height won't be as dense as at sea level (with an average height of the track).
 
I can't help but wonder now old this problem is!
 

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