SOLUTION: David rode his bike east at 20 mi/hr for 36 minutes with a tailwind. He then turned around to ride home along the same route against the wind. If the return trip took 45 minutes,

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Question 1114986: David rode his bike east at 20 mi/hr for 36 minutes with a tailwind. He then turned around to ride home along the same route against the wind. If the return trip took 45 minutes, find the wind speed and the speed at which David pedals with no wind (to the nearest mile per hour).
Found 3 solutions by stanbon, ikleyn, Alan3354:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
David rode his bike east at 20 mi/hr for 36 minutes with a tailwind. He then turned around to ride home along the same route against the wind. If the return trip took 45 minutes, find the wind speed and the speed at which David pedals with no wind (to the nearest mile per hour).
with-wind DATA:
rate = 20 mph ; time = 36/60 hrs ; distance = rate*time = 12 miles
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Against-wind DATA:
dist = 12 miles ; time = (45/60) = 3/4 hr ; rate = dist/time = 12/(3/4) = 16 mph
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Equations:
bike + wind = 20 mph
bike - wind = 16 mph
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Add and solve for bike speed::
2b = 36
bike speed = 18 mph
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Subtract and solve for wind speed
2w = 4
wind speed = 2 mph
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Cheers,
Stan H.
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Answer by ikleyn(52871) About Me  (Show Source):
You can put this solution on YOUR website!
.
The physics model


V%5Btailwind%5D        = V%5Bbike%5D + V%5Bwind%5D

V%5Bagainst_the_wind%5D = V%5Bbike%5D - V%5Bwind%5D


does not work. It is not applicable to "bike-wind" problems.


Therefore, solving this problem on the base of this model is INCORRECT from the Physics point of view.


So, the solution by @Stanbone makes no sense.


Even posing this model in the framework of elementary Math/Physics makes no sense, 
since these branches has no adequate models to approach such problem.


I just noticed it in this forum couple of times before in previous two years.


Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Bike speed is not linearly related to windspeed.
Not a reasonable question/problem.
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If you bike at 10 mi/hr with a 10 mi/hr crosswind, it does not make your ground track 45 degrees from your heading, right?
If you don't pedal at all, does the wind move you and the bicycle?
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The force from the wind is a function of the square of the windspeed.