SOLUTION: Hi I would realy appreciate any help I could get with this word problem (step by step). Could you please also explain how I would check the answer. Thanks for your time Airsp

Algebra ->  Customizable Word Problem Solvers  -> Travel -> SOLUTION: Hi I would realy appreciate any help I could get with this word problem (step by step). Could you please also explain how I would check the answer. Thanks for your time Airsp      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 284519: Hi I would realy appreciate any help I could get with this word problem (step by step). Could you please also explain how I would check the answer. Thanks for your time
Airspeed. A plane carries enough fuel for 20 hours of
flight at an airspeed of 150 miles per hour. How far can it
fly into a 30 mile per hour headwind and still have enough
fuel to return to its starting point? (This distance is called
the point of no return.)
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


Flying into the headwind, the ground speed is airspeed minus wind speed, so 150 - 30 is 120. Flying with a tail wind, the ground speed is airspeed plus wind speed, here: 180 mph.

Distance equals rate times time. Let represent the time flying into the wind. Then, since the total trip has to be no more than 20 hours, the time flying home must be . And the distance, , out has to be equal to the distance coming back.

So:

Outward trip:



Return trip:



But since :



Now solve for

Once you have a value for , substitute it into either of the two equations and solve for d.


John