SOLUTION: A plane flying the 3458-mile trip from New York City to London, has a 50 mph tail wind. The flight's point of no return is the point at which the flight time required to return to

Algebra ->  Customizable Word Problem Solvers  -> Travel -> SOLUTION: A plane flying the 3458-mile trip from New York City to London, has a 50 mph tail wind. The flight's point of no return is the point at which the flight time required to return to      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 491409: A plane flying the 3458-mile trip from New York City to London, has a 50 mph tail wind. The flight's point of no return is the point at which the flight time required to return to New York is the same as the time required to continue to London. If the speed of the plane in still air is 360 mph, how far is New York from the point of no return?
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
A plane flying the 3458-mile trip from New York City to London, has a 50 mph tail wind.
The flight's point of no return is the point at which the flight time required to return to New York is the same as the time required to continue to London.
If the speed of the plane in still air is 360 mph, how far is New York from the point of no return?
:
Let p = dist from N.Y of this point
then
(3458-p) = dist from London of this point
:
360 + 50 = 410 mph effective speed going east (towards London)
360 - 50 = 310 mph effective speed going west (returning to N.Y.
:
Write a time equation; time = dist/speed
:
return time = continue time
p%2F310 = %283458-p%29%2F410
Cross multiply
410p = 310(3458-p)
410p = 1071980 - 310p
310p + 410p = 1071980
720p = 1071980
p = 1071980%2F720
p = 1488.86 mi from N.Y. is the point of no return.
:
:
Check solution by finding the times, (3458 - 1488.86 = 1969.14 mi, dist to London)
1969.14%2F410 = %281488.86%29%2F310
4.8 = 4.8 hrs, confirms our solution