<PRE><B>Question 141077</B>
With the wind, a plane flew 1400 km in 4 hours. On the return trip, the pilot was forced to land after 1 h 30 min, having traveled only 450 km. Find the rate of the plane in still air and the rate of the wind.
:
Let x = speed of the plane in still air
Let y = speed of the wind
then
(x-y) = speed against the wind
(x+y) = speed with the wind
:
Write two distance equations; dist = time * speed:
;
4(x+y) = 1400
1.5(x-y) = 450
:
Simplify the 1st equation divide by 4
Simplify the 2nd equation divide by 1.5
:
x + y = 350
x - y = 300
-------------adding eliminates y
2x = 650
x = {{{650/2}}}
x = 325 mph plane speed in still air
:
Use x + y = 350 to find y
325 + y = 325
y = 350 - 325
y = 25 mph wind speed.  
:
:
Check solution in original 1st equation
4(325 + 25) =
4 * 350 = 1400, confirms our solutions

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