SOLUTION: With a north wind blowing at 20 mi/hr a helicopter flies due north for 30 miles and then returns. If the total traveling time was 48 mins find the helicopters speed in still air.

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Question 543154: With a north wind blowing at 20 mi/hr a helicopter flies due north for 30 miles and then returns. If the total traveling time was 48 mins find the helicopters speed in still air.
Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
Helicopter speed x mph
wind speed 20 mph

against wind x- 20 mph
with wind x+ 20 mph

Distance= 30 miles

Time against + time with = 0.8 hours
t=d/r
30 /( x + 20 ) + 30 /(x - 20 ) = 0.8

LCD = (x - 20 ) ( x + 20 )
30 *( x - 20 ) + 30 (x + 20 ) = 0.8
30 x - 600 + 30 x + 600 = 0.8 ( x ^2 - 400 )
60 x = 0.8 x ^2 - 320
0.8 x ^2 - -60 x - 320

Find the roots of the equation by quadratic formula

a= 0.8 , b= -60 , c= -320

b^2-4ac= 3600 + 1024
b^2-4ac= 4624 %09sqrt%28%094624%09%29=%0968%09
x=%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F%282a%29
x1=%28-b%2Bsqrt%28b%5E2-4ac%29%29%2F%282a%29
x1=( 60 + 68 )/ 1.6
x1= 80
x2=( 60 -68 ) / 1.6
x2= -5
Ignore negative value = 80 mph speed
m.ananth@hotmail.ca