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.

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) (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

x1=( 60 + 68 )/ 1.6
x1= 80
x2=( 60 -68 ) / 1.6
x2= -5
Ignore negative value = 80 mph speed
m.ananth@hotmail.ca

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