SOLUTION: A private Learjet 31A transporting passengers was flying with a tailwind and traveled 1120 mi in 2 h.Flying against the wind on the return trip,the jet was able to travel only 980m
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Question 815143: A private Learjet 31A transporting passengers was flying with a tailwind and traveled 1120 mi in 2 h.Flying against the wind on the return trip,the jet was able to travel only 980mi in 2 h. Find the speed of the jet in calm air and the rate of the wind. Answer by mananth(16946) (Show Source):
Distance against 980 miles distance with 1120 miles
t=d/r against wind -
980.00 / ( x - y )= 2.00
2.00 ( x - y ) = 2.00
2.00 x - 2.00 y = 980.00 ....................1
1120.00 / ( x + y )= 2.00
2.00 ( x + y ) = 1120.00
2.00 x + 2.00 y = 1120.00 ...............2
Multiply (1) by 1.00
Multiply (2) by 1.00
we get
2.00 x + -2.00 y = 980.00
2.00 x + 2.00 y = 1120.00
4.00 x = 2100.00
/ 4.00
x = 525.00 mph
plug value of x in (1) y
2.00 x -2.00 y = 980.00
1050.00 -2.00 -1050.00 = 980.00
-2.00 y = 980.00
-2.00 y = -70.00 mph
y = 35.00
Plane speed 525.00 mph
wind speed 35.00 mph
