A? twin-engined aircraft can fly 1064 miles from city A to city B in 4 hours with the wind and make the return trip in 7 hours against the wind. What is the speed of the? wind?
Accepted Solution
A:
Answer:Wind speed = 159.6 miles per hour.Step-by-step explanation:The distance is the same so some sort of basic equation can be made.GivensFrom A to Bt = 4 hoursv = p + wd = 1064===========From B to At1 = 7 hoursv = p - wd = 1064Equationt * (p + w) = t1 * ( p - w) Solution4(p + w) = 7(p - w)4p + 4w = 7 p - 7wAdd 7w to both sides.4p + 4w + 7w = 7p 4p + 11w = 7p Subtract 4p from both sides11w = 7p - 4p11w = 3pDivide by 3p = 11w/3====================4P + 4W = 106411w/3 + 3W = 1064w(3.6667 + 3) = 1064W*6.666667 = 1064W = 1064 / 6.666667W = 159.6 mph