Not counting the "turn around" , a helicopter carrying passengers between an airport and the roof of a downtown skycraper 20 miles away can make the round trip in 15 minutes. The helicopter flies 2 miles away with the wind when the time it flies 1 mike againstnthe wind. What is the wind speed?
vw=velocity of wind
vh=velocity of helo
distance=velocity*time
20=(vh+vw)*timethere
20=(vh-vw)*time back
but time there+time back=15min
and time there=2*timeback
or timethere=5min
and time back=10 min
finally
20=(vh+vw)*5min
20=(vh-vw)*10min
or
4=(vh+vw)
2=(vh-vw)
subtracting second from first equation
2=2vw
velocity wind=1mi/min a very stiiff breeze...a cat 1 hurricane.
Curiously, vh=3mi/min or 180 mph. Hmmm, never saw a helo do that. Such is math.
Check my work.
To solve this problem, we need to set up a system of equations. Let's define the speed of the helicopter as H and the speed of the wind as W.
We know that the distance from the airport to the skyscraper and back is 20 miles, which means that the total flying time is 15 minutes. We can set up the equation:
(20 / (H + W)) + (20 / (H - W)) = 15
However, we are given additional information that the time it takes the helicopter to fly 2 miles with the wind is the same as the time it takes to fly 1 mile against the wind. We can set up another equation based on this:
2 / (H + W) = 1 / (H - W)
Now we have a system of two equations that we can solve.
We can start by simplifying the second equation:
2(H - W) = H + W
2H - 2W = H + W
H = 3W
Now we substitute this value of H into the first equation:
(20 / (3W + W)) + (20 / (3W - W)) = 15
(20 / 4W) + (20 / 2W) = 15
(5 / W) + (10 / W) = 15
(15 / W) = 15
Now we can solve for W by cross-multiplying:
15W = 15
W = 1
Therefore, the wind speed is 1 mile per minute.