A man flies a small airplane from Fargo to Bismarck, North Dakota --- a distance of 180 miles. Because he is flying into a head wind, the trip takes him 2 hours. On the way back, the wind is still blowing at the same speed, so the return trip takes only 1 hour 12 minutes. What is his speed in still air, and how fast is the wind blowing?
120 mph
30 mph winds
Let
s = still-air speed
w = wind speed
1:12 = 1 1/5 = 6/5 hours
Since distance = time * speed, you have
2(s-w) = 180
6/5 (s+w) = 180
Now solve for s and w
To find the speed of the airplane in still air and the speed of the wind, we can use the concept of relative velocity.
Let's assume the speed of the airplane in still air is A mph, and the speed of the wind is W mph.
On the way from Fargo to Bismarck (against the wind), the total time taken is 2 hours. We can use the formula: Speed = Distance / Time.
So, (A - W) = 180 / 2
Simplifying the equation, we get:
A - W = 90
On the way back from Bismarck to Fargo (with the wind), the total time taken is 1 hour 12 minutes. Firstly, we need to convert 1 hour 12 minutes to hours by dividing it by 60. Therefore, the time becomes: 1 + 12/60 = 1.2 hours.
Using the formula: Speed = Distance / Time, we have:
(A + W) = 180 / 1.2
Simplifying the equation, we get:
A + W = 150
Now, we can solve the equations simultaneously to find the values of A and W.
Adding the two equations:
(A - W) + (A + W) = 90 + 150
Simplifying:
2A = 240
Dividing by 2:
A = 120
Substituting the value of A back into one of the equations, we find:
120 - W = 90
Simplifying:
W = 120 - 90
W = 30
Therefore, the speed of the airplane in still air is 120 mph, and the speed of the wind is 30 mph.