Question 630725: It takes 9.5 hours to fly from Baltimore to Munich and 11 hours to return. If an airplane averages 550 mph in still air, what is the average rate of the wind blowing in the direction from Baltimore to Munich? Assume the speed of the wind (jet stream) is constant and the same for both legs of the trip. Answer by rmnavalta(7) (Show Source):
This problem is a UNIFORM MOTION PROBLEM
We will be using distance= rate x time.
*Legend B-Baltimore M-Munich
Let x = average rate of wind blowing in the direction from B to M
Given: average rate of the airplane in still air = 550mph
Notes: With the wind, the rate of the airplane is equal to (550+x)mph.
-The plane moves faster.
Against the wind, the rate of the airplane is equal to (550-x) mph.
-The plane moves slower.
| time(h) | rate(mph) | distance(mi)= rate*time |
B to M | 9.5 | 550+x | 9.5(550+x) |
M to B | 11 | 550-x | 11(550-x) |
(Distance from B to M is the same as from M to B)
SET-UP THE EQUATION--> 9.5(550+x)=11(550-x)
Solve. 5225+9.5x = 6050-11x
5225+9.5x+11x = 6050-11x + 11x ---> Add 11x on both sides.
5225+20.5x-5225 = 6050-5225 ---> Subtract 5225 on both sides.
(20.5x)/20.5 = 825/20.5 ---> Divide both sides by 20.5.
x = 40.24 mph
Answer: The average rate of wind blowing in the direction from B to M is approximately 40.24 mph.