An aircraft flew 5 hours with the wind. The return trip took 6 hours against the wind. If the speed of the plane in still air is 250 miles per hour more than the speed of the wind, find the wind speed and the speed of the plane in still air.
Let's call the speed of the wind "w" (in miles per hour) and the speed of the plane in still air "p" (in miles per hour).
Since the plane flew 5 hours with the wind, the distance it traveled can be expressed as 5(p+w). Similarly, the distance traveled in 6 hours against the wind can be expressed as 6(p-w).
As per the problem, the speed of the plane in still air is 250 miles per hour more than the speed of the wind, which can be written as p = w + 250.
We can now solve the problem using these three equations:
1. Distance with the wind: 5(p+w) = D
2. Distance against the wind: 6(p-w) = D
3. Speed in still air: p = w + 250
Since the distance is the same in both cases, we can equate equation 1 and equation 2:
5(p+w) = 6(p-w)
Simplifying this equation, we get:
5p + 5w = 6p - 6w
11w = p
Now, we can substitute the value of p from equation 3 into this equation to solve for w:
11w = w + 250
10w = 250
w = 25
Substituting the value of w into equation 3, we get:
p = w + 250
p = 25 + 250
p = 275
Therefore, the wind speed is 25 miles per hour and the speed of the plane in still air is 275 miles per hour.
Let's assume the speed of the plane in still air is "x" miles per hour, and the speed of the wind is "w" miles per hour.
On the outbound trip with the wind, the plane's effective speed would be (x + w) miles per hour.
The total distance covered on this trip would be 5 * (x + w) miles.
On the return trip against the wind, the plane's effective speed would be (x - w) miles per hour.
The total distance covered on this trip would be 6 * (x - w) miles.
Since the distance covered on both trips is the same (it's a round trip), we can set up the following equation:
5 * (x + w) = 6 * (x - w)
Let's solve this equation step-by-step:
1. Distribute and simplify:
5x + 5w = 6x - 6w
2. Combine like terms:
5w + 6w = 6x - 5x
11w = x
3. Rearrange the equation to isolate x:
11w = x
x = 11w
So, the speed of the plane in still air is 11 times the speed of the wind.
Since we also know that the speed of the plane in still air is 250 miles per hour more than the speed of the wind, we can set up another equation:
x = w + 250
Now, substitute x = 11w into this equation:
11w = w + 250
10w = 250
w = 250/10
w = 25
The wind speed is 25 miles per hour.
To find the speed of the plane in still air, substitute w = 25 into x = 11w:
x = 11 * 25
x = 275
The speed of the plane in still air is 275 miles per hour.
To solve this problem, let's break it down step by step:
Step 1: Assign variables
Let's assign variables to the unknown values in the problem. Let's use:
- Speed of the plane in still air: P
- Speed of the wind: W
Step 2: Set up equations
Based on the problem, we know that:
- The plane flew 5 hours with the wind, so the distance covered is 5P.
- The return trip took 6 hours against the wind, so the distance covered is 6(P - W).
Step 3: Solve the equations
Since distance = speed × time, we can use the equations above to set up the following equation:
5P = 6(P - W)
Step 4: Simplify and solve for the variables
Let's distribute the 6 on the right side of the equation:
5P = 6P - 6W
Now, let's simplify the equation by combining like terms:
6W = P
We also know that the speed of the plane in still air is 250 miles per hour more than the speed of the wind:
P = W + 250
Now, we can substitute this value of P into the previous equation:
6W = W + 250
Step 5: Solve for the wind speed
Let's simplify and solve for W:
6W - W = 250
5W = 250
W = 50
Therefore, the wind speed is 50 mph.
Step 6: Solve for the speed of the plane in still air
Now that we know the wind speed, let's substitute it back into one of our equations to find the speed of the plane in still air:
P = W + 250
P = 50 + 250
P = 300
Therefore, the speed of the plane in still air is 300 mph.
To recap:
- The wind speed is 50 mph.
- The speed of the plane in still air is 300 mph.