An aircraft flew 6 hours with the wind. The return trip took 7 hours against the wind. If the speed of the plane in still air is 276 miles per hour more than the speed of the wind, find the wind speed and the speed of the plane in still air.
The wind soeed was ______ mph.
Distance = 6(256+w)=7(256-w)
Solve for w.
To solve this problem, we can use the concept of relative speed. Let's assume the speed of the wind is "w" mph, and the speed of the plane in still air is "p" mph.
Given that the plane flew 6 hours with the wind, we can calculate the total distance traveled during this time:
Distance = Speed × Time
Distance = (p + w) × 6
Similarly, for the return trip, which took 7 hours against the wind, the distance can be calculated as:
Distance = Speed × Time
Distance = (p - w) × 7
Since the distances for both the trips are the same (as it's a round trip), we can equate the two equations:
(p + w) × 6 = (p - w) × 7
Now, let's simplify this equation step by step:
6p + 6w = 7p - 7w
6w + 7w = 7p - 6p
13w = p
We are also given that the speed of the plane in still air is 276 mph more than the speed of the wind. So, the equation becomes:
p = w + 276
Substituting this value of p in the previous equation, we get:
13w = w + 276
12w = 276
w = 276/12
w = 23
Therefore, the wind speed is 23 mph.
To find the speed of the plane in still air, we can substitute the value of w in the equation:
p = w + 276
p = 23 + 276
p = 299
Therefore, the speed of the plane in still air is 299 mph.