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 hours. What is the plane's speed in still air, and how fast is the wind blowing?
Plane's speed --- x mph
wind speed ----- y mph
2(x-y) = 180
x-y = 90
1(x+y) = 180
x+y = 180
add them: 2x = 270
x = 135
back in x+y= 180
135+y=180
y = 45
state a proper conclusion.
To find the plane's speed in still air and the speed of the wind, we can set up a system of equations based on the given information.
Let's say the speed of the plane in still air is "P" (in miles per hour) and the speed of the wind is "W" (in miles per hour).
On the flight from Fargo to Bismarck:
The distance is 180 miles and the time taken is 2 hours.
Since the plane is flying into a headwind, the effective speed will be reduced.
The equation for this part of the journey can be written as:
180 = (P - W) * 2
On the return flight from Bismarck to Fargo:
The distance is still 180 miles, but now the time taken is 1 hour.
Since the plane is now flying with the same headwind, the effective speed will be increased.
The equation for this part of the journey can be written as:
180 = (P + W) * 1
Now we have a system of two equations:
1) 180 = 2(P - W)
2) 180 = 1(P + W)
We can solve this system of equations to find the values of P (the plane's speed in still air) and W (the speed of the wind).
We can solve equation 2) for P:
180 = P + W
P = 180 - W
Now substitute P in equation 1) with its value from equation 2):
180 = 2(180 - W - W)
180 = 2(180 - 2W)
180 = 360 - 4W
4W = 360 - 180
4W = 180
W = 180 / 4
W = 45
Now substitute the value of W back into equation 2) to find P:
180 = P + 45
P = 180 - 45
P = 135
Therefore, the plane's speed in still air is 135 miles per hour, and the speed of the wind is 45 miles per hour.
To solve this problem, we can use the concept of relative velocity. Let's assume the speed of the plane in still air is "P" and the speed of the wind is "W."
Step 1: Calculate the speed of the plane
On the way to Bismarck, the plane is flying into a headwind, which means it is flying against the wind. Therefore, the effective speed of the plane will be the speed in still air minus the speed of the wind.
Let's assume the speed of the plane on the way to Bismarck is P-W.
Given that the distance between Fargo and Bismarck is 180 miles and the time it takes to travel that distance is 2 hours, we can write the equation:
Distance = Speed x Time
180 = (P - W) x 2
Step 2: Calculate the speed of the plane on the way back
On the way back from Bismarck to Fargo, the plane is flying with the wind. Therefore, the effective speed of the plane will be the speed in still air plus the speed of the wind.
Let's assume the speed of the plane on the way back is P+W.
Given that the distance between Bismarck and Fargo is 180 miles and the time it takes to travel that distance is 1 hour, we can write the equation:
Distance = Speed x Time
180 = (P + W) x 1
Step 3: Solve the equations
From equation (1): 180 = (P - W) x 2
Simplify equation (1): P - W = 90 or P = 90 + W
From equation (2): 180 = (P + W) x 1
Simplify equation (2): P + W = 180
Step 4: Use the values obtained from step 3 to solve for P and W
Substitute the value of P from equation (1) into equation (2):
(90 + W) + W = 180
90 + 2W = 180
2W = 180 - 90
2W = 90
W = 90/2
W = 45
Substitute the value of W into equation (2):
P + 45 = 180
P = 180 - 45
P = 135
So, the speed of the plane in still air is 135 mph, and the speed of the wind is 45 mph.