Flying with the wind, a plane traveled 570 miles in 3 hours. Flying against the wind, the plane traveled the same distance in 5 hours. Find the rate of the plane in calm air and the rate of the wind.
Rate of plane ______mph
Rate of wind ________mph
Oh, flying against the wind sounds like such a drag! But don't worry, I'm here to help with some comedic calculations.
Let's say the rate of the plane in calm air is "P" mph and the rate of the wind is "W" mph. When the plane is flying with the wind, it travels at a speed of P + W mph, and when it's flying against the wind, it travels at a speed of P - W mph.
We're given that flying with the wind, the plane traveled 570 miles in 3 hours, so we can set up the equation:
570 = (P + W) * 3
And when the plane is flying against the wind, it covers the same distance in 5 hours:
570 = (P - W) * 5
Now, let's solve this puzzle!
Divide the first equation by 3 and the second equation by 5 to break it down for our mathematical minds:
190 = P + W
114 = P - W
Now we have a system of equations! Are you ready for the grand finale?
Add the two equations together because we never turn down an opportunity for addition:
190 + 114 = 2P
304 = 2P
Finally, divide both sides by 2 to gracefully unveil the answer:
P = 152 mph
Now we know the rate of the plane in calm air! Sit tight, the finish line is close.
Substitute the value of P back into either of the original equations. Let's choose the first one:
190 = P + W
190 = 152 + W
To discover the rate of the wind, we simply subtract:
W = 190 - 152
W = 38 mph
And now we know the rate of the wind too! We did it!
So, the rate of the plane in calm air is 152 mph, and the rate of the wind is 38 mph. Happy flying!
To find the rate of the plane in calm air and the rate of the wind, we can use the concept of relative speed.
Let's assume the rate of the plane in calm air is "x" mph, and the rate of the wind is "y" mph.
When the plane flies with the wind, its effective speed is increased by the speed of the wind. Thus, the speed of the plane is (x + y) mph.
Given that the plane traveled 570 miles in 3 hours with the wind, we can set up the equation:
(x + y) * 3 = 570
Simplifying the equation, we get:
3x + 3y = 570 ----(1)
Similarly, when the plane flies against the wind, its effective speed is decreased by the speed of the wind. Thus, the speed of the plane is (x - y) mph.
Given that the plane traveled 570 miles in 5 hours against the wind, we can set up another equation:
(x - y) * 5 = 570
Simplifying the equation, we get:
5x - 5y = 570 ----(2)
Now, we have a system of two equations:
3x + 3y = 570 ----(1)
5x - 5y = 570 ----(2)
We can solve this system of equations by either substitution or elimination method. Let's use the elimination method, multiplying equation (1) by 5 and equation (2) by 3 to eliminate the "y" terms:
15x + 15y = 2850 ----(3)
15x - 15y = 1710 ----(4)
Adding equations (3) and (4), we get:
30x = 4560
x = 152
Substituting the value of x in equation (1), we can find the value of y:
3(152) + 3y = 570
456 + 3y = 570
3y = 570 - 456
3y = 114
y = 38
Therefore, the rate of the plane in calm air is 152 mph, and the rate of the wind is 38 mph.
To find the rate of the plane in calm air and the rate of the wind, we can set up a system of equations based on the given information.
Let's assume the rate of the plane in calm air is 'x' mph, and the rate of the wind is 'y' mph.
When the plane is flying with the wind, its effective speed is increased, which allows it to cover the distance of 570 miles in 3 hours. This can be expressed as:
(x + y) * 3 = 570
Similarly, when the plane is flying against the wind, its effective speed is decreased, which causes it to cover the same distance of 570 miles in 5 hours. This can be expressed as:
(x - y) * 5 = 570
Now, we can solve this system of equations to find the values of x and y.
Let's solve it using the method of substitution:
From the first equation, we can express x in terms of y:
x = (570/3) - y
x = 190 - y
Substituting this value of x into the second equation:
(190 - y - y) * 5 = 570
(190 - 2y) * 5 = 570
950 - 10y = 570
-10y = 570 - 950
-10y = -380
y = -380 / -10
y = 38
Now, substitute the value of y back into the equation x = 190 - y:
x = 190 - 38
x = 152
Therefore, the rate of the plane in calm air is 152 mph, and the rate of the wind is 38 mph.
Eq1: Vp + Vw = 570/3 = 190
Eq2: Vp - Vw = 570/5 = 114
Sum: 2Vp = 304
Vp = 152 mi/h
In Eq1, replace Vp with 152:
152 + Vw = 190
Vw = 38 mi/h.