When a plane flies into the wind, it can travel 3000mi in 6 hr. When it flies with the wind, it can travel the same distance in 5 hr. Find the rate of the plane in still air and the rate of the wind.
x = Rate without wind
y = Rate of the wind
T = D/R
6 = 3000/(x - y)
6(x - y) = 3000
6x - 6y = 3000
5 = 3000/(x + y)
5(x + y) = 3000
5x + 5y = 3000
Solve simultaneously,
6x - 6y = 3000
5x + 5y = 3000
5 *(6x - 6y = 3000)= 30x - 30y = 15,000
6 *(5x + 5y = 3000)= 30x + 30y = 18,000
30x - 30y = 15,000
30x + 30y = 18,000
60x = 33,000
x = 550 rate of the plane in still air
5x + 5y = 3000
x = 550
5(550) + 5y = 3000
2750 + 5y = 3000
5y = 250
y = 50 rate of the wind
To find the rate of the plane in still air (let's call it P) and the rate of the wind (let's call it W), we can set up a system of equations based on the given information:
1. When the plane flies into the wind:
- Distance = 3000 miles
- Time = 6 hours
- Rate (P - W) = P (rate of the plane in still air) minus W (rate of the wind)
2. When the plane flies with the wind:
- Distance = 3000 miles
- Time = 5 hours
- Rate (P + W) = P (rate of the plane in still air) plus W (rate of the wind)
We can use these equations to solve for P and W.
Let's start with the first equation, where the rate of the plane (P - W) is 3000 miles / 6 hours:
(P - W) = 3000 / 6
P - W = 500 (Equation 1)
Now, let's move to the second equation, where the rate of the plane (P + W) is 3000 miles / 5 hours:
(P + W) = 3000 / 5
P + W = 600 (Equation 2)
Now we have a system of linear equations (Equation 1 and Equation 2) which we can solve simultaneously.
Adding Equation 1 and Equation 2, we eliminate the variable W:
(P - W) + (P + W) = 500 + 600
2P = 1100
P = 550
Substituting the value of P back into Equation 2, we can solve for W:
550 + W = 600
W = 600 - 550
W = 50
Therefore, the rate of the plane in still air is 550 mph, and the rate of the wind is 50 mph.