1) An airplane took 2 hours to fly 600 km against a head wind. The return trip with the wind took 1 2/3 hours. Find the speed of the plane in still air.
What was wrong with the way I showed you earlier?
oh, I just can't find the link right now.
1) An airplane took 2 hours to fly 600 km against a head wind. The return trip with the wind took 1 2/3 hours. Find the speed of the plane in still air.
could you post the equation and answer?
To find the speed of the plane in still air, we can use the concept of relative speed. Let's suppose the speed of the plane in still air is represented by 'x' km/h, and the speed of the wind is represented by 'y' km/h.
When the plane is flying against the headwind, its effective speed (relative to the ground) is reduced by the speed of the wind. Therefore, the speed against the wind is (x - y) km/h. We are given that the plane took 2 hours to fly 600 km against the headwind, so we can write the equation:
(x - y) * 2 = 600
When the plane is flying with the wind, its effective speed (relative to the ground) is increased by the speed of the wind. Therefore, the speed with the wind is (x + y) km/h. We are given that the plane took 1 2/3 hours (or 5/3 hours) to fly back, so we can write the equation:
(x + y) * (5/3) = 600
Now we have two equations with two unknowns (x and y). We can solve this system of equations to find the values of x and y.
Let's first solve the first equation for x:
2x - 2y = 600
x - y = 300 (Dividing both sides by 2)
Now let's solve the second equation for x:
(5/3)x + (5/3)y = 600
5x + 5y = 1800 (Multiplying both sides by 3)
x + y = 360 (Dividing both sides by 5/3)
Now we have a system of equations:
x - y = 300
x + y = 360
By adding the two equations together, we can eliminate the 'y' variable:
(2x + 0y) = (300 + 360)
2x = 660
x = 330
Therefore, the speed of the plane in still air is 330 km/h.