Two airplanes leave an airport at the same time, flying in the same direction. One of the plane is flying at twice the speed of the other. If after 4 hours they are 1800 km apart find the speed of each plane.
distance = speed * time, so if the speeds are s and 2s, then
4(2s) = 4s + 1800
To find the speed of each plane, we can set up a system of equations based on the given information.
Let's suppose the speed of the slower plane is x km/h.
Therefore, the speed of the faster plane would be 2x km/h.
Now, we know the distance traveled by both planes after 4 hours should add up to 1800 km.
Since distance = speed × time, we can write the following equations:
Distance covered by slower plane = x km/h × 4 hours = 4x km
Distance covered by faster plane = 2x km/h × 4 hours = 8x km
According to the problem, the sum of the distances covered by both planes equals 1800 km. Therefore:
4x + 8x = 1800
Combining like terms:
12x = 1800
To solve for x, we divide both sides of the equation by 12:
x = 1800/12
x = 150
So, the speed of the slower plane is 150 km/h, and the speed of the faster plane is twice that:
2x = 2 × 150
= 300 km/h
Hence, the speed of the slower plane is 150 km/h, and the speed of the faster plane is 300 km/h.