Two trains, Aand B , leave a train station at the same time but in different directions, one towards east and the other towards west. Train B travels 20km/h faster than train A. After 6 minutes, the two trains are 11 km apart. Find the average speed of each train.
speed of slower train --- x km/h
speed of faster train ---- x+20 km/h
6 minutes = 6/60 or 1/10 hour
(1/10)x + (1/10)(x+20) = 11
times 10
x + x+20 = 110
2x = 90
x = 45
the slower train goes at 45 km/h
the faster at 65 km/h
To find the average speed of each train, we can set up an equation based on the given information.
Let's assume the speed of train A is "x" km/h. Since train B is traveling 20 km/h faster than train A, the speed of train B would be "x + 20" km/h.
Now, let's consider the distances each train covers in 6 minutes (or 6/60 = 1/10 hours):
Train A covers a distance of (x * 1/10) km,
Train B covers a distance of ((x + 20) * 1/10) km.
According to the problem, the total distance covered by the two trains is 11 km. Therefore, we can write the equation:
(x * 1/10) + ((x + 20) * 1/10) = 11
To solve this equation, we can simplify it:
(x/10) + ((x + 20)/10) = 11
[(x + x + 20)/10] = 11
(2x + 20)/10 = 11
2x + 20 = 11 * 10
2x + 20 = 110
2x = 110 - 20
2x = 90
x = 90/2
x = 45
So, the average speed of train A is 45 km/h, and the average speed of train B is 45 + 20 = 65 km/h.