From points A and B, two buses left at the same time towards each other, and the first, having twice the speed

| August 11, 2021 | education

From points A and B, two buses left at the same time towards each other, and the first, having twice the speed, traveled all the way 1 hour faster than the second. How many minutes earlier would they meet if the speed of the 2nd increased to the speed of the 1st?

Suppose that the distance from A to B is equal to a km and the speed of the second bus is x km / h. Then the speed of the first bus is 2 * x km / h.

This means that the second bus travels the entire distance in a / x hours, and the first bus travels this distance in a / 2 * x hours.

We get the following equation:

a / x – a / 2 * x = 1,

(2 * a – a) / 2 * x = 1,

a = 2 * x,

x = a / 2 (km / h) – the speed of the second bus.

2 * a / 2 = a (km / h) – the speed of the first bus.

At this speed, the buses will meet through:

a: (a + a / 2) = a: (3 * a / 2) = 2/3 (h) = 4 (min).

If the second bus also travels at a speed of a km / h, then they will meet through:

a: (a + a) = 1/2 (h) = 30 (min).

This means that the buses will meet 40 – 30 = 10 minutes earlier.



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