A bus left the city for the village with an average speed of v1 = 50 km / h

A bus left the city for the village with an average speed of v1 = 50 km / h, after a time t1 = 1 hour a minibus followed it v2 = 60 km / h at what distance and how long will the bus catch up?

Given:

v1 = 50 km / h – average speed of the bus;

t1 = 1 hour – the time interval after which the minibus left after the bus;

v2 = 60 km / h – the average speed of the minibus.

It is required to determine the time t (hour) after which the minibus will catch up with the bus.

Let’s introduce a coordinate system, the origin of which will be the point of departure of the bus and minibus.

For the initial moment of time, we will take the departure of the minibus. Then the equation of the minibus motion will look like:

x2 (t) = v2 * t, and the bus:

x1 (t) = v1 * t1 + v1 * t – where v1 * t1 is the path that the bus managed to drive before the minibus exit.

To find the meeting time, let us equate both equations of motion (since at the meeting point their coordinates will be the same):

x1 (t) = x2 (t);

v1 * t1 + v1 * t = v2 * t;

v1 * t1 = v2 * t – v1 * t;

v1 * t1 = t * (v2 – v1);

t = v1 * t / (v2 – v1) = 50 * 1 / (60 – 50) = 50/10 = 5 hours.

Answer: the minibus will catch up with the bus in 5 hours.