physics,please make me understand briefly

Question

two towns a and b are connected by a regular bus service with a bus leaving in either direction every T minutes. A man cycling with a speed of 20 km/h in the direction a to b notices that a bus goes past him every 18 min in the direction of his motion, and every 6 min in the opposite direction. What is the period t of the bus service and with what speed (assumed constant) do the buses ply on the road?

Answer 1

Hello! I'd be happy to help with your question.

First, let the bus speed be v km/hour.

We know that buses leave every T minutes; the distance between two buses going the same way is:

spacing = v * T/60

The cyclist's speed is 20 km/h.

Therefore:

v* T/60 = 18/60 (v-20)

v * T/60 = 6/60 (v+20)

For buses going in the same direction, the relative speed is:

v - 20

They pass him every 18 minutes:

spacing = (v - 20) * 18/60

For buses going in the opposite direction, the relative speed is:

v + 20

They pass him every 6 minutes:

spacing = (v + 20) * 6/60

Set them equal to one another:

(v - 20) * 18/60 = (v + 20) * 6/60

Now we cancel 1/60:

18( v - 20) = 6( v + 20)

18v - 360 = 6v + 120

12v = 480

v = 40

So, the buses travel at 40km/h

Now we find T using the same direction equation:

v* T/60 = 18/60 * (v - 20)

Substitute v = 40:

40 * T/60 = 18/60 * (20)

40T = 360

T = 9

So, the bus service period is: 9 minutes

Final Answer:

Buses leave every 9 minutes, and their speed is 40km/h.