Two towns A and B are connected by regular bus service with a bus leaving in either direction every T minutes. A man cycling with a speed of 20 km h–1 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?

Solution

Given that

Speed of the cyclist = 20 km/h

A bus goes past him every 18 min in the direction of cyclist motion and 6 min in the opposite direction

Find out

We have to determine the period T of the bus service and speed of the buses.

Solution

Speed of each bus = V_b

Speed of the cyclist = V_c = 20 km/h

The relative velocity of the buses plying in the direction of motion of cyclist = V_b – V_c
The buses playing in the direction of motion of the cyclist go past him after every 18 minutes i.e.(18/60) s.

Distance covered = (V_b – V_c ) x 18/60

Since the buses are leaving every T minutes.

Therefore, the distance is equal to V_b x (T/60)

(V_b – V_c ) x 18/60 = V_b x (T/60) ——-(I)

The relative velocity of the buses plying in the direction opposite to the motion of cyclist is V_b + V_c
The buses go past the cyclist every 6 minutes i.e.(6/60) s.

Distance covered = (V_b + V_c ) x 6/60

Therefore, (V_b +V_c ) x 6/60 = V_b x (T/60)——(ii)

On dividing equation (ii) by (I) we get

(V_b – V_c ) 18/(V_b +V_c ) 6 = 1

(V_b – V_c )3 = (V_b +V_c )

On substituting the value of V_c

(V_b – 20 )3= (V_b + 20 )

3V_b – 60 = V_b + 20

2V_b = 80

V_b = 80/2

= 40 km/h

To determine the value of T, substitute the values of V_b and V_c in equation (i)

(V_b – V_c ) x 18/60 = V_b x (T/60)

(40 – 20) x (18/60) = 40 x (T/60)

T = (20 x 18) /40

= 9 minutes

Answer

Speed of the buses = 40km/h

Period T of bus service = 9 minutes