Question

Two towns A and B are connected by a regular bus service with a bus leaving in either directions either directions every T minutes. A man cycling with a speed of 20 kmh^-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?

Answer 1

Let v_b = Speed of each bus and v_c = speed of cyclist.

Relative speed of buses plying in the same direction of motion of cyclist =v_b -v_c.

The buses plying in the direction of the motion of the cyclist go past him after every 18 minutes, i.e., 18/60 hrs.

Distance covered is (v_b - v_c) x 18/60.

Since a bus leaves after avery T minutes, therefore distance is also equal to v_b x T/60.

Therefore, (v_b - v_c) x 18/60 = v_b x T/60.....(i)

Relative velocity of the buses plying opposite to the direction of motion of the cyclist is (v_b +v_c) after every 6 minutes.

Distance covered is (v_b +v_c) x 6/60

Thus, (v_b + v_c) x 6/60 = v_b x T/60 ......(ii)

Dividing (i) by (ii) we have

(v_b - v_c/v_b + v_c) = 18/6 = 1

On simplification, v_b = 2v_c but v_c = 20 kmh^-1

Hence, v_b = 40 km^-1.

From equation (i) (40 -20) x 18/60 = 40 x T/60

or T = 9 minutes.