Q) There is a regular bus service between 2 towns A and B,with a bus leaving town A and B every T minutes. A cyclist moving with a speed of 20 km h^-1 in the direction A to B notices that a bus goes past him every 18 minutes in the direction of his motion,and every 6 minutes in the opposite direction. Find the period T of the bus service and the speed with which the buses ply between A and B. Assume the speed of the bus to be constant.

A)

Let D = distance between A and B Let velocity of the buses = v Relative velocity of buses with respect to the cyclist (going from A to B) v-20 km/hr for buses going from A to B v+20 km/hr for Buses going from B to A Now use Time = distance/velocity Use the 2 equations to get d and v D/(v-20) = 18/60 ---------------------1 D/(v+20) = 6/60 -----------------------2 =>v=40 km/hr and D=6 km T = 6/40 hr = 9 mins Hence the speed of the buses is 40km/hr & the period of bus service is 9 mins . (A new bus is sent after every 9 mins)

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