Two towns `A` and `B` are connected by a regular bus service with a bus leaving in either direction every `T` min. `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?

Explanation

Let ` km h^(-1)` be the constant speed with which the buses pky between the torwns ` A and B` The relative velocity of the bus ( for the motion ` A to B` 0 with respect to the cylist ( i.e. in the dierction in which the cyclist is going = 9v - 20 0 km h^(-1)`. The relative velocity of the bus from ` B to A` with cyclist ` = ( v+ 20 0 km h^(-1)`
The distanc etravelled by the bus in time ` T ( minutes) = vT`
As per question ` ( vT)/(v- 20) = 18` or ` v T = 18 v- 18 xx 20 ` ...(i)
and ` ( vt) /( v+ 20) = 6 or ` v T = 6 v + 20 xx 6` (ii)
Equating (i) & (ii) we get
` 18 v- 18 xx 20 = 6 v+ 20 xx 6` or ` 12v = 20 6 + 18 xx 20 = 480 or v= 40 km h^(-1_`
Puttig this value of (v) in(i) we get
` 40 T= 18 xx 40 - 18 xx 20 = 18 xx 20 or ` T = 18 xx 20// 40 = 9 min`