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^(-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 v km `h^(-1)` be the constant speed with which the buses ply between the towns A and B.
The relative velocity of the bus (for the motion A to B) w.r.to the cyclist is going = (v-20) `kmh^(-1)`. The relative velocity of the bus from B to A w.r.to the cyclist = (v + 20) `kmh^(-1)`.
The distance travelled by the bus in time T(min) = `upsilonT`
As per question `(vt)/(v-20)=18` (Or) vT
`=18v-18 times 20" ... (i)"`
and `(vT)/(v+20)=6" (Or) "vT=6v+20 times 6" ..... (ii)"`
Equation (i) and (ii) we get
`18v-18 times 20=6v+20 times 6" (Or)"`
`12v=20 times 6+18 times 20=480`
(Or) `upsilon=40kmh^(-1)`
Putting this value of `upsilon` in (i) we get
`40 T=18 times 40-18 times 20=18 times 20`
(Or) `T=18 times 20//40=9` min.