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
We have to determine the period T of the bus service and speed of the buses.
Speed of each bus = Vb
Speed of the cyclist = Vc = 20 km/h
The relative velocity of the buses plying in the direction of motion of cyclist = Vb – Vc
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 = (Vb – Vc ) x 18/60
Since the buses are leaving every T minutes.
Therefore, the distance is equal to Vb x (T/60)
(Vb – Vc ) x 18/60 = Vb x (T/60) ——-(I)
The relative velocity of the buses plying in the direction opposite to the motion of cyclist is Vb + Vc
The buses go past the cyclist every 6 minutes i.e.(6/60) s.
Distance covered = (Vb + Vc ) x 6/60
Therefore, (Vb +Vc ) x 6/60 = Vb x (T/60)——(ii)
On dividing equation (ii) by (I) we get[(Vb – Vc ) x 18/60]/ [(Vb + Vc ) x 6/60 ]= [Vb x (T/60)] /[Vb x (T/60)]
(Vb – Vc ) 18/(Vb +Vc ) 6 = 1
(Vb – Vc )3 = (Vb +Vc )
On substituting the value of Vc
(Vb – 20 )3= (Vb + 20 )
3Vb – 60 = Vb + 20
2Vb = 80
Vb = 80/2
= 40 km/h
To determine the value of T, substitute the values of Vb and Vc in equation (i)
(Vb – Vc ) x 18/60 = Vb x (T/60)
(40 – 20) x (18/60) = 40 x (T/60)
T = (20 x 18) /40
= 9 minutes
Speed of the buses = 40km/h
Period T of bus service = 9 minutes