Two towns A and B are connected by 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?

Asked by Pragya Singh | 1 year ago |  77

##### Solution :-

Speed of each bus = Vb

Speed of the cyclist = V= 20 km/h

The relative velocity of the buses plying in the direction of motion of cyclist is 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 $$\dfrac{18}{60}$$

Since the buses are leaving every T minutes. Therefore, the distance is equal to Vb x (T/60)

(Vb – Vc ) x$$\dfrac{18}{60}$$ = Vb x ($$\dfrac{T}{60}$$) ——(1)

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.($$\dfrac{6}{60}$$) s.

Distance covered = (Vb + Vc ) x $$\dfrac{6}{60}$$

Therefore, (Vb +Vc ) x $$\dfrac{6}{60}$$ = Vb x ($$\dfrac{T}{60}$$)——(2)

Dividing (2) by (1)

$$\dfrac{(Vb – Vc )} {\dfrac{18}{60}}\times \dfrac{(Vb + Vc )} {\dfrac{6}{60}}$$

$$\dfrac{Vb} {\dfrac{T}{60}}\times \dfrac{Vb}{\dfrac{T}{60}}$$

$$\dfrac{ (V_b – V_c ) 18}{(V_b +V_c ) 6}=1$$

$$(V_b – V_c )^3$$ = (Vb +Vc )

Substituting  the value of Vc

(Vb – 20 )3= (Vb + 20 )

3Vb – 60 = Vb + 20

2Vb = 80

Vb = $$\dfrac{18}{2}$$ = 40 km/h

To find the value of T, substitute the values of Vb and Vc in equation (1)

(Vb – Vc ) x $$\dfrac{18}{60}$$ = Vb x ($$\dfrac{T}{60}$$)

(40 – 20) x ($$\dfrac{18}{60}$$) = 40 x ($$\dfrac{T}{60}$$)

T = $$\dfrac{(20 \times 18)}{40}$$ = 9 minutes

Answered by Abhisek | 1 year ago

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