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

### Related Questions

#### The velocity-time graph of a particle in one-dimensional motion is shown in the figure.

The velocity-time graph of a particle in one-dimensional motion is shown in the figure.

(a) $$x (t_2) = x (t_1) + v (t_1) (t_2 – t_1) +$$ ($$\dfrac{1}{2}$$$$a(t_2 – t_1)^2$$

(b) $$v(t_2) = v(t_1) + a(t_2 – t_1)$$

(c) Vaverage = $$\dfrac{ x(t_2) – x (t_1)}{(t_2 – t_1)}$$

(d)  aaverage =  $$\dfrac{v(t_2) – v (t_1) }{(t_2 – t_1)}$$

(e) $$x (t_2) = x (t_1) + v_{av} (t_2 – t_1) +$$ ($$\dfrac{1}{2}$$$$a_{av} (t_2 – t_1)^2$$

(f) $$x(t_2) – x (t_1)$$ = Area under the v-t curve bounded by t- axis and the dotted lines.

#### The speed-time graph of a particle moving along a fixed direction is shown in the figure.

The speed-time graph of a particle moving along a fixed direction is shown in the figure. Obtain the distance traversed by the particle between

(a) t = 0 s to 10 s,

(b) t = 2 s to 6 s.

#### Two stones are thrown up simultaneously from the edge of a cliff 200 m high with initial speeds of 15 m s–1

Two stones are thrown up simultaneously from the edge of a cliff 200 m high with initial speeds of $$15ms^{-1}$$ and $$30ms^{-1}$$Verify that the graph shown in given figure correctly represents the time variation of the relative position of the second stone with respect to the first. Neglect air resistance and assume that the stones do not rebound after hitting the ground. Take g = $$10ms^{-2}$$ Give the equations for the linear and curved parts of the plot.

#### On a long horizontally moving belt figure, a child runs to and fro with a speed 9 km h–1

On a long horizontally moving belt figure, a child runs to and fro with a speed 9 km h–1 (with respect to the belt) between his father and mother located 50 m apart on the moving belt. The belt moves with a speed of 4 km h–1. For an observer on a stationary platform outside, what is the

(a) speed of the child running in the direction of motion of the belt ?.

(b) speed of the child running opposite to the direction of motion of the belt?

(c) time taken by the child in (a) and (b)?

Which of the answers alter if motion is viewed by one of the parents?