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

Speed of each bus = V_{b}

Speed of the cyclist = V_{c }= 20 km/h

The relative velocity of the buses plying in the direction of motion of cyclist is V_{b} – V_{c}

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 = (V_{b} – V_{c} ) x \( \dfrac{18}{60}\)

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

(V_{b} – V_{c} ) x\( \dfrac{18}{60}\) = V_{b} x (\( \dfrac{T}{60}\)) ——(1)

The relative velocity of the buses plying in the

direction opposite to the motion of cyclist is V_{b} + V_{c}

The buses go past the cyclist every 6 minutes i.e.(\( \dfrac{6}{60}\)) s.

Distance covered = (V_{b} + V_{c} ) x \( \dfrac{6}{60}\)

Therefore, (V_{b} +V_{c} ) x \( \dfrac{6}{60}\) = V_{b} x (\( \dfrac{T}{60}\))——(2)

Dividing (2) by (1)

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

= \( \dfrac{Vb} {\dfrac{T}{60}}\times \dfrac{Vb}{\dfrac{T}{60}}\)

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

\( (V_b – V_c )^3\) = (V_{b} +V_{c} )

Substituting the value of V_{c}

(V_{b} – 20 )3= (V_{b} + 20 )

3V_{b} – 60 = V_{b} + 20

2V_{b} = 80

V_{b} = \( \dfrac{18}{2}\) = 40 km/h

To find the value of T, substitute the values of V_{b} and V_{c} in equation (1)

(V_{b} – V_{c} ) x \( \dfrac{18}{60}\) = V_{b} 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 agoThe 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.

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**(b)** t = 2 s to 6 s.

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

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