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Question

A bus starts from rest moving with an acceleration of $$2\ m/{s}^{2}$$. A cyclist, $$96\ m$$ behind the bus starts simultaneously towards the bus at $$20 \ m/s$$. After what time will he be able to overtake the bus:-


A
8 s
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B
10 s
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C
12 s
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D
1 s
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Solution

The correct option is A $$8\ s$$
Velocity of bus $$V_b= 0$$ , Velocity of cyclist $$V_c= 20 m/s$$, 
Acceleration of bus $$a_b= 2 m/s^2$$ Acceleration of cyclist $$a_c= 0$$
Relative velocity $$V_{cb}= V_c-V_b= 20 m/s $$
Relative acceleration $$a_{cb}= a_c-a_b= -2 m/s^2$$
Relative separation $$= 96 m$$
Using $$s=ut+\dfrac{1}{2} at^2$$
$$\Rightarrow 96= 20t-\dfrac{1}{2}2t^2$$   $$\Rightarrow t^2-20t+96=0$$
Solving the equation , we get $$t= 8 s$$ and $$t=12 s$$
Hence, after $$8s$$, cyclist will overtake the bus.

Physics

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