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Question

A student is standing at a distance of $$50$$ meters behind a bus. As soon as the bus starts with an acceleration of $$1 \ ms^{-2}$$, the student starts running towards the bus with a uniform velocity $$u$$. Assuming the motion to be along straight road the minimum value of $$u$$, so that the student is able to catch the bus is:


A
5ms1
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B
8ms1
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C
10ms1
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D
12ms1
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Solution

The correct option is C $$10\;ms^{-1}$$

For bus $$\displaystyle s=\dfrac{1}{2}at^{2}\Rightarrow s=\dfrac{1}{2}\times 1\times t^{2}$$......$$(i)$$

For student $$(50+s)=ut$$........$$(ii)$$

from $$(i)$$ and $$(ii)$$ $$\displaystyle 50+\dfrac{t^{2}}{2}=ut$$

$$\displaystyle \Rightarrow t^{2}-2ut+100=0$$

$$\displaystyle t=\dfrac{2u\pm\sqrt{4u^{2}-4\times 100}}{2}$$

t must be real so $$4u^{2}-400=0$$ for $$t$$ to be minimum

$$u=10\;m/s$$


Physics

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