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
B
8ms1
C
10ms1
D
12ms1

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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