question_answer

A 2 m wide truck is moving with a uniform speed \[{{v}_{0}}=8m{{s}^{-1}}\]along a straight horizontal road. A pedestrain starts to cross the road with a uniform speed v, when the truck is 4 m away from him. The minimum value of v so that he can cross the road safely is

A) \[2.62\,m{{s}^{-1}}\]

B)  \[4.6\,m{{s}^{-1}}\]

C)  \[0.89\,m{{s}^{-1}}\]

D)  \[1.414\,m{{s}^{-1}}\]

Correct Answer: C

Solution :

 Let the man start crossing the road at an angle \[\theta \] with the road side. For safe crossing, the condition is that the man must cross the road by the time, the truck describes the distance \[(4+2cot\theta )\] So, \[\frac{4+2\cot \theta }{2}=\frac{2/\sin \theta }{v}\]\[v=\frac{2}{2\sin \theta +\cos \theta }\] For minimum \[v,\frac{dv}{d\theta }=0\] \[\Rightarrow \] \[\frac{-2(2cos\theta -sin\theta )}{{{(2sin\theta +\cos \theta )}^{2}}}=0\] \[\Rightarrow \] \[2\cos \theta -\sin \theta =0\] \[\Rightarrow \]\[\tan \theta =2,\,\]so \[\sin \theta =\frac{2}{\sqrt{5}},\cos \theta =\frac{1}{\sqrt{5}}\] \[{{v}_{\min }}=\frac{2}{2(2/\sqrt{5})+(1/\sqrt{5})}=\frac{2}{\sqrt{5}}=0.89\,m/s\]