question_answer

A 2 m wide truck is moving with a uniform speed \[{{v}_{0}}=8\,\,m{{s}^{-1}}\] along a straight horizontal road. A pedestrian 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\[(4+2\cot \theta )\]. So,\[\frac{4+2\cot \theta }{2}=\frac{2/\sin \theta }{v}or\,\,v=\frac{2}{2\sin \theta +\cos \theta }\] For minimum\[v,\,\,\frac{dv}{d\theta }=0\] \[\Rightarrow \]               \[\frac{-2(2\cos \theta -\sin \theta )}{{{(2\sin \theta +\cos \theta )}^{2}}}=0\] \[\Rightarrow \]               \[2\cos \theta -\sin \theta =0\] \[\Rightarrow \]               \[\tan \theta =2,\,\,so\,\,\sin \theta =\frac{2}{\sqrt{5}},\,\,\cos \theta =\frac{1}{\sqrt{5}}\]      \[{{v}_{\min }}=\frac{2}{2(2/\sqrt{5})+(1/\sqrt{5})}=\frac{2}{\sqrt{5}}=0.89\,\,m/s\]