question_answer

A 2 m wide truck is moving with a uniform speed \[{{\text{v}}_{\text{0}}}\text{= 8 m/s}\] 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        

B)        4.6 m/s

C) 3.57 m/s        

D)        1.414 m/s

Correct Answer: C

Solution :

[c] Let the man starts crossing the road at an angle \[\theta \] as shown in figure. For safe crossing the condition is that the man must cross the road by the time the truck describes the distance 4 + AC or \[4+2cot\theta \]..

\[\therefore \frac{4+2\cot \theta }{\text{g}}=\frac{2/\sin \theta }{\text{v}}\text{ or v=}\frac{8}{2\sin \theta +\cos \theta }\,\,...\left( \text{i} \right)\]

For minimum v, \[\frac{\text{dv}}{\text{d}\theta }=0\]

\[\text{or}\,\,\,\frac{-\,8\left( 2\cos \theta -\sin \theta  \right)}{{{\left( 2\sin +cos\theta  \right)}^{2}}}=0\]

\[\text{or }\,\,\text{2}\,\text{cos}\theta -\sin \theta =0\text{ or tan}\theta \,\text{=2}\]

\[\text{From equation (i),}\] \[{{\text{v}}_{\min }}=\frac{8}{2\left( \frac{2}{\sqrt{5}} \right)+\frac{1}{\sqrt{5}}}=\frac{8}{\sqrt{5}}=3.57\text{ m/s}\]