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, \[\frac{\text{dv}}{\text{d}\theta }=0\] |
\[\text{or}\,\,\,\frac{-\,8\left( 2\cos \theta -\sin \theta \right)}{{{\left( 2\sin +cos\theta \right)}^{2}}}=0\] |
\[\text{or }\,\,\text{2}\,\text{cos}\theta -\sin \theta =0\text{ or tan}\theta \,\text{=2}\] |
\[\text{From equation (i),}\] \[{{\text{v}}_{\min }}=\frac{8}{2\left( \frac{2}{\sqrt{5}} \right)+\frac{1}{\sqrt{5}}}=\frac{8}{\sqrt{5}}=3.57\text{ m/s}\] |
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