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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Motion in a Straight Line
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A 2m wide truck is moving with a uniform speed of $v_0=8 m/s$ along a staright horizontal road. A pedestrian starts to cross the road with a uniform speed of v when the truck is 4m away from him. The minimum value of 'v' so that he can cross the road safely is

\[(a)\; 2.62 m/s \quad (b)\;4.6 m/s \quad (c)\;3.57 m/s \quad (d)1.414 m/s \]

 

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1 Answer

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Let the man crossing the road at an angle $\theta$. For safe crossing the man must cross the road by the time the truck describes a distance $4+AC$
or $4+2 \cot \theta$
$=>\large\frac{AC+4}{v_0}=\frac{BC}{v}$
$=\large\frac{4+2 \cot \theta}{8}=\frac{2 /\sin \theta}{v}$
$v= \large\frac{8}{2 \sin \theta+\cos \theta}$
for minimum time
$\large\frac{dv}{d\theta}$$=0=>\large\frac{-8(2 \cos \theta- \sin \theta)}{(2 \sin \theta+\cos \theta)^2}$$=0$
$2 \cos \theta-\sin \theta=0$
$\tan \theta=2$
$=>\sin \theta= \large\frac{2}{\sqrt 5}$
$=>\cos \theta= \large\frac{1}{\sqrt 5}$
Therefore $v= \large\frac{8}{2 \times \Large\frac{2}{\sqrt 5}-\frac{1}{\sqrt 5}}$
$=\large\frac{8}{\sqrt 5}$
$=3.57 m/s$
Hence c is the correct answer.
answered Jun 25, 2013 by meena.p
edited May 23, 2014 by lmohan717
 

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