As shown below AB represents an infinite wall tangential to a horizontal semi circular track O is point source of light on the ground at the center of the circle .A block moves along the circular track with speed v starting from the point where the wall touches the circle .If the velocity and acceleration of shadow along the length of the wall is respectively v and $α$ then

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$V=vcos (\frac{vt}{R})$

${V=v sec}^{2} (\frac{vt}{R})$

$a= (\frac{v^{2}}{R}) {sec}^{2} (\frac{vt}{R}) tan (\frac{vt}{R})$

$a= (\frac{{2v}^{2}}{R}) {sec}^{2} (\frac{vt}{R}) tan (\frac{vt}{R})$

answer is B.

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Detailed Solution

$θ=ωt$

$θ= \frac{V}{R} t$

$Tanθ= \frac{x}{R}$

$x=RTanθ=RTan \frac{Vt}{R}$

$\frac{d x}{d t} = R \sec^{2} \frac{V t}{R} . \frac{V}{R}$

$a_{S} = \frac{d V_{s}}{d t} = V .2 \sec (\frac{V t}{R}) . \sec (\frac{V t}{R}) T a n \frac{V t}{R} . \frac{V}{R}$

$a_{S} = \frac{2 V^{2}}{R} \sec^{2} (\frac{V t}{R}) T a n (\frac{V t}{R})$

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