A particle travels along the arc of a circle of radius $r$ . Its speed depends on the distance travelled $l$ as $v = a \sqrt{l},$ where $‘ a ’$ is a constant. The angle $α$ between the vectors of net acceleration and the velocity of the particle is

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$α = \cos^{- 1} (\frac{2 r}{l})$

$α = \sin^{- 1} (\frac{2 r}{l})$

$α = \tan^{- 1} (\frac{2 l}{r})$

$α = s e c^{- 1} (\frac{2 r}{l})$

answer is C.

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Detailed Solution

Let, when particle is at angular position $α$ then distance travelled $= l$

$v = a \sqrt{l};$ But $a_{t} = \frac{d v}{d t} = \frac{a}{2 \sqrt{l}} \frac{d l}{d t} = \frac{a v}{2 \sqrt{l}} = \frac{a^{2}}{2}$ and $a_{c} = \frac{v^{2}}{r} = (\frac{a^{2} l}{r})$

Angle between $a_{n e t}$ & $a_{t} :; α = \tan^{- 1} (\frac{2 l}{r})$

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