Question

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 $\alpha $ between the vectors of total acceleration and the velocity of the particle is

• A. $\alpha=\tan ^{-1}\left(\frac{2 l}{r}\right)$
• B. $\alpha=\cos ^{-1}\left(\frac{2 l}{T}\right)$
• C. $\alpha=\sin ^{-1}\left(\frac{2 l}{r}\right)$
• D. $\alpha=\cot ^{-1}\left(\frac{2 l}{T}\right)$

Answer 1

Answer: (A)

Let, when particle is at angular position $\theta $ 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 $a_{c} = \frac{v^{2}}{r} = \left(\right. \frac{a^{2} l}{r} \left.\right)$
Angle between $a_{n e t}$ &$a_{t}$ :
$\alpha=\tan ^{-1}\left(\frac{2 l}{T}\right)$