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

Answer 1

Answer: (A)

Let, when particle is at angular position $\theta$ then distance travelled $=l$
$v=a\sqrt{l}$
But $a_t=\frac{dv}{dt}=\frac{a}{2\sqrt{l}}\frac{dl}{dt}=\frac{av}{2\sqrt{l}}=\frac{a^2}{2}$
And $a_c=\frac{v^2}{r}=(\frac{a^{2}l}{r})$
Angle between $a_{net}$ &$a_t$ :
$\alpha ={\tan }^{-1}\left(\frac{2l}{T}\right)$