Question

A particle starts from rest at time $t=0$ and move on a circular path of radius $1m$ with tangential acceleration $\sqrt{3}m/s^2$.After $1s$, its acceleration makes an angle $\theta$ with its velocity. What is the value of $3{\tan }^2\theta ?$

Answer 1

Draw a rough diagram


Find $\tan \theta$

Given:

$a_t=\sqrt{3}m/s^2,t=as,r=1m$

We know that the linear velocity is

$v=a_{t}t$

From the diagram

$\tan \theta =\frac{a_c}{a_t}$

$\tan \theta =\frac{\left(v^2/r\right)}{a_t}=\frac{v^2}{a_{t}r}$

$=\frac{a_t^2{t}^2}{a_{t}r}=\frac{a_t{t}^2}{r}$

$=\sqrt{3}\times \frac{1^2}{1}=\sqrt{3}$

$\therefore 3{\tan }^2\theta =3\times 3=9$

Final Answer :$9$