Question

Angular position of a particle performing circular motion is given by $\theta =(t^2-2t)rad$. Consider the ads circular track $2m$ Angular distance of particle at $t=3s$

Answer 1

Angular distance of particle $\theta$ is given by $\theta =(t^2-3t)$ rad

Given radius of circular track $r=2$m

Angular distance$=r\theta$

$=2(t^2-2t)$

$=(2t^2-4t)m$

At $t=3s$

Angular distance$=2(3{)}^2-4(3)$

$=18-12=6$m

$\therefore$ Angular distance at $t=3s$ is $6$m.