Q 11-03-182JEE MainEasy
A particle starts from rest and moves on a circle of radius $2$ m with its speed increasing as $v = 2t$ ($v$ in m/s, $t$ in s). The distance it covers before its total acceleration makes an angle of $45^\circ$ with its velocity is
Answer: (D) $1$ m
Concept. The tangential acceleration is $\dfrac{dv}{dt}$ and the centripetal acceleration is $\dfrac{v^2}{R}$. The total acceleration is at $45^\circ$ to the velocity when the two are equal.
Step 1. Components.
$$a_t = 2\ \text{m/s}^2,\qquad a_c = \frac{(2t)^2}{2} = 2t^2$$
Step 2. Condition. $2t^2 = 2 \Rightarrow t = 1$ s.
Step 3. Distance. $s = \displaystyle\int_0^1 2t\,dt = 1$ m.
Answer. $1$ m (Option D).
Solution by Sreeraj P, M.Sc Physics