Question

# Tangential acceleration of a particle moving in a circle of radius $$1\ m$$ varies with time $$t$$ as (initial velocity of particle is zero). Time after which total acceleration of particle makes and angle of $$30^{o}$$ with radial acceleration is

A
4 sec
B
4/3 sec
C
22/3 sec
D
2 sec

Solution

## The correct option is C $$2^{2/3}\ sec$$using equation for a straight line $$(y=mx+c)$$ where,$$m=\tan 60^o=\sqrt{3}$$ (slop) $$C=0$$ from graphrelation b/w $$a_T$$ and $$t$$ is obtained as:-$$a_T=\tan 60^ot+0$$this tangential acceleration increases velocity (v) of the particless as.$$a_T =\dfrac{dv}{dt} \Rightarrow \dfrac{dv}{dt}=\sqrt{3}t$$$$\Rightarrow \int dv =\int \sqrt{3}+dt \Rightarrow v=\dfrac{\sqrt{3}t^2}{2}$$so, $$a_T$$ a particular time, centripetal accin  $$(a_c)$$ is given by :-$$a_c=\dfrac{v^2}{r}=\dfrac{3}{4}t^4 (r=1) \, \vec{a_T}$$let a time $$\vec{a}$$ masses angle $$30^o$$ with $$\vec{a_c}$$. so it makes angles $$60^o$$ with $$\vec{a_T}$$ ($$\vec{a_c} K \vec{a_T}$$ are perpendicular)$$\Rightarrow \vec{a_T}=\sqrt{3t^2+\dfrac{9}{16}+8}\dfrac{1}{2}$$ that given,$$\Rightarrow 3t^2=\dfrac{1}{11}(3t^2+\dfrac{9}{16}+8)$$$$\Rightarrow t=0$$ or $$t=2^{2/3}$$.Physics

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