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(C) From the given graph,the tangential acceleration $a_T$ is a linear function of time $t$ passing through the origin with a slope of $	an 60^{\circ

Vedclass · Accepted Answer

$2^{2/3} \, s$

Vedclass · Answer

(C) From the given graph,the tangential acceleration $a_T$ is a linear function of time $t$ passing through the origin with a slope of $	an 60^{\circ} = \sqrt{3}$.Thus,$a_T = \sqrt{3} t$.Since $a_T = \frac{dv}{dt}$,we have $\frac{dv}{dt} = \sqrt{3} t$.Integrating with respect to time (with initial velocity $v=0$ at $t=0$): $v = \int_{0}^{t} \sqrt{3} t \, dt = \frac{\sqrt{3}}{2} t^2$.The centripetal (radial) acceleration is $a_c = \frac{v^2}{r}$. Given $r = 1 \, m$,we have $a_c = \frac{(\frac{\sqrt{3}}{2} t^2)^2}{1} = \frac{3}{4} t^4$.The total acceleration vector $\vec{a}$ is the vector sum of the radial acceleration $\vec{a}_c$ and the tangential acceleration $\vec{a}_T$. Since $\vec{a}_c$ and $\vec{a}_T$ are perpendicular,the angle $	heta$ that the total acceleration makes with the radial acceleration $\vec{a}_c$ is given by $	an 	heta = \frac{a_T}{a_c}$.We are given $	heta = 30^{\circ}$,so $	an 30^{\circ} = \frac{a_T}{a_c}$.$\frac{1}{\sqrt{3}} = \frac{\sqrt{3} t}{\frac{3}{4} t^4} = \frac{4 \sqrt{3} t}{3 t^4} = \frac{4}{\sqrt{3} t^3}$.$t^3 = \frac{4 	imes \sqrt{3}}{\sqrt{3}} = 4$.$t = 4^{1/3} = (2^2)^{1/3} = 2^{2/3} \, s$.

Tangential acceleration of a particle moving in a circle of radius $1 \, m$ varies with time $t$ as shown in the graph (initial velocity of the particle is zero). The time after which the total acceleration of the particle makes an angle of $30^{\circ}$ with the radial acceleration is:

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