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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. The slope of the line is $	an

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$2^{2/3} \ s$

Vedclass · Answer

(C) From the given graph,the tangential acceleration $a_T$ is a linear function of time $t$ passing through the origin. The slope of the line is $	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 $v(0) = 0$,we get $v = \int_0^t \sqrt{3} t \ dt = \frac{\sqrt{3} t^2}{2}$.The centripetal (radial) acceleration is $a_c = \frac{v^2}{r}$. Given $r = 1 \ m$,$a_c = \frac{(\sqrt{3} t^2 / 2)^2}{1} = \frac{3 t^4}{4}$.The total acceleration vector $\vec{a}$ makes an angle $	heta = 30^{\circ}$ with the radial acceleration vector $\vec{a}_c$. Since $\vec{a}_c$ and $\vec{a}_T$ are perpendicular,we have $	an 	heta = \frac{a_T}{a_c}$.Substituting the values: $	an 30^{\circ} = \frac{\sqrt{3} t}{3 t^4 / 4} \Rightarrow \frac{1}{\sqrt{3}} = \frac{4 \sqrt{3} t}{3 t^4} = \frac{4}{t^3}$.Solving for $t$: $t^3 = 4 \sqrt{3} 	imes \sqrt{3} = 4 	imes 3 = 12$. This seems to contradict the options. Let's re-evaluate: $	an 30^{\circ} = \frac{a_T}{a_c} \Rightarrow \frac{1}{\sqrt{3}} = \frac{\sqrt{3} t}{3 t^4 / 4} = \frac{4 \sqrt{3} t}{3 t^4} = \frac{4}{\sqrt{3} t^3}$.Thus,$t^3 = 4 	imes 3 / 1 = 12$ is incorrect. Let's re-check: $	an 30^{\circ} = \frac{a_T}{a_c} = \frac{\sqrt{3} t}{3 t^4 / 4} = \frac{4 \sqrt{3} t}{3 t^4} = \frac{4}{\sqrt{3} t^3}$.Wait,$\frac{1}{\sqrt{3}} = \frac{4}{\sqrt{3} t^3} \Rightarrow t^3 = 4 \Rightarrow t = 4^{1/3} = 2^{2/3} \ s$.

The 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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