A particle moves in a circle with speed v var | vedclass

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(B) Given speed $v(t) = at$,where $a = 2 \ m/s^2$. In circular motion,$v = r \omega$,so $\omega = \frac{v}{r} = \frac{at}{r}$. Since $\omega = \frac{d

Vedclass · Accepted Answer

$2 \sqrt{1+64 \pi^2}$

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(B) Given speed $v(t) = at$,where $a = 2 \ m/s^2$. In circular motion,$v = r \omega$,so $\omega = \frac{v}{r} = \frac{at}{r}$. Since $\omega = \frac{d	heta}{dt}$,we have $d	heta = \frac{at}{r} dt$. Integrating for $n$ rounds,$	heta = 2\pi n = \int_0^t \frac{at}{r} dt = \frac{at^2}{2r}$. Thus,$t^2 = \frac{4\pi nr}{a}$. Radial acceleration $a_r = \frac{v^2}{r} = \frac{(at)^2}{r} = \frac{a^2 t^2}{r} = \frac{a^2}{r} \cdot \frac{4\pi nr}{a} = 4\pi na$. Tangential acceleration $a_t = \frac{dv}{dt} = a$. Total acceleration $A = \sqrt{a_t^2 + a_r^2} = \sqrt{a^2 + (4\pi na)^2} = a \sqrt{1 + (4\pi n)^2}$. Given $a = 2$ and $n = 2$,$A = 2 \sqrt{1 + (4 \cdot \pi \cdot 2)^2} = 2 \sqrt{1 + 64\pi^2}$.

$A$ particle moves in a circle with speed $v$ varying with time as $v(t) = 2t$. The total acceleration of the particle after it completes $2$ rounds of the cycle is:

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