(A) The normal acceleration is given by $a_n = \frac{v^2}{R} = k t^\alpha$,where $k$ is a constant.From this,the square of the velocity is $v^2 = R k

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(A) The normal acceleration is given by $a_n = \frac{v^2}{R} = k t^\alpha$,where $k$ is a constant.From this,the square of the velocity is $v^2 = R k t^\alpha$.The kinetic energy of the object is $K = \frac{1}{2} m v^2 = \frac{1}{2} m R k t^\alpha$.The power $P$ developed by all forces is equal to the rate of change of kinetic energy,$P = \frac{dK}{dt}$.$P = \frac{d}{dt} \left( \frac{1}{2} m R k t^\alpha \right) = \frac{1}{2} m R k \alpha t^{\alpha-1}$.Since $m$,$R$,$k$,and $\alpha$ are constants,we have $P \propto t^{\alpha-1}$.

Class 11 Physics 3-2.Motion in Plane Kinematics Circular Motion (Uniform Angular Accelaration)

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An object moves along a circle with normal acceleration proportional to $t^\alpha$,where $t$ is the time and $\alpha$ is a positive constant. The power developed by all the forces acting on the object will have time dependence proportional to

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$t^{\alpha-1}$
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$t^{\alpha / 2}$
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$t^{\frac{1+\alpha}{2}}$
D
$t^{2 \alpha}$

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3-2.Motion in Plane

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An object moves along a circle with normal acceleration proportional to $t^\alpha$,where $t$ is the time and $\alpha$ is a positive constant. The power developed by all the forces acting on the object will have time dependence proportional to

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If the equation for the angular displacement of a particle moving on a circular path is given by:
$\theta = 2t^3 + 0.5$
Where $\theta$ is in radians and $t$ is in seconds,then the angular velocity of the particle at $t = 2 \, s$ is: