(B) Given,$v = a\sqrt{s}$.Tangential acceleration $a_t = v \frac{dv}{ds}$.Since $\frac{dv}{ds} = a \cdot \frac{1}{2\sqrt{s}} = \frac{a}{2\sqrt{s}}$,we

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(B) Given,$v = a\sqrt{s}$.Tangential acceleration $a_t = v \frac{dv}{ds}$.Since $\frac{dv}{ds} = a \cdot \frac{1}{2\sqrt{s}} = \frac{a}{2\sqrt{s}}$,we have $a_t = (a\sqrt{s}) \cdot (\frac{a}{2\sqrt{s}}) = \frac{a^2}{2}$.Centripetal acceleration $a_c = \frac{v^2}{R} = \frac{(a\sqrt{s})^2}{R} = \frac{a^2 s}{R}$.The angle $\alpha$ between the total acceleration vector and the velocity vector is given by $\tan \alpha = \frac{a_c}{a_t}$.Substituting the values,$\tan \alpha = \frac{a^2 s / R}{a^2 / 2} = \frac{2s}{R}$.

Class 11 Physics 3-2.Motion in Plane Non-uniform Circular Motion (Centrifugal/Centripetal and tangential Accelaration)

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$A$ particle moves along an arc of a circle of radius $R$. Its velocity depends on the distance covered as $v = a\sqrt{s}$,where $a$ is a constant. Then the angle $\alpha$ between the vector of the total acceleration and the vector of velocity as a function of $s$ will be:

A
$\tan \alpha = \frac{R}{2s}$
B
$\tan \alpha = \frac{2s}{R}$
C
$\tan \alpha = \frac{2R}{s}$
D
$\tan \alpha = \frac{s}{2R}$

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3-2.Motion in Plane

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Non-uniform Circular Motion (Centrifugal/Centripetal and tangential Accelaration)

In the given figure,$a = 15 \, m s^{-2}$ represents the total acceleration of a particle moving in the clockwise direction in a circle of radius $R = 2.5 \, m$ at a given instant of time. The speed of the particle is ........ $m/s$.

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The force required to keep a body in uniform circular motion is:

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The magnitude of the centripetal force acting on a body of mass $m$ executing uniform circular motion in a circle of radius $r$ with speed $v$ is:

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$A$ car is travelling at $30 \,ms^{-1}$ speed on a circular road of radius $300 \,m$. If its speed is increasing at the rate of $4 \,ms^{-2}$, then its acceleration is (in $\,ms^{-2}$)

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$A$ particle moves on a circle of radius $0.1 \ m$ with a velocity $v = 1.0t$. The total acceleration at $t = 5 \ s$ will be ........ $m/s^2$.

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$A$ particle moves along an arc of a circle of radius $R$. Its velocity depends on the distance covered as $v = a\sqrt{s}$,where $a$ is a constant. Then the angle $\alpha$ between the vector of the total acceleration and the vector of velocity as a function of $s$ will be:

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In the given figure,$a = 15 \, m s^{-2}$ represents the total acceleration of a particle moving in the clockwise direction in a circle of radius $R = 2.5 \, m$ at a given instant of time. The speed of the particle is ........ $m/s$.

The force required to keep a body in uniform circular motion is:

The magnitude of the centripetal force acting on a body of mass $m$ executing uniform circular motion in a circle of radius $r$ with speed $v$ is:

$A$ car is travelling at $30 \,ms^{-1}$ speed on a circular road of radius $300 \,m$. If its speed is increasing at the rate of $4 \,ms^{-2}$, then its acceleration is (in $\,ms^{-2}$)

$A$ particle moves on a circle of radius $0.1 \ m$ with a velocity $v = 1.0t$. The total acceleration at $t = 5 \ s$ will be ........ $m/s^2$.

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