(B) In circular motion,the car experiences two types of acceleration:$1$. Radial (centripetal) acceleration: $a_r = \frac{v^2}{r}$$2$. Tangential acce

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$\sqrt {\frac{v^4}{r^2} + a^2}$

(B) In circular motion,the car experiences two types of acceleration:$1$. Radial (centripetal) acceleration: $a_r = \frac{v^2}{r}$$2$. Tangential acceleration: $a_t = a$ (given as the rate of change of speed)Since these two accelerations are perpendicular to each other,the resultant acceleration $a_{net}$ is given by:$a_{net} = \sqrt{a_r^2 + a_t^2}$Substituting the values:$a_{net} = \sqrt{\left(\frac{v^2}{r}\right)^2 + a^2} = \sqrt{\frac{v^4}{r^2} + a^2}$

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

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$A$ car is travelling with linear velocity $v$ on a circular road of radius $r$. If it is increasing its speed at the rate of $a \ m/s^2$,then the resultant acceleration will be

A
$\sqrt {\frac{v^4}{r^2} - a^2}$
B
$\sqrt {\frac{v^4}{r^2} + a^2}$
C
$\sqrt {\frac{v^2}{r^2} - a^2}$
D
$\sqrt {\frac{v^2}{r^2} + a^2}$

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3-2.Motion in Plane

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

$A$ car is moving with speed $30 \,m/s$ on a circular path of radius $500 \,m$. Its speed is increasing at a rate of $2 \,m/s^2$. What is the total acceleration of the car (in $\,m/s^2$)?

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For a particle in circular motion,the centripetal acceleration is

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$A$ car is travelling at $40\text{ m/s}$ on a circular path of radius $40\text{ m}$. It is increasing its speed at the rate of $2\text{ m/s}^2$. Its net acceleration is (in $\text{m/s}^2$) nearly (in $\text{ m/s}^2$)

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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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The acceleration of a particle performing non-uniform circular motion is shown in the diagram at point $P$. If the radius of the circular path is $2 \ m$,then what is the velocity at point $P$ in $m/s$?

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$A$ car is travelling with linear velocity $v$ on a circular road of radius $r$. If it is increasing its speed at the rate of $a \ m/s^2$,then the resultant acceleration will be

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$A$ car is moving with speed $30 \,m/s$ on a circular path of radius $500 \,m$. Its speed is increasing at a rate of $2 \,m/s^2$. What is the total acceleration of the car (in $\,m/s^2$)?

For a particle in circular motion,the centripetal acceleration is

$A$ car is travelling at $40\text{ m/s}$ on a circular path of radius $40\text{ m}$. It is increasing its speed at the rate of $2\text{ m/s}^2$. Its net acceleration is (in $\text{m/s}^2$) nearly (in $\text{ m/s}^2$)

$A$ 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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