(D) The magnitude of the change in velocity $\Delta \vec{v}$ for a particle moving with constant speed $v$ through an angle $\theta$ is given by the f

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(D) The magnitude of the change in velocity $\Delta \vec{v}$ for a particle moving with constant speed $v$ through an angle $\theta$ is given by the formula: $\Delta v = 2v \sin\left(\frac{\theta}{2}\right)$ Given: Speed $v = 10\,m/s$ Angle $\theta = 60^{\circ}$ Substituting the values: $\Delta v = 2 \times 10 \times \sin\left(\frac{60^{\circ}}{2}\right)$ $\Delta v = 20 \times \sin(30^{\circ})$ Since $\sin(30^{\circ}) = 0.5$: $\Delta v = 20 \times 0.5 = 10\,m/s$ Therefore,the magnitude of the change in velocity is $10\,m/s$.

Class 11 Physics 3-2.Motion in Plane Angular Variables and Basic of Uniform Circular Motion

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$A$ particle is moving along a circular path with a constant speed of $10\,m/s.$ What is the magnitude of the change in velocity of the particle,when it moves through an angle of $60^{\circ}$ around the centre of the circle? .......... $m/s$

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A
$10\sqrt{3}$
B
$0$
C
$10\sqrt{2}$
D
$10$

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3-2.Motion in Plane

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Angular Variables and Basic of Uniform Circular Motion

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As shown in the figure,a particle is moving with a constant speed $v = \pi \, m/s$ in a circular path. Considering its motion from point $A$ to point $B$,where the angle subtended at the center is $120^{\circ}$,the magnitude of the average velocity is:

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$A$ particle is moving along a circular path with a constant speed of $10\,m/s.$ What is the magnitude of the change in velocity of the particle,when it moves through an angle of $60^{\circ}$ around the centre of the circle? .......... $m/s$

Similar Questions

$A$ heavy stone is tied to the end of a string and whirled in a horizontal circular path of radius $20 \, cm$ with a constant angular speed. If the centripetal acceleration is $980 \, cm \, s^{-2}$,what is its angular speed?

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Assertion $(A)$ :- If a body is in a state of uniform circular motion,then its velocity and acceleration both are varying.Reason $(R)$ :- If the magnitude of velocity is $v$ and the radius of uniform circular motion is $r$,then the magnitude of acceleration is $v^2/r$.

As shown in the figure,a particle is moving with a constant speed $v = \pi \, m/s$ in a circular path. Considering its motion from point $A$ to point $B$,where the angle subtended at the center is $120^{\circ}$,the magnitude of the average velocity is:

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Assertion $(A)$ :- If a body is in a state of uniform circular motion,then its velocity and acceleration both are varying.
Reason $(R)$ :- If the magnitude of velocity is $v$ and the radius of uniform circular motion is $r$,then the magnitude of acceleration is $v^2/r$.