A particle is moving on a circular path with a constant speed v. It's change of velocity as it moves from A to B in the figure is
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In uniform circular motion, speed is constant but velocity changes due to change in direction.
Change in velocity \( \Delta\vec{v} = \vec{v}_f - \vec{v}_i \).
Magnitude of change: \( |\Delta\vec{v}| = \sqrt{v_f^2 + v_i^2 - 2v_f v_i \cos\phi} \), where \(\phi\) is the angle between \(\vec{v}_f\) and \(\vec{v}_i\).
For uniform circular motion, \(v_i=v_f=v\). The angle turned by the velocity vector is the same as the angle subtended at the center \(\theta\).
So, \( |\Delta\vec{v}| = \sqrt{v^2+v^2-2v^2\cos\theta} = \sqrt{2v^2(1-\cos\theta)} \).
Use \(1-\cos\theta = 2\sin^2(\theta/2)\).
The particle moves with constant speed \(v\) on a circular path. This means the magnitude of its velocity is always \(v\).
Let \(\vec{v}_A\) be the velocity vector at point A and \(\vec{v}_B\) be the velocity vector at point B.
Then \(|\vec{v}_A| = v\) and \(|\vec{v}_B| = v\).
The change in velocity is \( \Delta\vec{v} = \vec{v}_B - \vec{v}_A \).
We need to find the magnitude of this change: \( |\Delta\vec{v}| = |\vec{v}_B - \vec{v}_A| \).
The angle between the velocity vectors \(\vec{v}_A\) and \(\vec{v}_B\) is the same as the angle \(\theta\) subtended by the arc AB at the center, because velocity is always tangential and the radius is always perpendicular to the tangent. As the radius sweeps an angle \(\theta\), the tangent also turns by an angle \(\theta\).
Using the formula for the magnitude of the difference of two vectors:
\( |\vec{v}_B - \vec{v}_A|^2 = |\vec{v}_B|^2 + |\vec{v}_A|^2 - 2|\vec{v}_B||\vec{v}_A|\cos\theta \)
\( |\Delta\vec{v}|^2 = v^2 + v^2 - 2(v)(v)\cos\theta = 2v^2 - 2v^2\cos\theta = 2v^2(1-\cos\theta) \).
Using the trigonometric identity \(1-\cos\theta = 2\sin^2(\theta/2)\):
\( |\Delta\vec{v}|^2 = 2v^2 (2\sin^2(\theta/2)) = 4v^2\sin^2(\theta/2) \).
Taking the square root:
\( |\Delta\vec{v}| = \sqrt{4v^2\sin^2(\theta/2)} = 2v |\sin(\theta/2)| \).
Since \(\theta\) is likely an angle \(0 \le \theta \le 2\pi\), \(\theta/2\) is in \(0 \le \theta/2 \le \pi\), so \(\sin(\theta/2) \ge 0\).
Thus, \( |\Delta\vec{v}| = 2v \sin(\theta/2) \).
This matches option (a).
\[ \boxed{2v \sin \frac{\theta}{2}} \]
