A particle describes an angle $θ$ in a circular path with a constant speed $ν$ . Find the (a) change in the velocity of the particle and (b) average acceleration of the particle during the motion in the curve (circle).

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Solution

a. As the particle moves from P to Q, the velocity turns through an angle $Θ$ . Then,
$| Δ \to v | = \sqrt{v_{1}^{2} + v_{2}^{2} - 2 v_{1} v_{2} c o s Θ}$

$= \sqrt{v^{2} + v^{2} - 2 v v c o s Θ}$ = $2 v s i n \frac{Θ}{2}$

b. The time of motion is $Δ t = R Θ / v$ .Then, average acceleration is $| {\to a}_{a v} | = \to v Δ t$ .
Substituting $| Δ \to v$ and $| Δ v$ , we have $| {\to a}_{a} v | = \frac{2 v s i n \frac{Θ}{2}}{\frac{R Θ}{v}} = \frac{2 v^{2}}{R Θ} s i n \frac{Θ}{2}$

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