Particle Moves A to B as shown in figure. Find average acceleration?[tex]\sf a) \: 2 \sqrt{2} \: \dfrac{ {v}^{2} }{\pi r} \\ [/tex][tex]\sf b) \: zero \\ [/tex][tex]\sf c) \sqrt{2} v \\ [/tex][tex]\sf d) \: \sqrt{2} \: \dfrac{ {v}^{2} }{\pi r} [/tex]. Created by MystifiedBoy. Physics

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Particle Moves A to B as shown in figure.

Average acceleration ?

First of all, average acceleration is defined as the change in velocity divided by the time.

In uniform circular motion, change in velocity for a particular angular displacement $\theta$ is given as:

$\boxed{\sf\Delta v = 2v \sin( \dfrac{ \theta}{2} )}$

Let average acceleration be "a" :

$\implies \sf \:a = \dfrac{\Delta v}{t}$

$\implies \sf \:a = \dfrac{2v \sin( \frac{ \theta}{2} ) }{t}$

$\implies \sf \:a = \dfrac{2v \sin \bigg( \dfrac{ \theta}{2} \bigg) }{ \bigg( \dfrac{ \frac{2\pi r}{4} }{v} \bigg) }$

Put $\theta$ = 90°

$\implies \sf \:a = \dfrac{2v \sin \bigg( \dfrac{ {90}^{ \circ} }{2} \bigg) }{ \bigg( \dfrac{ \frac{\pi r}{2} }{v} \bigg) }$

$\implies \sf \:a = \dfrac{2v \sin \bigg( {45}^{ \circ}\bigg) }{ \bigg( \dfrac{\pi r}{2v} \bigg) }$

$\implies \sf \:a = \dfrac{2v \times \dfrac{1}{ \sqrt{2} } }{ \bigg( \dfrac{\pi r}{2v} \bigg) }$

$\implies \sf \:a = \dfrac{4v \times \dfrac{1}{ \sqrt{2} } }{ \bigg( \dfrac{\pi r}{v} \bigg) }$

$\implies \sf \:a = \dfrac{4 {v}^{2} \times \dfrac{1}{ \sqrt{2} } }{ \pi r }$

$\implies \sf \:a = \dfrac{2 \sqrt{2} {v}^{2} }{ \pi r }$

So, final answer is :

$\boxed{ \bf \:a = \dfrac{2 \sqrt{2} {v}^{2} }{ \pi r } }$

MystifiedBoy Thank you so much Bhaiya ! :)

nirman95 Welcome :)