A) \[\frac{{{v}^{2}}}{r}\cdot \sin \left( \frac{2v}{R} \right)\]
B) \[\frac{2{{v}^{2}}}{R}\cdot \sin \left( \frac{\theta }{2} \right)\]
C) \[\frac{2{{v}^{2}}}{R\theta }\cdot \cos \left( \frac{\theta }{2} \right)\]
D) \[\frac{{{v}^{2}}}{2}\cdot \sin \left( \frac{2v}{R} \right)\]
Correct Answer: D
Solution :
As particle moves by an angle\[\theta .\] |
Then, change in velocity is \[\Delta v={{v}_{2}}-{{v}_{1}}.\] |
Magnitude of change in velocity is |
\[|\Delta v|=\sqrt{{{({{v}_{2}}-{{v}_{1}})}^{2}}}\] |
\[\Rightarrow \] \[|\Delta v|=\sqrt{v_{1}^{2}+v_{2}^{2}-2{{v}_{1}}{{v}_{2}}\cos \theta }\] |
Here, \[{{v}_{1}}={{v}_{2}}=v\] |
\[\Rightarrow \] \[|\Delta v|=\sqrt{{{v}^{2}}+{{v}^{2}}-2{{v}^{2}}\cos \theta }\] |
\[=\sqrt{2{{v}^{2}}(1-\cos \theta })\] |
\[=\sqrt{2{{v}^{2}}\left[ 1-\left( 1-2{{\sin }^{2}}\frac{\theta }{2} \right) \right]}\] |
\[=\sqrt{2{{v}^{2}}\left( 1-1+2{{\sin }^{2}}\frac{\theta }{2} \right)}\] |
\[=\sqrt{2{{v}^{2}}\left( 2{{\sin }^{2}}\frac{\theta }{2} \right)}=\sqrt{4{{v}^{2}}{{\sin }^{2}}\frac{\theta }{2}}\] |
\[\therefore \]\[|\Delta v|=2v\sin \frac{\theta }{2}\] |
Time duration of motion is given 4s |
So, \[\Delta t=4=\frac{R\theta }{v}\]\[\Rightarrow \]\[\theta =\frac{4v}{R}\] |
So, average acceleration of the particle is |
\[|{{a}_{avg}}|=\frac{|\Delta v|}{\Delta t}=\frac{2v\cdot \sin \frac{\theta }{2}}{4}=\frac{v}{2}\sin \left( \frac{2v}{R} \right)\] |
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