A) Acceleration is along - R
B) Magnitude of acceleration vector is\[\frac{{{v}^{2}}}{R}\], where v is the velocity of particle
C) Magnitude of the velocity of particle is \[8m/s\]
D) Path of the particle is a circle of radius
Correct Answer: C
Solution :
(i) The position vector of a particle R as a function of time is given by \[\mathbf{R}=4\sin (2\pi t)\mathbf{\hat{i}}+4cos(2\pi t)\mathbf{\hat{j}}\] x-axis component, \[x=4\sin 2\pi t\] ?(i) y-axis component, \[y=4\cos 2\pi t\,\] ?(ii) Squaring and adding both equations, we get \[{{x}^{2}}+{{y}^{2}}={{4}^{2}}[si{{n}^{2}}(2\pi t)+co{{s}^{2}}(2\pi t)]\] i.e. \[{{x}^{2}}+{{y}^{2}}={{4}^{2}}\] i.e. equation of circle and radius is 4 m. (ii) Acceleration vector, \[\mathbf{a}=\frac{{{v}^{2}}}{R}(-\mathbf{\hat{R}})\], while v is velocity of a particle. (iii) Magnitude of acceleration vector, \[a=\frac{{{v}^{2}}}{R}\] (iv) As, we have \[{{v}_{x}}=+4(cos2\pi t)2\pi \] and \[{{v}_{y}}=-4(sin2\pi t)2\pi \] Net resultant velocity, \[v=\sqrt{v_{x}^{2}+v_{y}^{2}}\] \[=\sqrt{{{(8\pi )}^{2}}(co{{s}^{2}}2\pi t+si{{n}^{2}}2\pi t)}\] \[v=8\pi \] \[[\because \,co{{s}^{2}}2\pi t+si{{n}^{2}}\pi t=1]\] So. Option (c) is incorrect.
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