| The position vector of a particle R as a function of time is given by |
| \[\mathbf{R}=4\sin (2\pi t)\,\mathbf{\hat{i}}+4\cos (2\pi t)\,\mathbf{\hat{j}}\] where R is in metre, t is in seconds and \[\mathbf{\hat{i}}\] and \[\mathbf{\hat{j}}\] denote unit vectors along x and y-directions, respectively. Which one of the following statements is wrong for the motion of particle? [NEET 2015 (Re)] |
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 \[8\,m/s\]
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 |
| \[\mathbf{R}=4\sin \,(2\pi t)\,\mathbf{\hat{i}}+4\,\cos \,(2\pi t)\,\mathbf{\hat{j}}\] |
| 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 |
| \[{{x}^{2}}+{{y}^{2}}={{4}^{2}}[{{\sin }^{2}}(2\pi t)+{{\cos }^{2}}(2\pi t)]\] |
| 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 \[{{v}_{x}}=+\,4\,(\cos \,2\,\pi t)\,2\pi \] and \[{{v}_{y}}=-\,4\,(\sin 2\,\pi t)\,2\pi \] |
| Net resultant velocity, |
| \[v=\sqrt{v_{x}^{2}+v_{y}^{2}}\] |
| \[=\sqrt{{{(8\pi )}^{2}}({{\cos }^{2}}2\pi t+{{\sin }^{2}}2\pi t)}\] |
| \[v=8\pi \] \[[\because \,\,{{\cos }^{2}}2\pi t+{{\sin }^{2}}\pi t=1]\] |
| So. Option is incorrect. |
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