A) Path of the particle is a circle of radius 4 meter
B) Acceleration vectors is along \[-\vec{R}\]
C) Magnitude of acceleration vector is \[\frac{{{v}^{2}}}{R}\] where v is the velocity of particle.
D) Magnitude of the velocity of particle is 8 meter/second
Correct Answer: D
Solution :
\[\vec{R}=4\sin (2\pi t)\hat{i}+4\cos (2\pi t)\hat{j}=x\hat{i}+yj\] Now, \[{{x}^{2}}+{{y}^{2}}={{4}^{2}},\]which is equation of circle of radius R. So, the motion is uniform circular motion with speed \[V=8\pi \sqrt{2}\,m/s\]You need to login to perform this action.
You will be redirected in
3 sec