Question

# The position vector of a particle $$\vec{R}$$ as a function of time is given by $$\vec{R}=4\sin(2\pi t)\hat{i}+4\cos (2\pi t)\hat{j}$$ Where R is in meters, t is in seconds and $$\hat{i}$$ and $$\hat{j}$$ denote unit vectors along x and y-directions, respectively. Which one of the following statements is wrong for the motion of particle?

A
Path of the particle is a circle of radius 4 meter
B
Acceleration vectors is along R
C
Magnitude of acceleration vector is v2R where v is the velocity of particle
D
Magnitude of the velocity of particle is 8 meter/second

Solution

## The correct option is B Acceleration vectors is along $$-\vec{R}$$$$R=4sin\left( 2\pi t \right) { i }^{ \curlywedge }+cos\left( 2\pi t \right) { j }^{ \curlywedge }$$this is the vector equation $$V=\dfrac { dR }{ dt }$$$$=8\pi cos\left( 2\pi t \right) { i }^{ \curlywedge }+8\pi sin\left( 2\pi t \right) { j }^{ \curlywedge }$$$$a=\dfrac { dV }{ dt }$$$$=-16{ \pi }^{ 2 }sin\left( 2\pi t \right) { i }^{ }-16{ \pi }^{ 2 }cos\left( 2\pi t \right) { j }^{ }$$$$a=-{ 4\pi }^{ 2 }\quad 4sin\left( 2\pi t \right) { i }^{ \curlywedge }+4cos\left( 2\pi t \right) { j }^{ \curlywedge }$$$$a=-{ 4\pi }^{ 2 }\quad \left( \overline { R } \right)$$hence the acceleration vector is along $$-\overline { R }$$Physics

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