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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
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B
Acceleration vectors is along R
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C
Magnitude of acceleration vector is v2R where v is the velocity of particle
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D
Magnitude of the velocity of particle is 8 meter/second
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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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