The position vector of a particle → R as a function of time is given by: → R = 4 (2 ) ^ i + 4 ( 2 ) ^ j Where R is in meters, t is in seconds and ^ i and ^ j denote unit vectors along x- and y-directions, respectively.

Correct answer: B. Magnitude of the velocity of particle is 8 meter/second x = 4 (2π t) y = 4 ( 2 ) Squaring and adding x ^ 2 + y ^ 2 = [4 (2 )] ^ 2 + [4 (2 )] ^ 2 x ^ 2 + y ^ 2 = 16 This means the particle is moving in a circle V = r w = (2π)(4) = 8π

NEET 2015

Exam: NEET
Year: 2015
Subject: Physics
Chapter: Motion in One Dimension

NEET 2015 · Physics · Motion in One Dimension

The position vector of a particle → R as a function of time is...

The position vector of a particle

\overset{\to}{R}

as a function of time is given by:

\overset{\to}{R} = 4\sin \left(2\pit\right)\overset{^}{i} + 4\cos ( 2\pit ) \overset{^}{j}

Where R is in meters, t is in seconds and

\overset{^}{i}

and

\overset{^}{j}

denote unit vectors along x- and y-directions, respectively. Which one of the following statements is wrong for the motion of particle?

Options

AMagnitude of acceleration vector is $\dfrac{{v}^{2}}{R}$ , where $v$ is the velocity of particle
BMagnitude of the velocity of particle is 8 meter/second
CPatch of the particle is a circle of radius 4 meter
DAcceleration vector is along $- \overset{\to}{R}$

Correct answer

B. Magnitude of the velocity of particle is 8 meter/second

Step-by-step solution

$x = 4\sin \left(2\pi t\right)$
$y = 4\cos ( 2\pit )$

Squaring and adding

${x}^{2} + {y}^{2} = {\left[4\sin \left(2\pit\right)\right]}^{2} + {\left[4\cos \left(2\pit\right)\right]}^{2}$

${x}^{2} + {y}^{2} = 16$

This means the particle is moving in a circle
$V = r w = \left(2\pi\right)\left(4\right) = 8\pi$

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