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Motion in A Plane question

2015 · Q130
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Motion in A Plane question

2015 · Q130

NEETPhysicsMotion in A PlaneMCQ+4 / −1
The positions vector of a particle R→\overrightarrow RR as a function of time is given by R→\overrightarrow RR = 4sin(2π\piπt)i^\widehat ii + 4cos(2π\piπt)j^\widehat jj​. Where R is in meters, t is in seconds and i^\widehat ii and j^\widehat jj​ denote unit vectors along x-and y-directions, respectively. Which one of the following statements is wrong for the motion of particle?
  1. A
    Magnitude of the velocity of particle is 8 meter/second.
  2. B
    Path of the particle is a circle of radius 4 meter.
  3. C
    Acceleration vector is along −-−R→\overrightarrow RR.
  4. D
    Magnitude of acceleration vector is v2R{{{v^2}} \over R}Rv2​
    where v is the velocity of particle.
View written solutionFree

Correct answer: A

Here, R→=4sin⁡(2πt)i^+4cos⁡(2πt)j^\overrightarrow R = 4\sin \left( {2\pi t} \right)\widehat i + 4\cos \left( {2\pi t} \right)\widehat jR=4sin(2πt)i+4cos(2πt)j​

The velocity of the particle is
v→=dR→dt=ddt[4sin⁡(2πt)i^+4cos⁡(2πt)j^]\overrightarrow v = {{d\overrightarrow R } \over {dt}} = {d \over {dt}}\left[ {4\sin \left( {2\pi t} \right)\widehat i + 4\cos \left( {2\pi t} \right)\widehat j} \right]v=dtdR​=dtd​[4sin(2πt)i+4cos(2πt)j​]
=8πcos(2πt)i^−8πsin(2πt)j^ = 8\pi cos\left( {2\pi t} \right)\widehat i - 8\pi sin\left( {2\pi t} \right)\widehat j=8πcos(2πt)i−8πsin(2πt)j​

Its magnitude is

∣v→∣=(8πcos⁡(2πt))2+(−8πsin⁡(2πt))2\left| {\overrightarrow v } \right| = \sqrt {{{\left( {8\pi \cos \left( {2\pi t} \right)} \right)}^2} + {{\left( { - 8\pi \sin \left( {2\pi t} \right)} \right)}^2}} ​v​=(8πcos(2πt))2+(−8πsin(2πt))2​

=64π2cos⁡2(2πt)+64π2sin⁡2(2πt)= \sqrt {64{\pi ^2}{{\cos }^2}\left( {2\pi t} \right) + 64{\pi ^2}{{\sin }^2}\left( {2\pi t} \right)}=64π2cos2(2πt)+64π2sin2(2πt)​

=64π2[cos⁡2(2πt)+sin⁡2(2πt)]= \sqrt {64{\pi ^2}\left[ {{{\cos }^2}\left( {2\pi t} \right) + {{\sin }^2}\left( {2\pi t} \right)} \right]}=64π2[cos2(2πt)+sin2(2πt)]​

=64π2= \sqrt {64{\pi ^2}}=64π2​     (as sin2θ\theta θ + cos2θ\theta θ = 1)

= 8π\pi π m/s

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