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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Motion in a Straight Line
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A particle is moving in x-y plane with $y=\large\frac{x}{2}$ and $v_x=4-2t$. The displacement versus time of the particle would be

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$y=\large\frac{x}{2}$ =>  the particle is moving in a straight line passing through orgin.
$v_x=4-2t=>u_x=4\;a_x=-2$
since $y=\large\frac{x}{2}$ we have
$v_y=\large\frac{v_x}{2}$$=>v_y$=2- t
=>$a_y=-1$ is also negative
$v_x\;and \;v_y$ are positive
$a_x \; and\;a_y$ are negative
=>motion is first retarded then accelerated
Hence c is the correct answer.

 

answered Jun 25, 2013 by meena.p
edited Jul 28, 2014 by thagee.vedartham
 

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