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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Motion in a Straight Line
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Velocity of a particle moving in a straight line varies with its displacement as $v=(\sqrt {4 +4s}) m/s$. Displacement of particle at time t=0 is s=0. The displacement of particle at time t=2s is

\[(a)\;4 m \quad (b)\;8 m \quad (c)\;6 m \quad (d)\;10 m\]

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$v=(\sqrt {4+4s)}$
Comparing with
We get $a=2 m/s^2$ and $u=2 m/s$
Therefore $S=ut+\frac{1}{2}at^2\; at\;t=2 sec$
$\qquad\qquad S=2 \times 2 +\frac{1}{2} \times 2 \times 2^2$
$\qquad\qquad S=8m$
Hence b is the correct answer.


answered Jul 2, 2013 by meena.p
edited Jan 26, 2014 by meena.p

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