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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Motion in a Straight Line
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A body is moving due north at 10 m/s. In 1 second the velocity changes to 10 m/s due east. Find the average acceleration of the body.

\[(a)\;zero\;ms^{-2} \quad (b)\;20\;ms^{-2} \quad (c)\;\frac{10}{\sqrt 2} ms^{-2}\quad (d)\;10 \sqrt 2 ms^{-2}\]

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$acceleration=\large\frac{\overrightarrow v-\overrightarrow u}{t}$
$\qquad\qquad\quad=\large\frac{10 ms^{-1}\;East-10 m^{-1} due\; north}{1}$
$\qquad\qquad\quad=\large\frac{10 ms^{-1}\;East+10 m^{-1} due \;south}{1}$
$\qquad\qquad\quad=10 \sqrt 2 south\;east\;ms ^{-2}$
Hence d is the correct answer. 


answered Jun 20, 2013 by meena.p
edited Jan 21, 2014 by meena.p

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