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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Motion in a Straight Line
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A body travelling with uniform acceleration crosses two points A and B with velocities $20\;ms^{-1}$ and $30 \;ms^{-1}$ respectively . The speed of the body at the mid-point of A and B is

$\begin{array}{1 1} 24\;ms^{-1} \\ 25\;ms^{-1} \\ 25.5\;ms^{-1} \\10 \sqrt 6 \;ms^{-1}\end{array} $

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Answer : $ 25.5\;ms^{-1}$
$30^2-20^2=2al, 30^2-v^2=2a \large\frac{l}{2}$
$30^2-20^2 =2 \times 30^2 -2v^2$
or $2v^2= 2\times 30^2 -30^2 +20^2$
$\qquad= 900 +400 =1300 $
or $v^2 =650$
or $v= \sqrt {650}\;ms^{-1}$
$\qquad= 25.5 \;ms^{-1}$
answered Aug 19, 2014 by meena.p

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