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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Motion in a Straight Line
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A body travels 2 m in the first two seconds and $2.2 \;m$ in the next four seconds. What will be its velocity at the end of the seventh second from the start, if acceleration is constant .

$\begin{array}{1 1}(A)\;0.30ms^{-1}\\(B)\;0.10ms^{-1} \\(C)\;0.9 ms^{-1} \\(D)\;6.7 ms^{-1} \end{array} $

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Since in $t=2$s, distance covered $s=2m$ we have ,
$2= u \times 2+\large\frac{1}{2}$$a \times 2^2$
or $u+a=1$-----(i)
As the distance covered in $t=2$s to $6\;s$ is $2.2 m$ We have
$2.2=u(6-2)+\large\frac{1}{2}$$a (6^2-2^2)$
From equation (i) and (ii)
$a= -0.15 ms^{-2}$
From the relation $v=u+at$, We have on putting,
$u=1.15 ms^{-1},a = -0.15 ms^{-2}$ and $ t=7\;s$
$v= 1.15 +(-0.15) \times 7$
$\quad=0.10 ms^{-1}$
Hence B is the correct answer.
answered Jul 25, 2014 by meena.p

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