Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Motion in a Straight Line
0 votes

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} $

Can you answer this question?

1 Answer

0 votes
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

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App