logo

Ask Questions, Get Answers

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

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

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

Can you answer this question?
 
 

1 Answer

0 votes
$v=(\sqrt {4+4s)}$
$v^2=4+4s$
Comparing with
$v^2=u^2+2as$
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
 

Related questions

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