## In This post you will get answer of Q.

A particle moves in a straight line with a constant acceleration. It changes its velocity from $10 ms^{-1}$ to $20ms^{-1}$ while passing through a distance 135 m in t second. The value of t is

Solution:

## Let $u$ and $v$ be the first and final velocities of particle and $a$ and $s$ be the constant acceleration and distance covered by it. From third equation of motion

$v^2=u^2+2as$

$Rightarrow$ $(20)^2=(10)^2+2a times 135$

or $a=frac{300}{2 times 135}=frac{10}{9}ms^{-2}$

Now using first equation of motion,

v=u+at

or $t=frac{v-u}{a}=frac{20-10}{(10/9)}=frac{10 times 9}{10}=9s$