A particle moves in a straight line with a constant acceleration....

Key Idea : The problem requires kinematics equations of motion. Let $u$ and $v$ be the first and final velocities of particle and $a \& s$ be the constant acceleration and distance covered by it. From third equation of motion $$\begin{array}{l} y^2 =u^2+2 a s \\ \Rightarrow (20)^2 =(10)^2+2 a \times 135 \\ \text{or } a =\frac{300}{2 \times 135}=\frac{10}{9} \mathrm{~ms}^{-2} \end{array}$$ Now using first equation of motion, $$\begin{aligned} & v=u+a t \\ & \text{or } t=\frac{v-u}{a}=\frac{20-10}{(10 / 9)}=\frac{10 \times 9}{10}=9 \mathrm{~s} \end{aligned}$$