a body starts with initial speed u and moves in a straight line with constant acceleration its has speed v after covering distance 2s its speed just after covering distance s

Dear student

Given that the initial speed is u and after covering the 2s distance its speed is v so using the third equation of the motion we get 

$$\begin{gathered} v^2=u^2+2a·2s \\ \Rightarrow as=\frac{v^2-u^2}{4} \end{gathered}$$
where a is the constant acceleration of the body.
Now let the speed of the body after covering the distance s is given by 

$v{\text{'}}^2=u^2+2as$
putting the value of as we get 
$$\begin{gathered} \Rightarrow v{\text{'}}^2=u^2+2·\frac{v^2-u^2}{4} \\ \Rightarrow v{\text{'}}^2=u^2+\frac{v^2-u^2}{2} \\ \Rightarrow v{\text{'}}^2=\frac{2u^2+v^2-u^2}{2} \\ \Rightarrow v{\text{'}}^2=\frac{u^2+v^2}{2} \\ \Rightarrow v\text{'}=\sqrt{\frac{u^2+v^2}{2}} \end{gathered}$$
Corrct option is (4)