1. Given:

   • First distance $= x$ with velocity $v_1$
   • Second distance $= x$ with velocity $v_2$
   • Both are in the same direction

2. Average velocity formula: $v=\frac{\text{total displacement}}{\text{total time}}$

   Since motion is in the same direction, total displacement is $x+x=2x$

3. Time taken in each part:

   • For first part, $t_1=\frac{x}{v_1}$
   • For second part, $t_2=\frac{x}{v_2}$

   So total time is $t=t_1+t_2=\frac{x}{v_1}+\frac{x}{v_2}$

4. Compute average velocity: $v=\frac{2x}{\frac{x}{v_1}+\frac{x}{v_2}}$

   Factor out $x$ from denominator: $v=\frac{2x}{x\left(\frac{1}{v_1}+\frac{1}{v_2}\right)}$

   Cancel $x$: $v=\frac{2}{\frac{1}{v_1}+\frac{1}{v_2}}$

5. Rearranging, $\frac{2}{v}=\frac{1}{v_1}+\frac{1}{v_2}$

6. Check options:

   • A: $V=V_1+V_2$ ❌
   • B: $V=\frac{v_1+v_2}{2}$ ❌ (this is not valid for equal distances)
   • C: $\frac{1}{v}=\frac{1}{v_1}+\frac{1}{v_2}$ ❌
   • D: $\frac{2}{V}=\frac{1}{v_1}+\frac{1}{v_2}$ ✅

Therefore, the correct relation is $\boxed{\frac{2}{v}=\frac{1}{v_1}+\frac{1}{v_2}}$ So, the correct option is D.