An object moves with speed $$v_1, v_2$$ and $$v_3$$ along a line segment $$AB, BC$$ and $$CD$$ respectively as shown in figure. Where $$AB = BC$$ and $$AD = 3AB$$, then average speed of the object will be 

Given:

• Speeds on segments:$$v_1\ on\ AB\ ,\ v_2\ ​on\ BC,\ v_3\ on\ CD$$
• AB=BC
• AD=3AB

$$SinceAD=AB+BC+CD=3AB$$

$$LetAB=BC=L$$

Then:

$$AD=L+L+CD=3L\Rightarrow CD=L$$

So all three segments are equal:

$$AB=BC=CD=L$$

Total distance:

$$=L+L+L=3L$$

Time taken:

$$t=\frac{L}{v_1}+\frac{L}{v_2}+\frac{L}{v_3}$$

Average speed:

Cancel L:

$$v_{avg}=\frac{3}{\frac{1}{v_1}+\frac{1}{v_2}+\frac{1}{v_3}}$$

On simplifying : -

$$\frac{3v_1 v_2 v_3}{v_1 v_2 + v_2 v_3 + v_3 v_1}$$