An object moves with speed v _ 1 , v _ 2 and v _ 3 along a line...

Speed of an object is defined as the distance travelled by the object per unit time.

Let's assume that, $AB = d$. So, $BC = d, AD = 3d$. Also, assume that the time required by the object to travel $AB, BC, CD$ are ${t}_{1}, {t}_{2}, {t}_{3}$ respectively.

Now,

$\begin{cases}AD & = & 3AB \\ AB + BC + CD & = & 3AB \\ AB + AB + CD & = & 3AB \\ CD & = & AB\end{cases}$

Hence,

$\begin{cases}{t}_{1} & = & \dfrac{AB}{{v}_{1}} \\ & = & \dfrac{d}{{v}_{1}} \\ {t}_{2} & = & \dfrac{d}{{v}_{2}} \\ {t}_{3} & = & \dfrac{d}{{v}_{3}}\end{cases}$

The formula to calculate the average speed $\left({v}_{a}\right)$ of the object is given by

${v}_{a} = \dfrac{AD}{{t}_{1} + {t}_{2} + {t}_{3}} . . . . . . . . . . . . . . . . . . . . . . . . ( 1 )$

Substitute the expressions for the parameters into equation (1) and simplify to obtain the required average speed of the object.

$\begin{cases}{v}_{a} & = & \dfrac{3d}{\dfrac{d}{{v}_{1}} + \dfrac{d}{{v}_{2}} + \dfrac{d}{{v}_{3}}} \\ & = & \dfrac{3{v}_{1}{v}_{2}{v}_{3}}{{v}_{1}{v}_{2} + {v}_{2}{v}_{3} + {v}_{3}{v}_{1}}\end{cases}$