Question

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

• A. $\frac{\left(v_1+v_2+v_3\right)}{3}$
• B. $\frac{3 v_1 v_2 v_3}{\left(v_1 v_2+v_2 v_3+v_3 v_1\right)}$
• C. $\frac{\left(v_1+v_2+v_3\right)}{3 v_1 v_2 v_3}$
• D. $\frac{v_1 v_2 v_3}{3\left(v_1 v_2+v_2 v_3+v_3 v_1\right)}$

Answer 1

The Correct Option is B

\(AB=x\)
\(BC=x\)
\(2x+CD=3x\)
\(CD=x\)
\(⟨v⟩= \frac{3x}{\frac{x}{v_1}+\frac{x}{v_2}+\frac{x}{v_3}}=\frac{3 v_1 v_2 v_3}{\left(v_1 v_2+v_2 v_3+v_1 v_3\right)}\)

Answer 2

Given: \( AB = x \), \( BC = x \), \( AD = 3x \) \[ \text{Total Distance} = 3x \] \[ \text{Total Time} = \frac{x}{v_1} + \frac{x}{v_2} + \frac{x}{v_3} \] \[ \text{Average Speed} = \frac{3x}{\frac{x}{v_1} + \frac{x}{v_2} + \frac{x}{v_3}} = \frac{3v_1 v_2 v_3}{v_1 v_2 + v_2 v_3 + v_3 v_1} \]