Question

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

• A. \( \frac{v_1 v_2 v_3}{3(v_1 v_2 + v_2 v_3 + v_3 v_1)} \)
• B. \( \frac{3 v_1 v_2 v_3}{v_1 v_2 + v_2 v_3 + v_3 v_1} \)
• C. \( \frac{(r_1^2 + r_2^2 + r_3^2)}{3} \)
• D. \( \frac{(v_1 + v_2 + v_3)}{3 v_1 v_2 v_3} \)

Hint:

For average speed calculations when covering equal distances at different speeds, use: \[ V_{{avg}} = \frac{{Total Distance}}{{Total Time}} \] which simplifies to a harmonic mean formula in cases of multiple segments.

Answer 1

The Correct Option is B

Step 1: Understanding the given conditions. Given: - \( AB = BC = x \) - \( CD = AB = x \) - \( AD = 3x \) - Speeds: \( v_1 \), \( v_2 \), and \( v_3 \) along \( AB \), \( BC \), and \( CD \) respectively. 
Step 2: Compute the time taken for each segment. Using \( {Time} = \frac{{Distance}}{{Speed}} \), \[ t_1 = \frac{x}{v_1}, \quad t_2 = \frac{x}{v_2}, \quad t_3 = \frac{x}{v_3} \] Total time taken: \[ T = t_1 + t_2 + t_3 = \frac{x}{v_1} + \frac{x}{v_2} + \frac{x}{v_3} \] 
Step 3: Compute the average speed. \[ V_{{avg}} = \frac{{Total Distance}}{{Total Time}} = \frac{3x}{\frac{x}{v_1} + \frac{x}{v_2} + \frac{x}{v_3}} \] Simplifying: \[ V_{{avg}} = \frac{3}{\frac{1}{v_1} + \frac{1}{v_2} + \frac{1}{v_3}} \] Rewriting in terms of a common denominator: \[ V_{{avg}} = \frac{3 v_1 v_2 v_3}{v_1 v_2 + v_2 v_3 + v_3 v_1} \] 
Final Answer: \[ \boxed{\frac{3 v_1 v_2 v_3}{v_1 v_2 + v_2 v_3 + v_3 v_1}} \]