JEE Main 2024

Question

An object moves with speed $v_1, v_2$ and $v_3$ along a line segment AB, BC and CD respectively. If AB = BC and AD = 3AB, 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). $$\frac{v_1 + v_2 + v_3}{3}$$ D). $$\frac{v_1 + v_2 + v_3}{3v_1 v_2 v_3}$$

Edvaya · Accepted Answer

The correct answer is $$\frac{3 v_1 v_2 v_3}{v_1 v_2 + v_2 v_3 + v_3 v_1}$$. **Solution:** Let the length of segment $	ext{AB} = x$. Given $	ext{AB} = 	ext{BC} = x$ and the total length $	ext{AD} = 3	ext{AB} = 3x$. The length of $	ext{CD} = 	ext{AD} - (	ext{AB} + 	ext{BC}) = 3x - (x + x) = x$. Thus, the three segments are of exactly equal length $x$. Total distance travelled $= 3x$. Total time taken $= t_1 + t_2 + t_3$ $ = \frac{x}{v_1} + \frac{x}{v_2} + \frac{x}{v_3}$. Average speed $$\langle v angle = \frac{	ext{Total Distance}}{	ext{Total Time}}$$ $$ = \frac{3x}{\frac{x}{v_1} + \frac{x}{v_2} + \frac{x}{v_3}}$$ $$ = \frac{3}{\frac{1}{v_1} + \frac{1}{v_2} + \frac{1}{v_3}} = \frac{3v_1 v_2 v_3}{v_2 v_3 + v_1 v_3 + v_1 v_2}$$.

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Physics > Mechanics > Motion In A Straight Line And Plane

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