An object moves with speeds v_1, v_2,and v_3 | vedclass

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Question

(C) Let $AB = x$.Given $AB = BC$,so $BC = x$.Given $AD = 3 AB$,so $AD = 3x$.Since $AD = AB + BC + CD$,we have $3x = x + x + CD$,which implies $CD = x$

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let $AB = x$.Given $AB = BC$,so $BC = x$.Given $AD = 3 AB$,so $AD = 3x$.Since $AD = AB + BC + CD$,we have $3x = x + x + CD$,which implies $CD = x$.The total distance covered is $D_{total} = AB + BC + CD = x + x + x = 3x$.The total time taken is $T_{total} = t_1 + t_2 + t_3 = \frac{AB}{v_1} + \frac{BC}{v_2} + \frac{CD}{v_3} = \frac{x}{v_1} + \frac{x}{v_2} + \frac{x}{v_3}$.The average speed is defined as $v_{avg} = \frac{D_{total}}{T_{total}} = \frac{3x}{\frac{x}{v_1} + \frac{x}{v_2} + \frac{x}{v_3}}$.Simplifying the expression,we get $v_{avg} = \frac{3x}{x(\frac{1}{v_1} + \frac{1}{v_2} + \frac{1}{v_3})} = \frac{3}{\frac{v_2 v_3 + v_1 v_3 + v_1 v_2}{v_1 v_2 v_3}} = \frac{3 v_1 v_2 v_3}{v_1 v_2 + v_2 v_3 + v_3 v_1}$.

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

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