JEE MAIN - Physics (2023 - 1st February Morning Shift - No. 9)
An object moves with speed $$v_1,v_2$$ and $$v_3$$ along a line segment AB, BC and CD respectively as shown in figure. Where AB = BC and AD = 3AB, then average speed of the object will be:
_1st_February_Morning_Shift_en_9_1.png)
$${{{v_1}{v_2}{v_3}} \over {3({v_1}{v_2} + {v_2}{v_3} + {v_3}{v_1})}}$$
$${{({v_1} + {v_2} + {v_3})} \over 3}$$
$${{({v_1} + {v_2} + {v_3})} \over {3{v_1}{v_2}{v_3}}}$$
$${{3{v_1}{v_2}{v_3}} \over {({v_1}{v_2} + {v_2}{v_3} + {v_3}{v_1})}}$$
Explanation
$A B=B C=C D$
$$ \begin{aligned} \Rightarrow \text { Average speed } & =\frac{\text { Distance }}{\text { Time }} \\\\ & =\frac{A D}{\frac{A B}{V_{1}}+\frac{A B}{V_{2}}+\frac{A B}{V_{3}}} \\\\ & =\frac{3 V_{1} V_{2} V_{3}}{V_{1} V_{2}+V_{2} V_{3}+V_{1} V_{3}} \end{aligned} $$
$$ \begin{aligned} \Rightarrow \text { Average speed } & =\frac{\text { Distance }}{\text { Time }} \\\\ & =\frac{A D}{\frac{A B}{V_{1}}+\frac{A B}{V_{2}}+\frac{A B}{V_{3}}} \\\\ & =\frac{3 V_{1} V_{2} V_{3}}{V_{1} V_{2}+V_{2} V_{3}+V_{1} V_{3}} \end{aligned} $$
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