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JEE Main 2023 (1 Feb, Shift 1) · Distance, Displacement & Average Speed

Distance, Displacement & Average Speed question from JEE Main 2023 (1 Feb, Shift 1)

Q13JEE Main 2023 · 1 FEB · SHIFT 1Medium

Distance, Displacement & Average Speed

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:

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Answer: (d)

$AB = BC$ and $AD = 3AB$ , so $AB = BC = CD = d$ .
Total time $= \frac{d}{v_1} + \frac{d}{v_2} + \frac{d}{v_3} = \frac{d(v_2 v_3 + v_3 v_1 + v_1 v_2)}{v_1 v_2 v_3}$
$v_{avg} = \frac{3d}{t} = \frac{3v_1 v_2 v_3}{v_1 v_2 + v_2 v_3 + v_3 v_1}$
So the answer is (d).

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