A bead is moving in medium 1 with a uniform s | vedclass

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(C) Step $1$: Distance traveled in medium $1$ $(d_1)$:$d_1 = 	ext{speed} 	imes 	ext{time} = 1\, m/s 	imes 2.5\, s = 2.5\, m$.Step $2$: Motion in a

Vedclass · Accepted Answer

$14$

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(C) Step $1$: Distance traveled in medium $1$ $(d_1)$:$d_1 = 	ext{speed} 	imes 	ext{time} = 1\, m/s 	imes 2.5\, s = 2.5\, m$.Step $2$: Motion in air:The bead enters the air with an initial velocity $u = 1\, m/s$. It falls freely for a distance $h = 2\, m$ under gravity $(g = 10\, m/s^2)$.Using the equation $v^2 - u^2 = 2gh$:$v^2 - (1)^2 = 2 	imes 10 	imes 2$$v^2 - 1 = 40$$v^2 = 41$$v = \sqrt{41} \approx 6.4\, m/s$.Step $3$: Distance traveled in medium $2$ $(d_2)$:The bead enters medium $2$ with velocity $v = \sqrt{41}\, m/s$ and moves with this uniform speed for $t = 1.5\, s$.$d_2 = v 	imes t = \sqrt{41} 	imes 1.5 \approx 6.403 	imes 1.5 = 9.6045\, m$.Step $4$: Total distance $(D)$:$D = d_1 + h + d_2 = 2.5 + 2 + 9.6045 = 14.1045\, m$.Rounding to one decimal place, the total distance is $14.1\, m$.

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