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(D) Let $L$ be the length of the rigid rod $AB$. Let $x$ be the distance of $A$ from the corner and $y$ be the distance of $B$ from the corner.From th

Vedclass · Accepted Answer

$5.8$

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(D) Let $L$ be the length of the rigid rod $AB$. Let $x$ be the distance of $A$ from the corner and $y$ be the distance of $B$ from the corner.From the geometry,we have $L^2 = x^2 + y^2$.Differentiating both sides with respect to time $t$,we get:$0 = 2x \frac{dx}{dt} + 2y \frac{dy}{dt}$$x \frac{dx}{dt} + y \frac{dy}{dt} = 0$Given that the velocity of $A$ is $v_A = -\frac{dx}{dt} = 10\; m/s$ (since $x$ is decreasing).So,$\frac{dx}{dt} = -10\; m/s$.Substituting this into the equation:$x(-10) + y \frac{dy}{dt} = 0$$y \frac{dy}{dt} = 10x$$\frac{dy}{dt} = 10 \left( \frac{x}{y} ight)$From the triangle,$\frac{x}{y} = \cot(\alpha)$.At $\alpha = 60^{\circ}$,$\frac{x}{y} = \cot(60^{\circ}) = \frac{1}{\sqrt{3}}$.Therefore,the velocity of $B$ is $v_B = \frac{dy}{dt} = 10 	imes \frac{1}{\sqrt{3}} = \frac{10}{1.732} \approx 5.77\; m/s$.Rounding to one decimal place,we get $5.8\; m/s$.

Two particles $A$ and $B$ are connected by a rigid rod $AB$. The rod slides along perpendicular rails as shown in the figure. The velocity of $A$ to the left is $10\; m/s$. What is the velocity of $B$ (in $m/s$) when the angle $\alpha = 60^{\circ}$?

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