Two particles A and B are connected by rigid | vedclass

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Question

$L ^{2}=x^{2}+y^{2}$

On differentiating we get,

$0=2 x \frac{ dx }{ dt }+2 y \frac{ dy }{ dt }$

The Velocity of $A =\frac{ dx }{ dt }=-10 m /

Vedclass · Accepted Answer

$5.8$

Vedclass · Answer

$L ^{2}=x^{2}+y^{2}$

On differentiating we get,

$0=2 x \frac{ dx }{ dt }+2 y \frac{ dy }{ dt }$

The Velocity of $A =\frac{ dx }{ dt }=-10 m / s$

The velocity of particle $B=\frac{ d y}{ dt }$

$\frac{ d y}{ dt }=\frac{- x }{ y } \frac{ dx }{ dt }=\cot 60^{0} 	imes 10=\frac{1}{\sqrt{3}} 	imes 10=5.8 m / s$

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

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