Question

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

• A. 10 m/s
• B. 9.8 m/s
• C. 5.8 m/s
• D. 17.3 m/s

Answer 1

Answer: (C)

$V$ elocity $=10m/s\alpha =60^{\circ }$
$L^2=x^2+y^2$
On differentiating we get,
$0=2x\frac{dx}{dt}+2y\frac{dy}{dt}$
The Velocity of $A=\frac{dx}{dt}=-10m/s$
Thus the velocity of particle $B$ is:
$\frac{dy}{dt}=\frac{-x}{y}\frac{dx}{dt}=\cot 60^0\times 10=\frac{1}{\sqrt{3}}\times 10=5.8m/s$