The velocity of point A on the rod is 2 $m s^{- 1}$ (leftwards) at the instant shown in fig. The velocity of the point B on the rod at this instant is

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1 ms^-1

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Solution

The correct option is B
1 ms^-1

Here y is constant $\frac{d l}{d t} = v_{B}$
$l^{2} = y^{2} + x^{2}$
$\Rightarrow 2 l \frac{d l}{d t} = 2 x \frac{d x}{d t}$
$\Rightarrow v_{B} = \frac{x}{l} v_{A} = v_{A} c o s 60^{o} = 1 m s^{- 1}$

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The velocity of point A on the rod is 2 ms−1 (leftwards) at the instant shown in fig. The velocity of the point B on the rod at this instant is

The correct option is B 1 ms-1 Here y is constant dldt=vB l2=y2+x2 ⇒2ldldt=2xdxdt ⇒vB=xlvA=vA cos60o=1 ms−1

The velocity of point A on the rod is 2 $m s^{- 1}$ (leftwards) at the instant shown in fig. The velocity of the point B on the rod at this instant is

The correct option is B
1 ms^-1

Here y is constant $\frac{d l}{d t} = v_{B}$
$l^{2} = y^{2} + x^{2}$
$\Rightarrow 2 l \frac{d l}{d t} = 2 x \frac{d x}{d t}$
$\Rightarrow v_{B} = \frac{x}{l} v_{A} = v_{A} c o s 60^{o} = 1 m s^{- 1}$