Two particles are moving along two long straight lines, in the same plane with same speed equal to $20 cm / s$ . The angle between the two lines is $60 °$ and their intersection point is $O$ . At a certain moment, the two particles are located at distances $3 m$ and $4 m$ from $O$ and are moving towards $O$ . Subsequently, the shortest distance between them will be

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$40 \sqrt{2} cm$

$50 \sqrt{2} cm$

$50 \sqrt{3} cm$

$50 cm$

answer is D.

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Detailed Solution

$v_{Q} = - 20 \hat{i}$

Two particles are moving along two long straight lines, in the same plane with same speed equal to 20 cm//s. The angle between the two linse is 60^@ and their intersection point

$v_{p} = - 20 \cos 60 \hat{i} - 20 \sin 60 ° \hat{j}$

$= - 10 \hat{i} - 10 \sqrt{3} \hat{j}$
Assuming $P$ to be at rest,
$v_{O P} = v_{Q} - v_{P}$
$= - 10 \hat{i} + 10 \sqrt{3} \hat{j}$
Now, $\tan θ = \frac{10 \sqrt{3}}{10} = \sqrt{3}$
or $θ = 60 °$
where, $θ$ is the angle of $v_{Q P}$ from $x$ -axis towards positive $y$ -axis.