Two ships are $10$ km apart on a line from south to north. The one farther north is moving towards the west at $40$ kmph and the other is moving towards the north at $40$ kmph. Then distance of their closest approach is

10 km

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10 \sqrt{2} k m

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\frac{10}{\sqrt{2}}

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20 km

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Solution

The correct option is C $\frac{10}{\sqrt{2}}$
Consider the first ship to be placed at the origin and everything will be solved in the frame of reference of this ship. So, it is considered to be at rest.
So the position of the second ship is -10j and is moving with a velocity 40i + 40j
If we plot the motion of the second ship with respect to the first ship, it forms a right angled isosceles triangle. Thus, dropping a perpendicular from the first ship to the line of motion of the second ship will give us the shortest distance. The length of this perpendicular is $10 s i n 45^{o}$ or $\frac{10}{\sqrt{2}}$

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Two ships are 10 km apart on a line from south to north. The one farther north is moving towards the west at 40 kmph and the other is moving towards the north at 40 kmph. Then distance of their closest approach is

Two ships are $10$ km apart on a line from south to north. The one farther north is moving towards the west at $40$ kmph and the other is moving towards the north at $40$ kmph. Then distance of their closest approach is