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Relative Velocity in 2D

Two ships are...

Question

Two ships are 10 km apart on a line joining south to north. The one farther north is steaming west at $20 {km h}^{- 1}$ . The other is steaming north at $20 {km h}^{- 1}$ . What is their distance of closest approach? How long do they take to reach it?

5 \sqrt{2} km, 15 min

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10 km, 12 min

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5 \sqrt{2} km, 12 min

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5 \sqrt{2} km, 10 \sqrt{2} min

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Solution

The correct option is A $5 \sqrt{2} km, 15 min$

The velocity component of both ships 1 and 2 are equal. Hence, the angle between the two vectors is 45 degrees.
Solving from the frame of particle - 1
We get $d_{s h o r t} = 10 sin 45^{\circ} = \frac{10}{\sqrt{2}} = 5 \sqrt{2} km$
$t = \frac{10 sin 45}{| {\to v}_{21} |} = \frac{10 \times \frac{1}{\sqrt{2}}}{20 \sqrt{2}} = \frac{1}{4} = 15 min$

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Two ships are 10 km apart on a line joining south to north. The one farther north is steaming west at 20 km $h^{- 1}$ . The other is steaming north at 20 km $h^{- 1}$ . What is their distance of closest approach ?

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Two ships are 10 km apart on a line joining south to north. The one farther north is steaming west at 20 km h−1. The other is steaming north at 20 km h−1. What is their distance of closest approach? How long do they take to reach it?

The correct option is A 5√2 km, 15 min The velocity component of both ships 1 and 2 are equal. Hence, the angle between the two vectors is 45 degrees. Solving from the frame of particle - 1 We get dshort=10sin45∘=10√2=5√2 km t=10sin45|→v21|=10×1√220√2=14=15 min

Two ships are 10 km apart on a line joining south to north. The one farther north is steaming west at $20 {km h}^{- 1}$ . The other is steaming north at $20 {km h}^{- 1}$ . What is their distance of closest approach? How long do they take to reach it?