Minimum Distance Between Two Ships Using Vectors

❓ Question

Ship A is sailing towards north-east with velocity

$$\vec v=30\hat i+50\hat j \text{ km/hr}$$

where:

$$\hat i \rightarrow \text{east}, \quad \hat j \rightarrow \text{north}$$

Ship B is at a distance of:

$$80 \text{ km east}$$

and

$$150 \text{ km north}$$

of Ship A and is sailing towards west at:

$$10 \text{ km/hr}$$

Ship A will be at minimum distance from B in:

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🖼 Question Image

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✍️ Short Explanation

This problem is based on:

👉 Relative velocity
👉 Closest approach concept
👉 Vector method in motion.

Main idea:

Minimum distance occurs when relative position becomes perpendicular to relative velocity.

$$\vec r \cdot \vec v_{rel}=0$$

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🔷 Step 1 — Initial Relative Position 💯

Take Ship A as origin initially.

Ship B is:

$$80 \text{ km east}$$

and

$$150 \text{ km north}$$

So:

$$\vec r_0=80\hat i+150\hat j$$

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🔷 Step 2 — Relative Velocity

Velocity of Ship A:

$$\vec v_A=30\hat i+50\hat j$$

Velocity of Ship B:

$$\vec v_B=-10\hat i$$

Relative velocity of B with respect to A:

$$\vec v_{BA}=\vec v_B-\vec v_A$$
$$=(-10-30)\hat i+(0-50)\hat j$$
$$=-40\hat i-50\hat j$$

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🔷 Step 3 — Relative Position After Time $t$

$$\vec r=\vec r_0+\vec v_{BA}t$$
$$=(80-40t)\hat i+(150-50t)\hat j$$

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🔷 Step 4 — Condition for Minimum Distance

At minimum distance:

$$\vec r \cdot \vec v_{BA}=0$$

So:

$$[(80-40t)\hat i+(150-50t)\hat j] \cdot (-40\hat i-50\hat j)=0$$

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🔷 Step 5 — Solve Equation

$$-40(80-40t)-50(150-50t)=0$$
$$-3200+1600t-7500+2500t=0$$
$$4100t=10700$$
$$t=\frac{107}{41}$$
$$t\approx2.6\text{ hr}$$

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🔷 Step 6 — JEE Trap Alert 🚨

❌ Relative velocity opposite sign se lena

❌ Minimum distance formula bhool jaana

❌ Initial position vector wrong lena

Remember:

$$\boxed{\vec r \cdot \vec v_{rel}=0}$$

for closest approach.

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✅ Final Answer

$$\boxed{2.6\text{ hr}}$$

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