Relative Motion of Two Ships | Minimum Distance Concept | JEE Physics Problem with Solution

 

🧭 Relative Motion of Ships – Minimum Distance Concept | JEE Physics | Doubtify JEE

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📌 Question:

Ship A is sailing towards north-east with velocity
v = 30 î + 50 ĵ km/h, where î points east and ĵ points north.
Ship B is at a distance of 80 km east and 150 km north of Ship A and is sailing west at 10 km/h.

🔍 Find the time at which Ship A is at minimum distance from Ship B.

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🖼️ Question Image:

🧠 Solution Image:

✍️ Concept & Explanation:

This is a Relative Motion in 2D question – commonly asked in JEE Mains & Advanced exams.

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🧭 Step-by-Step Breakdown:

Let initial position of Ship A = origin = (0, 0)
Initial position of Ship B = (80, 150) km

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Velocities:

• Ship A: vₐ = 30 î + 50 ĵ

• Ship B: v_b = –10 î (towards west)

Relative velocity of B w.r.t. A:
v_rel = v_b – vₐ = (–10 – 30) î + (0 – 50) ĵ = –40 î – 50 ĵ

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Let position of B relative to A at time t hours:
r_rel(t) = (80 – 40t) î + (150 – 50t) ĵ

To find minimum distance → minimize |r_rel(t)|

Use:

$$D(t)^2 = (80 – 40t)^2 + (150 – 50t)^2$$

Differentiate D²(t), set derivative to 0 to find t at minimum distance:

$$\frac{d}{dt}D(t)^2 = 2(80 – 40t)(–40) + 2(150 – 50t)(–50) = 0$$

Solving gives:

$$t = 2 \, \text{hours}$$

✅ Final Answer:

⏱️ 2 hours

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🔍 Why This Question Is Important:

• Helps develop clear understanding of vector motion

• Tests application of relative motion + minimization techniques

• Appears frequently in JEE Main and Advanced under Kinematics

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🎥 Video Solution:

🧠 Pro Tip:

Always shift to relative frame of reference when dealing with two moving objects. In vector motion, break components carefully and visualize on the graph.

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