Ship A is sailing towards north-east with velocity) → v = 30 ^ i + 50 ^ j km h ^ - 1 where ^ i points east and ^ j , north. The ship B is at a distance of 80km east and 150km north of Ship A and is sailing towards the

Correct answer: C. 2.6h → v _ rel = → v _ 1 - v _ 2 = 40i + 50j → r _ rel = → r _ 1 - → r _ 2 = - 80i - 150j t _ = | → v rel · → r rel | → v rel| ^ 2 | = | - 3200 - 7500 ( (40) ^ 2 + (50) ^ 2 ) ^ 2 | t _ = (10700)/(4100) = (107)/(41) = 2.6h

JEE Main 20198 AprilMorning Shift

Exam: JEE Main
Year: 2019
Date: 8 April
Shift: Morning Shift
Subject: Physics
Chapter: Mathematics in Physics

JEE Main 2019 · 8 April · Morning Shift · Physics · Mathematics in Physics

Ship A is sailing towards north-east with velocity) → v = 30 ^ i +...

Ship

A

is sailing towards north-east with velocity)

\overset{\to}{v} = 30\overset{^}{i} + 50\overset{^}{j}km{h}^{- 1}

where

\overset{^}{i}

points east and

\overset{^}{j},

north. The ship

B

is at a distance of

80km

east and

150km

north of Ship

A

and is sailing towards the west at

10km{h}^{- 1} .

A

will be at the minimum distance from

B

in:

Options

A $4.2h$
B $3.2h$
C $2.6h$
D $2.2h$

Correct answer

C. $2.6h$

Step-by-step solution

{\overset{\to}{v}}_{rel} = {\overset{\to}{v}}_{1} - {v}_{2} = 40i + 50j

{\overset{\to}{r}}_{rel} = {\overset{\to}{r}}_{1} - {\overset{\to}{r}}_{2} = - 80i - 150j

{t}_\min = \left|\dfrac{\overset{\to}{v}rel \cdot \overset{\to}{r}rel}{{\left|\overset{\to}{v}rel\right|}^{2}}\right| = \left|\dfrac{- 3200 - 7500}{{\left(\sqrt{{\left(40\right)}^{2} + {\left(50\right)}^{2}}\right)}^{2}}\right|

{t}_\min = \dfrac{10700}{4100} = \dfrac{107}{41} = 2.6h

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