Two ships A and B are sailing straight away from a fixed point O along routes such that AOB is always 120^ . At a certain instance, OA=8 ~km, OB=6 ~km and the ship A is sailing at the rate of 20 ~km / hr while the ship

Correct answer: A. 260 37 Let OA=x ~km, OB=y ~km, AB=R (AB)^2=(OA)^2+(OB)^2-2(OA)(OB) 120^ R^2=x^2+y^2-2 x y(-(1)/(2))=x^2+y^2+x y R at x=6 ~km , and y=8 ~km R= 6^2+8^2+6 × 8 =2 37 Differentiating equation (1) with respect to t gathered 2 R (d R)/(d t)=2 (d x)/(d t)+2 y (d y)/(d t)+(x (d y)/(d t)+y (d x)/(d t)) =(1)/(2 R)[2 × 8 × 20+2 × 6 × 30+(8 × 30+6 × 20)] (d R)/(d t)= 1 2 × 2 37 [1040]= 260 37 gathered

JEE Main 201411 AprilOnline

Exam: JEE Main
Year: 2014
Date: 11 April
Mode: Online
Subject: Mathematics
Chapter: Heights and Distances

JEE Main 2014 · 11 April · Online · Mathematics · Heights and Distances

Two ships A and B are sailing straight away from a fixed point O...

Two ships A and B are sailing straight away from a fixed point

\mathrm{O}

along routes such that

\angle \mathrm{AOB}

is always

120^{\circ}

. At a certain instance,

\mathrm{OA}=8 \mathrm{~km}, \mathrm{OB}=6 \mathrm{~km}

and the ship

\mathrm{A}

is sailing at the rate of

20 \mathrm{~km} / \mathrm{hr}

while the ship B sailing at the rate of

30 \mathrm{~km} / \mathrm{hr}

. Then the distance between

\mathrm{A}

and

\mathrm{B}

is changing at the rate (in

\mathrm{km} / \mathrm{hr}

Two ships A and B are sailing straight away from a fixed point O along routes such that AOB is always 120^ . At a

Options

A $\frac{260}{\sqrt{37}}$
B $\frac{260}{37}$
C $\frac{80}{\sqrt{37}}$
D $\frac{80}{37}$

Correct answer

A. $\frac{260}{\sqrt{37}}$

Step-by-step solution

Let

\mathrm{OA}=x \mathrm{~km}, \mathrm{OB}=y \mathrm{~km}, \mathrm{AB}=\mathrm{R}

(\mathrm{AB})^2=(\mathrm{OA})^2+(\mathrm{OB})^2-2(\mathrm{OA})(\mathrm{OB}) \cos

120^{\circ}

\mathrm{R}^2=x^2+y^2-2 x y\left(-\frac{1}{2}\right)=x^2+y^2+x y

\mathrm{R}

x=6 \mathrm{~km}

, and

y=8 \mathrm{~km}

\mathrm{R}=\sqrt{6^2+8^2+6 \times 8}=2 \sqrt{37}

Differentiating equation (1) with respect to

t

\begin{gathered} 2 R \frac{d R}{d t}=2 \frac{d x}{d t}+2 y \frac{d y}{d t}+\left(x \frac{d y}{d t}+y \frac{d x}{d t}\right) \\ =\frac{1}{2 R}[2 \times 8 \times 20+2 \times 6 \times 30+(8 \times 30+6 \times 20)] \\ \frac{d R}{d t}=\frac{1}{2 \times 2 \sqrt{37}}[1040]=\frac{260}{\sqrt{37}} \end{gathered}

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