Question

A ship $A$ is moving Westwards with a speed of $10km{h}^{-1}$ and a ship $B100km$ South of $A$, is moving Northwards with a speed of $10km{h}^{-1}$.
The time after which the distance between them becomes shortest, is

• A. $5\sqrt{2}h$
• B. $10\sqrt{2}h$
• C. $0h$
• D. $5h$

Answer 1

Answer: (D)

Given situation is shown in the figure.

Velocity of ship A
$v_A=10km{h}^{-1}$ towards west
Velocity of ship B
$v_B=10km{h}^{-1}$ towards north
$OS=100km$
OP = shortest distance
Relative velocity between $A$ and $B$ is
$v_{AB}=\sqrt{v_A^2+v_B^2}=10\sqrt{2}km{h}^{-1}$
cos $45^{\circ }=\frac{OP}{OS};\frac{1}{\sqrt{2}}=\frac{OP}{100}$
OP = $\frac{100}{\sqrt{2}}=\frac{100\sqrt{2}}{2}=50\sqrt{2}$ km
The time after which distance between them equals to $OP$ is given by
$t=\frac{OP}{v_{AB}}=\frac{50\sqrt{2}}{10\sqrt{2}}$
$\Rightarrow 1=5h$