Question

A ship $A$ is moving westward with a speed of $10$ km/s. and ship $B$ $100$ km south of $A$ is moving northward with a speed of $10$ km/hr. The time after which distance between then becomes shortest is:

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

Answer 1

Answer: (A)

$5$ hr

According to the problem

In diagram, the shortest distance b/w the ship A and B is QM.

$\sin {45}^0=\frac{MQ}{OQ}$

Implies that

Hence Also,

$VAB=\sqrt{\left({V_A}^2-{V_B}^2\right)}=\sqrt{{10}^2+{10}^2}10\sqrt{2}km/hr$

For $t$,

$t=\frac{MQ}{VAB}=\frac{50\sqrt{2}}{10\sqrt{2}}=5hr$