Question

A ship A is moving Westwards with a speed of $$10km/h$$ and a ship B $$100km$$ South of A is moving northwards with a speed of $$10km/h$$. The time after which the distance between them becomes shortest is:
 $$\begin{array}{rl}& A)0h\\ & B)5h\\ & C)5\sqrt{2}h\\ & D)10\sqrt{2}h\end{array}$$

Answer 1

Hint: Here, the velocities of ships A and B are given. From this, we can find the velocity of A with respect to B. Then, draw a vector diagram and mark the distances and angles. Consider the triangle made by the resultant velocity and the path of ship B and find out all the sides and thereby the time taken by the ships to reach their shortest distance can be calculated.
Formula used:
 $$\text{time = }{\displaystyle \frac{\text{distance}}{\text{velocity}}}$$

Complete step by step answer:

 Velocity of A with respect to B is,
 $${\overrightarrow{v}}_{AB}={\overrightarrow{v}}_A-(-{\overrightarrow{v}}_B)={\overrightarrow{v}}_A+{\overrightarrow{v}}_B$$
 Then,
 $$v_{AB}=\sqrt{v_A+v_B}$$ ------- 1
 Given that,
 Velocity of A, $$v_A=10km/h$$
Velocity of B, $$v_B=10km/h$$
 Substitute the velocities of ships A and B in equation 1 we get,
 Velocity of A with respect to B, $$v_{AB}=\sqrt{{10}^2+{10}^2}=10\sqrt{2}$$
 From the above diagram the shortest distance between ships A and B is PQ.
Consider triangle POQ,
 $$\sin 45={\displaystyle \frac{PQ}{OQ}}$$
 $$OQ=100km$$
 Then,
 $$PQ=OQ\times \sin 45=100\times {\displaystyle \frac{1}{\sqrt{2}}}=50\sqrt{2}m$$
 We have, $$\text{time = }{\displaystyle \frac{\text{distance}}{\text{velocity}}}$$
Substitute the values of PQ and $$v_{AB}$$in the above equation, we get,
Then, the time taken to reach the shortest distance,
$$t={\displaystyle \frac{PQ}{v_{AB}}}={\displaystyle \frac{50\sqrt{2}}{10\sqrt{2}}}=5h$$

Therefore, the answer is option B.

Note:
Alternate method to solve the question

Given,
Velocity of A, $$v_A=10km/h$$
Velocity of B, $$v_B=10km/h$$
Given that, both ships A and B are travelling with the same velocity. Then, at any instant $$t$$ the distance traveled by the ships will be the same.
i.e.,
Distance travelled by ship A = Distance travelled by ship B $$=10t$$
Then, the remaining distance for B is, $$100-10t$$.
Then, considering triangle ROQ,
$$d^2=P{O}^2+O{Q}^2$$
$$d^2={\left(10t\right)}^2+{(100-t)}^2$$
$$d=\sqrt{100t^2+10000+100t^2-200t}=\sqrt{200t^2+10000-2000t}$$
 Differentiating both sides with respect to time,
$${\displaystyle \frac{d\left(d\right)}{dt}}=0$$
$$\Rightarrow 0=400t-2000$$
$$\Rightarrow 400t=2000$$
$$\Rightarrow t=5hr$$