Answer

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**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:**

\[\text{time = }\dfrac{\text{distance}}{\text{velocity}}\]

**Complete step by step answer:**

Velocity of A with respect to B is,

\[{{\vec{v}}_{AB}}={{\vec{v}}_{A}}-\left( -{{{\vec{v}}}_{B}} \right)={{\vec{v}}_{A}}+{{\vec{v}}_{B}}\]

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,

\[\sin 45=\dfrac{PQ}{OQ}\]

\[OQ=100km\]

Then,

\[PQ=OQ\times \sin 45=100\times \dfrac{1}{\sqrt{2}}=50\sqrt{2}m\]

We have, \[\text{time = }\dfrac{\text{distance}}{\text{velocity}}\]

Substitute the values of PQ and \[{{v}_{AB}}\]in the above equation, we get,

Then, the time taken to reach the shortest distance,

\[t=\dfrac{PQ}{{{v}_{AB}}}=\dfrac{50\sqrt{2}}{10\sqrt{2}}=5h\]

**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,

\[{{d}^{2}}=P{{O}^{2}}+O{{Q}^{2}}\]

\[{{d}^{2}}={{\left( 10t \right)}^{2}}+{{\left( 100-t \right)}^{2}}\]

\[d=\sqrt{100{{t}^{2}}+10000+100{{t}^{2}}-200t}=\sqrt{200{{t}^{2}}+10000-2000t}\]

Differentiating both sides with respect to time,

\[\dfrac{d\left( d \right)}{dt}=0\]

\[\Rightarrow 0=400t-2000\]

\[\Rightarrow 400t=2000\]

\[\Rightarrow t=5hr\]

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