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Question

Answers

A. North east

B. South east

C. North west

D. South west

Answer

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We have given that the helicopter is moving towards south with a speed of \[50\,{\text{km}} \cdot {{\text{h}}^{ - 1}}\] and the train is moving towards east with the same speed of \[50\,{\text{km}} \cdot {{\text{h}}^{ - 1}}\].

Hence, the velocity of the helicopter can be written as

\[{\vec v_H} = - 50\,{\text{\hat j}}\]

The velocity of the train can be written as

\[{\vec v_T} = 50\,{\text{\hat i}}\]

The relative velocity \[{\vec v_{HT}}\] of the train with respect to the helicopter is given by

\[{\vec v_{HT}} = {\vec v_H} - {\vec v_T}\]

Substitute \[ - 50\,{\text{\hat j}}\] for \[{\vec v_H}\] and \[50\,{\text{\hat i}}\] for \[{\vec v_T}\] in the above equation.

\[{\vec v_{HT}} = - 50\,{\text{\hat j}} - 50\,{\text{\hat i}}\]

\[{\vec v_{HT}} = - 50\,{\text{\hat j}} + \left( { - 50\,{\text{\hat i}}} \right)\]

The diagram representing the directions of the velocity of helicopter, velocity of train and relative velocity of the train with respect to helicopter is as follows:

The angle between the relative velocity of the train with respect to helicopter and the velocity of the train and helicopter is given by

\[\theta = {\tan ^1}\left( {\dfrac{{{v_H}}}{{{v_T}}}} \right)\]

Substitute \[ - 50\,{\text{km}} \cdot {{\text{h}}^{ - 1}}\] for \[{v_H}\] and \[ - 50\,{\text{km}} \cdot {{\text{h}}^{ - 1}}\] for \[{v_T}\] in the above equation.

\[\theta = {\tan ^1}\left( {\dfrac{{ - 50\,{\text{km}} \cdot {{\text{h}}^{ - 1}}}}{{ - 50\,{\text{km}} \cdot {{\text{h}}^{ - 1}}}}} \right)\]

\[ \Rightarrow \theta = {\tan ^1}\left( 1 \right)\]

\[ \therefore \theta = 45^\circ \]

Therefore, the angle between the velocity of the train with respect to the helicopter with the velocity of the train and helicopter is \[45^\circ \].Thus, the direction of the helicopter seen by the passengers in the train is south west.