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Question: A helicopter is flying south with a speed of \[50\,{\text{km}} \cdot {{\text{h}}^{ - 1}}\]. A train ...

A helicopter is flying south with a speed of 50kmh150\,{\text{km}} \cdot {{\text{h}}^{ - 1}}. A train is moving at the same speed towards the east. The relative velocity of the helicopter as seen by the passengers in the train will be towards
A. North east
B. South east
C. North west
D. South west

Explanation

Solution

Rewrite the velocities of the helicopter and the train in the vector form. Then calculate the relative velocity of the train with respect to the helicopter. Rearrange the equation for this relative velocity and then calculate the angle between the two directions of the helicopter and the train. Hence, determine the direction of the helicopter seen by the passengers in the train.

Complete step by step answer:
We have given that the helicopter is moving towards south with a speed of 50kmh150\,{\text{km}} \cdot {{\text{h}}^{ - 1}} and the train is moving towards east with the same speed of 50kmh150\,{\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 vHT{\vec v_{HT}} of the train with respect to the helicopter is given by
vHT=vHvT{\vec v_{HT}} = {\vec v_H} - {\vec v_T}
Substitute - 50\,{\text{\hat j}} for vH{\vec v_H} and 50\,{\text{\hat i}} for vT{\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
θ=tan1(vHvT)\theta = {\tan ^1}\left( {\dfrac{{{v_H}}}{{{v_T}}}} \right)
Substitute 50kmh1 - 50\,{\text{km}} \cdot {{\text{h}}^{ - 1}} for vH{v_H} and 50kmh1 - 50\,{\text{km}} \cdot {{\text{h}}^{ - 1}} for vT{v_T} in the above equation.
θ=tan1(50kmh150kmh1)\theta = {\tan ^1}\left( {\dfrac{{ - 50\,{\text{km}} \cdot {{\text{h}}^{ - 1}}}}{{ - 50\,{\text{km}} \cdot {{\text{h}}^{ - 1}}}}} \right)
θ=tan1(1)\Rightarrow \theta = {\tan ^1}\left( 1 \right)
θ=45\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 4545^\circ .Thus, the direction of the helicopter seen by the passengers in the train is south west.

Hence, the correct option is D.

Note: The students should keep in mind that the direction of the velocity of the train is towards east. But when the values are substituted in the formula for the relative velocity of the train with respect to the helicopter, we can consider the direction of the train towards west. Hence, the direction of the helicopter seen by the passengers in the train is south west.