Question

An aeroplane flies 400m towards North, 300m towards west and then 1200m vertically upwards. Then, its displacement from the initial position is:
$\begin{align}
  & (a)1400m \\
 & (b)1500m \\
 & (c)1300m \\
 & (d)1200m \\
\end{align}$

Answer 1

Hint: It has been given to us the, the displacement of the plane in three different directions. So, in order to proceed what we do is we assume the cartesian axes along these directions. Then, we write the displacement vectors in terms of unit vectors. Once, we get the final displacement vector, we can very easily find its magnitude to get the displacement from the initial position.

Complete answer:
Let the displacement along the North direction be denoted by ${{d}_{N}}$ .
Let the axis along the North be the X-axis. Then the displacement along the north direction could be written as:
 $\Rightarrow \overrightarrow{{{d}_{N}}}=400\widehat{i}$
Now, let the displacement along the West direction be denoted by ${{d}_{W}}$. And the axis along the West is the Y-axis.
Then the displacement along the west direction could be written as follows:
$\Rightarrow \overrightarrow{{{d}_{W}}}=300\widehat{j}$
Now, let the displacement along the vertically upward direction be denoted by ${{d}_{V}}$
In the last part of the aero plane’s journey the flight flew vertically upwards for a total of 1200 meters. Let this vertically upward direction be the Z-axis. Then, the displacement along this direction could be written as:
$\Rightarrow \overrightarrow{{{d}_{V}}}=1200\widehat{k}$
Now, the net displacement vector of the plane (say $\overrightarrow{d}$) can be written as:
$\Rightarrow \overrightarrow{d}=\overrightarrow{{{d}_{N}}}+\overrightarrow{{{d}_{W}}}+\overrightarrow{{{d}_{V}}}$
Putting the values of these terms in the above equation, we get:
$\Rightarrow \overrightarrow{d}=400\widehat{i}+300\widehat{j}+1200\widehat{k}$
Hence, the magnitude of net displacement vector is equal to:
$\begin{align}
  & \Rightarrow \left| \overrightarrow{d} \right|=\sqrt{{{(400)}^{2}}+{{(300)}^{2}}+{{(1200)}^{2}}} \\
 & \Rightarrow \left| \overrightarrow{d} \right|=\sqrt{160000+90000+1440000} \\
 & \Rightarrow \left| \overrightarrow{d} \right|=1300m \\
\end{align}$
Hence, the net displacement of the aero plane from its initial position is 1300 meters.

Hence, option (c) is the correct option.

Note:
Here, we calculated the net displacement of the aeroplane by breaking it into components of 3-Dimensional Vector. In this way we got easy to solve equations. These types of questions can be easily solved using vectors and we should try to avoid using Trigonometry as much as possible as it makes our equations cumbersome to solve.