Chapter 5, Problem 60A

To determine

Magnitude and direction of resultant displacement.

Answer to Problem 60A

Magnitude of displacement will be 134.16 m and direction will be $45°$ south of east.

Explanation of Solution

Given:

First displacement of 30 due south and then 30 m due east.

Formula Used:

The resultant of two vectors is given by,

  $R=\sqrt{P^2+Q^2}$

Here, P and Q are two vectors which are perpendicular to each other.

Calculation:

It is given first displacement is 30 m due south.

Let $P=30\text{ m}$

And then 30 m due east.

Let $Q=30\text{ m}$

From the vector principle when two vectors are perpendicular to each other then their resultant magnitude is given by,

  $\begin{array}{c}R=\sqrt{P^2+Q^2}\\ =\sqrt{{30}^2+{30}^2}\\ =\sqrt{18000}\\ =\text{ 134}\text{.16 m}\end{array}$

Magnitude of displacement by graphical method is shown below.

It is given first displacement is 30 m due south.

Therefore,

  

Then 30 m due east.

Therefore,

  

So the resultant displacement will be,

  

Now direction of displacement will be,

  $\begin{array}{c}\theta ={\tan }^{-}\left(\frac{1}{1}\right)\\ =45°\end{array}$

Therefore direction of displacement will be $45°$ south of east.

As the magnitude of displacement is same by both mathematical and by graphical method.

Conclusion:

Therefore, magnitude of displacement will be 134.16 m and direction will be $45°$ south of east.