Question

A car goes 5 km east, 3km south, 2km west and 1km north. The magnitude of the resultant displacement will be ___ km at an angle of $ta{n}^{-1}$(- / -) with the positive x-axis (where x axis is in the east direction) in the clock wise direction.

• A. $3.6km,ta{n}^{-1}\left(\frac{2}{3}\right)$
• B. $3.6km,ta{n}^{-1}\left(\frac{3}{2}\right)$
• C. $Cannotbecalculated$
• D. $3.6km,ta{n}^{-1}\left(\frac{5}{3}\right)$

Answer 1

The correct option is A $3.6km,ta{n}^{-1}\left(\frac{2}{3}\right)$

The situation can be depicted as

These are 4 vectors and $\overline{AE}$ will give the net displacement.
Net Displacement along horizontal (east-west) direction: 5 Km east - 2 Km west = 3Km east.
Along vertical direction: 3Km south - 1 Km north = 2 Km south.
So situation will be like this

As per the triangle law for vector addition $\overline{AC}$ will give the net displacement
Using Pythagoras theorem in triangle ABC $|\overline{AC}|$ = $\sqrt{3^2+2^2}=3.64Km$
$tan\theta =\frac{2}{3}or\theta =ta{n}^{-1}\frac{2}{3}$.
This is the angle the net displacement vector makes with the positive x-axis in the clock wise direction.