Q.Represent graphically a displacement of 40 km, $30^\circ$ west of south.

A displacement is drawn as an arrow: its length (to a chosen scale) shows the magnitude, and its slant shows the direction.

Step 1 — Reference direction. "$30^\circ$ west of south" means: face south, then turn $30^\circ$ towards the west.

Step 2 — Scale. Take $1\ \text{cm} = 10\ \text{km}$, so $40\ \text{km}$ is a $4\ \text{cm}$ arrow.

Step 3 — Draw. From a point $O$, draw the due-south line (straight down). Rotate $30^\circ$ from it towards the west (to the left of the southward line) and draw the arrow $4\ \text{cm}$ long with the arrowhead at the tip. Mark the $30^\circ$ angle at $O$ and label the arrow "$40\ \text{km}$".

Reading the direction: "west of south"

In a phrase like "$30^\circ$ west of south", the second word is the reference you start from, and the first tells you which way to swing. So you face south first, then rotate $30^\circ$ towards the west. The final arrow lies in the south-west region, but nearer to south than to west.

Construction

1. Choose a scale. A displacement of $40\ \text{km}$ is convenient at $1\ \text{cm} = 10\ \text{km}$, giving a $4\ \text{cm}$ arrow.

2. Set up the compass directions. Mark a point $O$. Take north as up, south as down, east as right and west as left. Draw the due-south line vertically downward from $O$.

3. Measure the angle. Place a protractor at $O$ with its $0^\circ$ along the south line and mark $30^\circ$ on the west (left) side.

4. Draw the arrow. From $O$, draw a straight $4\ \text{cm}$ segment through the $30^\circ$ mark and put an arrowhead at the far end. Label the length "$40\ \text{km}$" and show the $30^\circ$ angle between the arrow and the south line.

What the diagram shows

The arrow points down and slightly to the left of straight-down: it makes a $30^\circ$ angle with due south, opening towards the west. Its $4\ \text{cm}$ length stands for the $40\ \text{km}$ magnitude.

Method: Representing a displacement vector graphically

Use this for any "draw a displacement of magnitude $M$, direction stated in compass words" problem.

Steps

Step 1: Decode the compass phrase "$\alpha$ of Y".

The second word ($Y$) is the reference direction you start from; the first word tells you which way to rotate through the angle $\alpha$. So "$30^\circ$ west of south" means: face south, then swing $30^\circ$ toward the west.

Step 2: Choose a convenient scale.

Pick a scale that turns the magnitude into a comfortable arrow length, e.g. $1\text{ cm}=10\text{ km}$ makes $40\text{ km}$ a $4\text{ cm}$ arrow. State the scale explicitly.

Step 3: Draw and annotate.

Mark north-up/south-down, draw the reference (south) line, use a protractor to lay off $\alpha$ on the correct side, then draw the scaled arrow with the head at the tip. Label the magnitude and mark the angle — the length encodes magnitude, the slant encodes direction.

Common Mistakes

Mistake 1: Confusing "$30^\circ$ west of south" with "$30^\circ$ south of west".

Why it's wrong: the second word is the reference direction, so here you measure $30^\circ$ from the south line toward the west — "south of west" would measure from the west line and point differently. Correct approach: start from the second-word direction and rotate through the angle toward the first word.

Mistake 2: Drawing the arrow without stating a scale.

Why it's wrong: a displacement's magnitude is encoded by the arrow's length, which is meaningless without a stated scale. Correct approach: fix a scale (e.g. $1\text{ cm}=10\text{ km}$), draw the arrow to that scale ($4\text{ cm}$), and label both the length and the $30^\circ$ angle.