Expressing a bearing velocity in exact component form

Key concept: This question combines bearings with vector components. The main skill is converting a direction measured from north into east-west and north-south components.

Step by step

What the bearing means

A bearing of $030^\circ$ is measured clockwise from north. That means the direction is 30 degrees east of north, not 30 degrees above the east axis.

Because $i$ points due east and $j$ points due north, we need the east and north components of a 12 m/s velocity vector.

Breaking the velocity into components

For a direction $30^\circ$ east of north:

• the north component is the adjacent side: $12\cos 30^\circ$
• the east component is the opposite side: $12\sin 30^\circ$

So the vector is

$\vec{v} = 12\sin 30^\circ\, i + 12\cos 30^\circ\, j.$

Now evaluate the exact values:

$12\sin 30^\circ = 12\cdot \frac12 = 6,$

$12\cos 30^\circ = 12\cdot \frac{\sqrt{3}}{2} = 6\sqrt{3}.$

Therefore,

$\vec{v} = 6i + 6\sqrt{3}\,j.$

Why this is exact form

The question asks for exact component form, so the answer should remain in surds rather than decimals. That is why $6\sqrt{3}$ is preferred over a rounded number.

The orientation is also important: since the bearing is in the northeast quadrant, both components are positive.

Final answer

$\boxed{\begin{pmatrix} 6 \\ 6\sqrt{3} \end{pmatrix}\text{ m/s}}$

or equivalently

$\boxed{6i + 6\sqrt{3}\,j\text{ m/s}}.$

Pitfall alert

The biggest mistake is treating a bearing as a standard angle from the positive x-axis. Bearings are measured clockwise from north, so the trig roles of sine and cosine swap compared with the usual unit-circle setup. Another common error is putting the wrong sign on the components. Since $030^\circ$ points northeast, both components must be positive. Finally, do not round $\sqrt{3}$ unless the question explicitly allows an approximation; exact form is usually required for full marks.

Try different conditions

If the bearing were $300^\circ$, the direction would be 60 degrees west of north, so the east component would be negative and the north component would still be positive. If the speed changed to 20 m/s on a bearing of $030^\circ$, the same method gives $10i + 10\sqrt{3}j$. The structure stays the same: identify whether the bearing is east or west of north, then use sine for the east-west part and cosine for the north-south part.

Further reading

bearing angle, vector components, unit vectors