WORKED EXAMPLE

5) An aircraft is flying due north with a velocity of $200\,m\,{s^{ - 1}}$. A side wind of velocity $50\,m\,{s^{ - 1}}$ is blowing due east. What is the aircraft’s resultant velocity (give the magnitude and direction)?
Here, the two velocities are at ${90^ \circ }$. A sketch diagram and Pythagoras’s theorem are enough to solve the problem.
Step 1: Draw a sketch of the situation – this is shown in Figure $1.16a$.
Step 2: Now sketch a vector triangle. Remember that the second vector starts where the first one ends. This is shown in Figure $1.16b$.

Step 3: Join the start and end points to complete the triangle.
Step 4: Calculate the magnitude of the resultant vector v (the hypotenuse of the right-angled triangle).
$\begin{array}{l}
{v^2} = {200^2} + {50^2} = 40000 + 2500 = 42500\\
v = \sqrt {42500}  \approx 206m\,{s^{ - 1}}
\end{array}$
Step 5: Calculate the angle $\theta $:
$\begin{array}{l}
\tan \theta  = \frac{{50}}{{200}} = 0.25\\
\theta  = {\tan ^{ - 1}}(0.25) \approx {14^ \circ }
\end{array}$
So the aircraft’s resultant velocity is $206\,m\,{s^{ - 1}}$ at ${14^ \circ }$ east of north, ${076^ \circ }$ or $N{76^ \circ }E$.