Question

An airplane whose air speed is 650km/h is supposed to fly in straight path 35° north of east. But a steady 95 km/h wind is blowing from north. In what direction should the plane head and what is the plane resultant speed? [6]

Answer 1

Draw the situation:

We see that the plane will not move at 35 degrees, the angle will decrease. Therefore, the plane's velocity vector must be oriented more toward North so that the resulting pale blue vector looks 35 degrees above E-axis. To do this, write the condition that would provide 35-degree angle of the resulting vector above E-axis:

$$\text{tan}35^\circ=\frac{V_N}{V_E},$$

where $V_N$ and $V_E$ are components of the resulting velocity vector along N- and E-axes respectively:

$$\text{tan}35^\circ=\frac{v_\text{p}\text{sin}\alpha-v_\text{w}}{v_p\text{cos}\alpha},$$

where $v_\text{p}$ and $v_\text{w}$ - magnitudes of plane and wind velocity vectors. We know the speeds (650 km/h for the plane and 95 km/h for the wind), therefore, we need to find the angle $\alpha$:

$$\alpha=41.9^\circ.$$

Now, we have the following situation: the plane is oriented 42 degrees North of East (green vector), the resulting velocity vector (thick pale green) is at 35 degrees.

The magnitude of the new plane velocity vector:

$$v=\sqrt{(v_\text{p}\text{sin}\alpha-v_\text{w})^2+(v_\text{p}\text{cos}\alpha)^2}=591\text{ km/h}.$$