Question

An airplane flies at a speed of 425 km/h in an easterly direction with an angle of 45 with respect to north. A wind blows from the south at a speed of 100 km/h. Graphically determine the real trajectory of the plane then calculate its real speed .

Answer 1

Solution

The graph above is the graphical representation of the airplane movement

For angle A

`A=90+45=135^(\circ)`

Note: Cosine Rule Formula

Usimg the cosine rule, we have

`\begin{gathered} a^2=b^2+c^2-2bccosC \\ a^2=425^2+100^2-2(425)(100)cos(135) \\ a^2=250729.0764 \\ a=√(250729.0764) \\ a=500.7285456 \\ a=500.73kmh^(-1) \end{gathered}`

Therefore, the real speed is

`\begin{equation*} 500.73kmh^(-1) \end{equation*}`

We are left with getting the real trajectory

The real trajectory is angle C

`C=45+\theta`

We will find theta by using the sine rule

`\begin{gathered} (a)/(sinA)=(c)/(sin\theta) \\ (500.7285456)/(sin135)=(100)/(sin\theta) \\ sin\theta=(100sin135)/(500.7285456) \\ sin\theta=0.1412155922 \\ \theta=8.11819343 \\ \theta=8.118^(\circ) \end{gathered}`

Thus,

`\begin{gathered} C=45+8.118 \\ C=53.118^(\circ) \end{gathered}`

Therefore, the answer is

`\begin{gathered} Real\text{ }Trajectory=53.118^(\circ) \\ Real\text{ }Speed=500.73kmh^(-1) \end{gathered}`