91Ó°ÊÓ

Velocity is the rate of change in position of an item in motion as seen from a specific frame of reference and measured by a specific time standard.

The speed of the plane relative to the ground is

$$\begin{gathered} V=\frac{d}{t} \\ V=\frac{3000km}{1.5hr} \\ V=2000\raisebox{1ex}{$km$}\!\left/ \!\raisebox{-1ex}{$hr$}\right. \end{gathered}$$

Velocity in meter

$$\begin{gathered} V=2000×\frac{3600}{1000} \\ V=556\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right. \end{gathered}$$

The speed of the plane relative to the earth is $556$m/s.

Hence the velocity can be calculated by the following equation:

$V_{ae}=V_{ap}+V_{pe}$

The component table will be as below:

Here we need to find the x component and the y component.

$$\begin{gathered} Xcomponent \\ {V}_{pex}=V_i\cos \mathrm{θ} \\ {V}_{pex}=\left(556\right)\cos 5^{°} \\ {V}_{pex}=554 \end{gathered}$$

$$\begin{gathered} Ycomponent \\ {V}_{pey}=V_i\sin \mathrm{θ} \\ {V}_{pey}=\left(-556\right)\sin 5^{°} \\ {V}_{pey}=-48.5 \end{gathered}$$

Hence the resultant vector will be

$$\begin{gathered} \stackrel{â‡\P Ä}{R}=\sqrt{{\left(X\right)}^2+{\left(Y\right)}^2} \\ \stackrel{â‡\P Ä}{R}=\sqrt{{\left(274\right)}^2+{\left(-48.5\right)}^2} \\ \stackrel{â‡\P Ä}{R}=278\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right. \end{gathered}$$

The velocity of the air relative to the earth is$278$$\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$.

The direction of the velocity will be:

$\begin{array}{rcl}\tan \mathrm{θ}& =& \frac{Y}{X}\\ \tan \mathrm{θ}& =& \frac{48.5}{274}\\ \mathrm{θ}& =& 10.0^{°}\end{array}$

The resultant vector has the direction of$10.0^{°}SE$.

c) The airspeed is very high at $556$m/s, which is more than the speed of sound. The wind speed is also very high at $276$m/s. Hence the problem is unreasonable.

d) The distance may be larger, or the time calculated be smaller.