Question

Unreasonable Results A commercial airplane has an air speed of m/s due east and flies with a strong tailwind. It travels km in a direction south of east in h.

(a) What was the velocity of the plane relative to the ground?

(b) Calculate the magnitude and direction of the tailwind’s velocity.

(c) What is unreasonable about both of these velocities?

(d) Which premise is unreasonable?

Short answer

a. $556\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$m/s.

b. $278\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$And $10.0^{°}SE$.

c. The air speed is very high at m/s, which is more than the speed of sound. The wind speed is also very high at m/s. Hence the problem is unreasonable.

d. Either the distance is larger or the time calculated is smaller.

Step-by-step solution

Definition of velocity

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 velocity of the plane relative to the earth

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

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

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

Velocity of the air relative to the earth

Hence the velocity can be calculated by the following equation:

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

The component table will be as below:

	X	Y
$V_{ap}$	-280	0
$V_{pe}$	554	-48.5
Resultant	274	-48.5

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

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

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

Determine the resultant velocity

Hence the resultant vector will be

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

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:

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

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.