An airplane in flight is subject to an air resistance force proportional to the square of its speed v. But there is an additional resistive force because the airplane has wings. Air flowing over the wings is pushed down and slightly forward, so from Newton’s third law the air exerts a force on the wings and airplane that is up and slightly backward (Fig. P6.94). The upward force is the lift force that keeps the airplane aloft, and the backward force is called induced drag. At flying speeds, induced drag is inversely proportional to ${\mathit{v}}^{\mathbf{2}}$, so the total air resistance force can be expressed by ${\mathit{F}}_{\mathbf{a}\mathbf{i}\mathbf{r}}\mathbf{=}\mathit{\alpha }{\mathit{v}}^{\mathbf{2}}\mathbf{+}\mathit{\beta }\mathbf{/}{\mathit{v}}^{\mathbf{2}}$, where$\mathit{\alpha }$ and$\mathit{\beta }$ are positive constants that depend on the shape and size of the airplane and the density of the air. For a Cessna150 , a small single-engine airplane, $\mathit{\alpha }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{30}\mathbf{}\mathbf{N}\mathbf{·}{\mathbf{s}}^{\mathbf{2}}\mathbf{/}{\mathbf{m}}^{\mathbf{2}}$and $\mathit{\beta }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{30}\mathbf{}\mathbf{N}\mathbf{·}{\mathbf{s}}^{\mathbf{2}}\mathbf{/}{\mathbf{m}}^{\mathbf{2}}$. In steady flight, the engine must provide a forward force that exactly balances the air resistance force. (a) Calculate the speed (in km/h) at which this airplane will have the maximum range (that is, travel the greatest distance) for a given quantity of fuel. (b) Calculate the speed (in km/h) for which the airplane will have the maximum endurance (that is, remain in the air the longest time).

It is given that the airplane is moving with a constant speed, and the total air resistance force is given by,

$F_{air}=\alpha v^2+\frac{\beta }{v^2}......\left(1\right)$

Here, role="math" localid="1667893949876" $\alpha =0.3\mathrm{N}·{\mathrm{s}}^2/{\mathrm{m}}^2$, and$\beta =3.5\times {10}^5\mathrm{N}·{\mathrm{s}}^2/{\mathrm{m}}^2$ are positive constants.

The expression of work done is given by,

$\mathit{W}\mathbf{=}\mathit{F}\mathit{s}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }$ ..........(1)

Where F is applied force, S is distance move in the direction of applied force and$\theta$ is angle between applied force and distance direction.

The amount of work done is given by,

$$\begin{gathered} W=FS \\ =S\left(\alpha v^2+\frac{\beta }{v^2}\right)..........\left(2\right) \end{gathered}$$

Here, S is the maximum range that can be travelled.

For maximum range or greatest distance,

$$\begin{gathered} \frac{dW}{dv}=0 \\ S\frac{d}{dv}\left(\alpha v^2+\frac{\beta }{v^2}\right)=0 \\ S\left[2\alpha v-2\frac{\beta }{v^3}\right]=0 \\ 2S\left[\alpha v-\frac{\beta }{v^3}\right]=0 \end{gathered}$$

Simplify further.

$$\begin{gathered} \left[\alpha v-\frac{\beta }{v^3}\right]=0 \\ \alpha v=\frac{\beta }{v^3} \\ {v}^4=\left(\frac{\beta }{\alpha }\right) \\ v={\left(\frac{\beta }{\alpha }\right)}^{\frac{1}{4}}........\left(3\right) \end{gathered}$$

Substitute $\alpha =0.3\mathrm{N}·\mathrm{s}/{\mathrm{m}}^2$, and$\beta =3.5\times {10}^5\mathrm{N}·\mathrm{s}/{\mathrm{m}}^2$ in equation (3).

role="math" localid="1667895034337" $$\begin{gathered} V={\left(\frac{3.5\times {10}^5\mathrm{N}\cdot {\mathrm{m}}^2/{\mathrm{s}}^2}{0.3\mathrm{N}\cdot {\mathrm{s}}^2/{\mathrm{m}}^2}\right)}^{\frac{1}{4}} \\ =32.9\mathrm{m}/\mathrm{s} \\ =(32.9\mathrm{m}/\mathrm{s})\left(\frac{60\times 60}{{10}^3}\right) \\ \approx 118.3\mathrm{km}/\mathrm{h} \end{gathered}$$

Therefore, the required velocity is $118.3\mathrm{km}/\mathrm{h}$.

Write the range in terms of$t$ as follows.

$S=vt$

From equation (2),

$$\begin{gathered} W=vt\left(\alpha v^2+\frac{\beta }{v^2}\right) \\ =t\left(\alpha v^3+\frac{\beta }{v}\right) \end{gathered}$$

For the maximum endurance or longest time $\frac{dW}{dv}=0$.

$$\begin{gathered} \frac{dW}{dv}=0 \\ t\frac{d}{dv}\left(\alpha v^3+\frac{\beta }{v^2}\right)=0 \\ t\left[3\alpha v^2-\frac{\beta }{v^2}\right]=0 \\ \left[3\alpha v^2-\frac{\beta }{v^3}\right]=0 \end{gathered}$$

Simplify further.

$$\begin{gathered} 3\alpha v^2=\frac{\beta }{v^3} \\ {v}^4=\frac{\beta }{3\alpha } \\ v={\left(\frac{\beta }{3\alpha }\right)}^{\frac{1}{4}}.........\left(4\right) \end{gathered}$$

Substitute $\alpha =0.3\mathrm{N}·\mathrm{s}/{\mathrm{m}}^2$, and$\beta =3.5\times {10}^5\mathrm{N}·\mathrm{s}/{\mathrm{m}}^2$ in equation (4).

$$\begin{gathered} V={\left(\frac{3.5\times {10}^5\mathrm{N}\cdot {\mathrm{m}}^2/{\mathrm{s}}^2}{3\times 0.3\mathrm{N}\cdot {\mathrm{s}}^2/{\mathrm{m}}^2}\right)}^{\frac{1}{4}} \\ =24.97\mathrm{m}/\mathrm{s} \\ =(24.97\mathrm{m}/\mathrm{s})\left(\frac{60\times 60}{{10}^3}\right) \\ \approx 89.9\mathrm{km}/\mathrm{h} \end{gathered}$$

Therefore, the required velocity is $89.9\mathrm{km}/\mathrm{h}$.