91Ó°ÊÓ

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

$F_{air}=\mathrm{Î\pm }{v}^2+\frac{\mathrm{β}}{v^2}......\left(1\right)$

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

The expression of work done is given by,

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

Where F is applied force, S is distance move in the direction of applied force and$\mathrm{θ}$ is angle between applied force and distance direction.

The amount of work done is given by,

$$\begin{gathered} W=FS \\ =S\left(\mathrm{Î\pm }{v}^2+\frac{\mathrm{β}}{v^2}\right)..........\left(2\right) \end{gathered}$$

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

For maximum range or greatest distance,

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

Simplify further.

$$\begin{gathered} \left[\mathrm{Î\pm }v-\frac{\mathrm{β}}{v^3}\right]=0 \\ \mathrm{Î\pm }v=\frac{\mathrm{β}}{v^3} \\ {v}^4=\left(\frac{\mathrm{β}}{\mathrm{Î\pm }}\right) \\ v={\left(\frac{\mathrm{β}}{\mathrm{Î\pm }}\right)}^{\frac{1}{4}}........\left(3\right) \end{gathered}$$

Substitute $\mathrm{Î\pm }=0.3\mathrm{N}·\mathrm{s}/{\mathrm{m}}^2$, and$\mathrm{β}=3.5×{10}^5\mathrm{N}·\mathrm{s}/{\mathrm{m}}^2$ in equation (3).

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

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),

$$\begin{gathered} W=vt\left(\mathrm{Î\pm }{v}^2+\frac{\mathrm{β}}{v^2}\right) \\ =t\left(\mathrm{Î\pm }{v}^3+\frac{\mathrm{β}}{v}\right) \end{gathered}$$

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

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

Simplify further.

$$\begin{gathered} 3\mathrm{Î\pm }{v}^2=\frac{\mathrm{β}}{v^3} \\ {v}^4=\frac{\mathrm{β}}{3\mathrm{Î\pm }} \\ v={\left(\frac{\mathrm{β}}{3\mathrm{Î\pm }}\right)}^{\frac{1}{4}}.........\left(4\right) \end{gathered}$$

Substitute $\mathrm{Î\pm }=0.3\mathrm{N}·\mathrm{s}/{\mathrm{m}}^2$, and$\mathrm{β}=3.5×{10}^5\mathrm{N}·\mathrm{s}/{\mathrm{m}}^2$ in equation (4).

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

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