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+ <meta name="description" content="What is the duration of a plane's flight from Novosibirsk to Moscow and back in a straight line, if the wind blows at an angle \alpha to the track at a speed u during the entire flight? The speed of the aircraft relative to the air v, the length of the route L. In which wind direction is the maximum flight duration?">
+ <meta name="author" content="Aliaksandr Melnichenka">
+ <meta name="date" content="2023-10" scheme="YYYY-MM">
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+ <title>What is the duration of a plane's flight from Novosibirsk to Moscow and back in a straight line, if the wind blows at an angle \alpha to the track at a speed u during the entire flight? The speed of the aircraft relative to the air v, the length of the route L. In which wind direction is the maximum flight duration?</title>
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+ <h2>Solutions of Savchenko Problems in Physics</h2>
+ <p class="author">
+ Aliaksandr Melnichenka <br/>
+ October 2023
+ </p>
+ </header>
+
+ <h3 id="back-link"><a href="../../#1.4">$\leftarrow$Back</a></h3>
+
+ <h3> Statement </h3>
+ <p>
+ $1.4.7^*.$ What is the duration of a plane's flight from Novosibirsk to Moscow and back in a straight line, if the wind blows at an angle $\alpha$ to the track at a speed $u$ during the entire flight? The speed of the aircraft relative to the air $v$, the length of the route $L$. In which wind direction is the maximum flight duration?
+ </p>
+
+ <h3>Solution</h3>
+ <p>
+
+ <center>
+ <figure>
+ <img src="https://savchenkosolutions.com/1/1.4.7/draw.png"
+ loading="lazy" width="200" />
+ </figure>
+ </center>
+<p >
+For the plane to fly on course, the following conditions must be met
+</p>
+<p class="exp">
+$$u \sin \alpha = v \sin \beta$$
+</p>
+<p >
+Where from
+</p>
+<p class="exp">
+$$\cos \beta = \sqrt{1 - u^2 \sin ^2 \alpha / v^2}$$
+</p>
+<p >
+And the total time there and back
+</p>
+<p class="exp">
+$$ t_1 = \frac{L}{v \cos \beta + u \cos \alpha} $$
+</p>
+ <p class="exp">
+$$ t_2 = \frac{L}{v\cos \beta - u \cos \alpha} $$
+</p>
+<p >
+We find the full time as
+</p>
+<p class="exp">
+$$t=t_1+t_2$$
+</p>
+<p >
+Substitute the value of $\cos \beta$:
+</p>
+<div class="scroll-wrapper">
+<p class="exp">
+$$ t=\frac{L}{\sqrt{v^2 - u^2 \sin ^2 \alpha } + u \cos \alpha} + \frac{L}{\sqrt{v^2 - u^2 \sin ^2 \alpha} - u \cos \alpha} $$
+</p>
+ <p class="exp">
+$$ t=L\frac{\sqrt{v^2 - u^2 \sin ^2 \alpha }+\sqrt{v^2 - u^2 \sin ^2 \alpha} }{(\sqrt{v^2 - u^2 \sin ^2 \alpha} + u \cos \alpha)(\sqrt{v^2 - u^2 \sin ^2 \alpha} - u \cos \alpha)} $$
+</p>
+ <p class="exp">
+$$ t=\frac{2L\sqrt{v^2 - u^2 \sin ^2 \alpha }}{(\sqrt{v^2 - u^2 \sin ^2 \alpha} + u \cos \alpha)(\sqrt{v^2 - u^2 \sin ^2 \alpha} - u \cos \alpha)} $$
+</p>
+</div>
+
+<p >
+Expressing the required time:
+</p>
+<p class="exp">
+$$ \fbox{$t=\frac{2L\sqrt{v^2 - u^2 \sin ^2 \alpha }}{v^{2}-u^{2}}$} $$
+</p>
+ </p>
+
+ <h4>Answer</h4>
+ <p>
+ $$t=\frac{2L\sqrt{v^{2}-u^{2}\operatorname{sin}^{2}\alpha}}{v^{2}-u^{2}}.\textrm{ Along the highway}.$$
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