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<h2>Solutions of Savchenko Physics Textbook</h2>
<p class="author">
Aliaksandr Melnichenka <br/>
October 2023
</p>
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− <article>
− <h3 id="back-link"><a href="../">$\leftarrow$Back</a></h3>
−
− <h4>Statement</h4>
− <p>
− $1.1.2.$ A radar determines the coordinates of a flying airplane by measuring the angle between the direction to the North Pole and the direction to the airplane and the distance from the radar to the airplane. At some point in time, the position of the airplane was determined by the coordinates: angle $\alpha_1 = 44^{\circ}$, distance $R_1 = 100\;km$. At a time interval of $5\;s$ after this moment, the coordinates of the airplane on the radar: angle $\alpha_2 = 46^{\circ}$, distance $R_2 = 100\;km$. In a Cartesian coordinate system with the $y$-axis pointing north and with radar at the origin, represent the position of the airplane at both moments of time; determine the modulus and direction of its velocity. Count the angle in a clockwise direction.
− </p>
+ <h3 id="back-link"><a href="../">$\leftarrow$Back</a></h3>
+
+ <h3> Statement </h3>
+
+ <p>
+ $1.1.2.$ A radar determines the coordinates of a flying airplane by measuring the angle between the direction to the North Pole and the direction to the airplane and the distance from the radar to the airplane. At some point in time, the position of the airplane was determined by the coordinates: angle $\alpha_1 = 44^{\circ}$, distance $R_1 = 100\;km$. At a time interval of $5\;s$ after this moment, the coordinates of the airplane on the radar: angle $\alpha_2 = 46^{\circ}$, distance $R_2 = 100\;km$. In a Cartesian coordinate system with the $y$-axis pointing north and with radar at the origin, represent the position of the airplane at both moments of time; determine the modulus and direction of its velocity. Count the angle in a clockwise direction.
+ </p>
+
<h4>Solution</h4>
− <p class="TxtSolutions">
+
+ <p>
$O_1$ is the initial position of the airplane. $O_2$ is the final position of the airplane. In time $t$ the airplane will fly the distance:
− </p>
− <p class="TxtSolutions" style="text-align: center;">
− $S = v_0 \cdot t$
− </p>
− <p class="TxtSolutions">
+ $$S = v_0 \cdot t$$
We find the distance $S$ from the isosceles triangle $AO_1O_2$, where the angle $O_1AO_2 = 2^{\circ}$. Then
− </p>
− <p class="TxtSolutions" style="text-align: center;">
− $S = 2R \cdot sin1^{\circ}$,
− </p>
− <p class="TxtSolutions">
+ $$S = 2R \cdot sin1^{\circ}$$
Where $R_1=R_2=R$
− </p>
− <p class="TxtSolutions" style="text-align: center;">
− $v_0 \cdot t = 2R \cdot sin1^{\circ}$,
− </p>
− <p class="TxtSolutions">
+ $$v_0 \cdot t = 2R \cdot sin1^{\circ}$$
Desired speed
− </p>
− <p class="TxtSolutions" style="text-align: center;">
− $v_0 = \frac{2R \cdot sin1^{\circ}}{t}$,
− </p>
− <p class="TxtSolutions">
+ $$v_0 = \frac{2R \cdot sin1^{\circ}}{t}$$
At small angle $sin\alpha \approx \alpha$ expressed in radians, i.e. $1^{\circ} = \frac{\pi}{180}$
− </p>
− <p class="TxtSolutions" style="text-align: center;">
− $\boxed{v_0 = \frac{2 \cdot 10^5 \cdot \pi}{5 \cdot 180} = 698\;m/s.}$
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+ $$\boxed{v_0 = \frac{2 \cdot 10^5 \cdot \pi}{5 \cdot 180} = 698\;m/s.}$$
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