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+ <h2>Solutions of Savchenko Physics Textbook</h2>
+ <p class="author">
+ Aliaksandr Melnichenka <br/>
+ October 2023
+ </p>
+ </header>
+
+ <h3 id="back-link"><a href="../">$\leftarrow$Back</a></h3>
+
+ <h3> Statement </h3>
+ <p>
+ $1.1.3.$ A bug flew into the room through an open window. The distance from the bug to the ceiling changed
+with a speed of 1 m/s, the distance to the wall opposite to the window changed with a speed of 2 m/s, to
+the side wall - with a speed of 2 m/s. After 1 s of flight the bug hit the corner between the ceiling and the
+side wall of the room. Determine the speed of the bug’s flight and the position in the window through
+which it flew into the room. The room is 2.5 m high, 4 m wide, and 4 m long.
+ </p>
+
+ <h3>Solution</h3>
+ <p>
+ <p>
+ Let the beetle hit the corner $B$. It is known that after one second of flight the beetle has hit point $B$, so the window is at a distance of $1$ m from the ceiling, since with respect to the ceiling its velocity has changed by $1$ m/s.
+ </p>
+ <p>
+ Relative to the side wall, its velocity changed by $2$ m/s. Therefore, it flew at a distance of $2$ m from the side wall. Point $A$ is the place where the beetle flew into the room.
+ </p>
+ <p>
+ The coordinates of point $A$: $1$ m from the ceiling and $2$ m from the side wall. Knowing the change of velocities along the three directions $XYZ$, let's find the flight velocity of the beetle:
+ </p>
+ <p style="text-align: center;">
+ $$v = \sqrt{v_x^2 + v_y^2 + v_z^2}, \;(1)$$
+ </p>
+ <p>
+ Let's substitute the values of velocity projections into $(1)$
+ </p>
+ <p style="text-align: center;">
+ $$v = \sqrt{2^2 + 2^2 + 1^2} = 3 \; \frac{m}{s}$$
+ </p>
+ </p>
+
+ <h4>Answer</h4>
+ <p>
+ A beetle flew into the room at point $A$ located $1$ m from the ceiling and $2$ m from the side wall with a velocity of $3\, m/s$
+ </p>
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