$1.1.3.$ 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.
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.
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.
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:
$$v = \sqrt{v_x^2 + v_y^2 + v_z^2}, \;(1)$$
Let's substitute the values of velocity projections into $(1)$
$$v = \sqrt{2^2 + 2^2 + 1^2} = 3 \; \frac{m}{s}$$
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 $3\, m/s$