$1.3.22^*.$ Planes fly in a straight line towards each other at the same speed $v$. The maximum range of detection of each other by them $l$. One plane, after detecting the other, makes a $U$-turn without changing the speed module, and flies parallel to the second plane. At what constant acceleration will the planes lose sight of each other at the end of the turn?
For the planes to lose sight of each other, the distance between them by the end of the turn must be $l$
Along the $Oy$ axis, the second plane must shift by the value of the double radius-curvature of the turn trajectory, and the first plane must shift in time $t$ by the value $vt$ along the $Ox$ axis.
Let's write the condition for losing sight of each other
$$ (2r)^2 + (l-vt)^2 = l^2\;(1)$$
Since the speed remains unchanged $v = \text{сonst}$
$$ vt = \pi r\;(2)$$
Substitute $(2)$ into $(1)$
$$ (4+\pi^2)r^2 - 2l \pi r = 0\;(3)$$
Express $r$
$$ r = l \frac{2 \pi}{4+\pi^2}\;(4) $$
Then, the acceleration with which the second plane is moving
$$ a = \frac{v^2}{r} = \frac{v^2}{l} \frac{4+\pi^2}{2 \pi}\;(5) $$
$$v = \sqrt{gR}=8\text{ km/s}$$