Solution

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

$$(2r)^2 + (l-vt)^2 = l^2\tag{1}$$

Since the speed remains unchanged $v = \text{сonst}$

$$vt = \pi r\tag{2}$$

Substitute $(2)$ into $(1)$

$$(4+\pi^2)r^2 - 2l \pi r = 0\tag{3}$$

Express $r$

$$r = l \frac{2 \pi}{4+\pi^2}\tag{4}$$

Then, the acceleration with which the second plane is moving

$$a = \frac{v^2}{r} = \frac{v^2}{l} \frac{4+\pi^2}{2 \pi}\tag{5}$$