$1.1.7.$ A submarine diving vertically and uniformly emits sound pulses of duration $\tau_0$. The duration of the pulse reflected from the bottom is $\tau$. The speed of sound in water is $c$. At what speed does the submarine dive?
Let's consider what the change of signal duration is related to. For this purpose, it is convenient to pass from the signal duration to its length.
Let the signal emitted by the sonar on the boat have the form shown in the figure, i.e. its intensity jumps from zero to some value at time $t$, remains constant during time $\tau_{0}$ and again jumps to zero at time $t + \tau_{0}$.
If the boat is stationary, the extent of the radiated signal (the distance between the leading and trailing edges) is $l_{0} = V \tau_{0}$. If the boat is sinking, the extent of the signal becomes smaller.
Indeed, let $AC = u \tau_{0}$, where $u$ is the submergence velocity of the boat.
The leading edge of the signal will move to point $B$ in time $\tau_{0}$, having travelled a distance $l_{0}$.
Hence, the signal length is equal to
$$l_{1} = BC = ( V - u ) \tau_{0}$$
This signal reaches the bottom, discharges from it without changing its extent, and travels towards the boat.
At some point in time, the leading edge of the signal will align with the boat.
The rear front of the signal continues to move with velocity $V + u$ relative to the boat, therefore, the duration of the signal received on the boat is equal to
$$\tau = \frac{l_{1} }{V + u} = \frac{(V - u) \tau_{0} }{V + u}$$
From here
$$u = V \frac{ \tau_{0} - \tau }{ \tau_{0} + \tau }$$
$$v = c(\tau _{0} − τ)/(\tau _{0} + τ)$$