1168. A bullet enters a board 10 cm thick at a speed of $200 \mathrm{m} /$ sec, and exits the board, having penetrated it, at a speed of 50 m/sec. Find how long the bullet's movement through the board lasted, if the resistance of the board to the bullet's motion is proportional to the square of its speed.
Solution. Let $m$ be the mass of the bullet, $s$ - the distance traveled by it in time $t$, measured from the moment it enters the board. Then the differential equation of the bullet's motion through the board is

$$
m \frac{d^{2} s}{d t^{2}}=-k\left(\frac{d s}{d t}\right)^{2} \text { or } \frac{d^{2} s}{d t^{2}}=-a\left(\frac{d s}{d t}\right)^{2}, a=\frac{k}{m}
$$

Assuming in this second-order equation $\frac{d s}{d t}=v$, we get a first-order equation with separable variables

$$
\frac{d v}{d t}=-a v^{2}, \quad \frac{d v}{v^{2}}=-a d t
$$

Integrating this, we find

$$
\frac{1}{v}=a t+c_{1} \text { or } v=\frac{1}{a t+c_{1}}
$$

From the initial condition $v=200$ when $t=0$, given in the problem, we determine the constant $c_{1}$ :

$$
200=\frac{1}{c_{1}}, \quad c_{1}=\frac{1}{200}
$$

Therefore, the dependence of the bullet's velocity on time as it moves through the board will be

$$
v=\frac{200}{1+200 a t}
$$

Assuming in the last equation $v=\frac{d s}{d t}$, separating variables and integrating, we get

$$
d s=\frac{200 d t}{1+20(1 a t}, \quad s=\frac{1}{a} \ln (1+200 a t)+c_{2}
$$

From the condition $s=0$ when $t=0$, it follows that $c_{2}=0$.

Therefore, the dependence of the distance traveled by the bullet in the board on time will be

$$
s=\frac{1}{a} \ln (1+200 a t)
$$

Assuming $v=50$ in equation (7) and $s=0.1$ (m) in equation (8), we get a system of equations with unknowns $t$ and $a$ :

$$
50=\frac{200}{1+200 a t} ; \quad 0.1=\frac{1}{a} \ln (1+200 a t)
$$

Solving the first equation for $1+200 a t=4$ and substituting into the second, we find the value of $a$ :

$$
0.1=\frac{1}{a} \ln 4 ; \quad a=10 \ln 4
$$

Finally, substituting the value of $a$ into the first equation of the system and solving for the required time of the bullet's flight through the board:

$$
50=\frac{200}{1+(200 \cdot 10 \ln 4) t} ; \quad t=\frac{3}{2000 \ln 4} \approx 0.001 \text { sec. }
$$