1167. A locomotive is moving along a horizontal section of the track c) at a speed of 72 km/hour. In what time and at what distance will it be stopped by the brake, if the resistance to motion after the start of braking is equal to 0.2 of its weight.
Solution. According to Newton's second law in mechanics, the differential equation of motion for the locomotive will be

$$
m \frac{d^{2} s}{d t^{2}}=-0.2 m g
$$

where $s$ is the distance traveled in time $t$, $m$ is the mass of the locomotive, and $g$ is the acceleration due to gravity.

Multiplying both sides of this equation by $d l$ and then integrating twice, we get

$$
\frac{d s}{d t}=-0.2 g t + c_{1}, \quad s=-0.1 g t^{2} + c_{1} t + c_{2}
$$

The values of the constants $c_{1}$ and $c_{2}$ are determined by the initial conditions: at $t=0$, $s=0$, and $\frac{d s}{d t}=72 \kappa \text{ m/h} = 20 \text{ m/s}$.

From the first condition, we have $c_{2}=0$. From the second condition, it follows that $c_{1}=20$.

Therefore, the equations of motion for the locomotive will be:

$$
\begin{gathered}
\frac{d s}{d t}=v=20-0.2 g t (\mu ; \text{sec}) \\
s=20 t-0.1 g t^{2} (\mu)
\end{gathered}
$$

Setting $v=0$ in equation (5), we find the braking time during which the locomotive will be stopped by the brake:

$$
t=\frac{20}{0.2 g} \approx \frac{100}{9.8} \approx 10.2 \text{ sec}
$$

Setting $t \approx 10.2$ in equation (6), we find the braking distance:

$$
s \approx 20 \cdot 10.2 - 0.1 \cdot 9.8 \cdot 10.2^{2} \approx 102 \text{ m}
$$