5. The engine of a car traveling at a speed of $v_{0}=72 \mathrm{km} / \mathbf{h}$ operates with a power of $P=50$ kW. Determine the distance from the point of engine shutdown at which the car will stop, if the resistance force is proportional to the car's speed. The mass of the car is m=1500 kg. (15

## points)
Answer: $240 m$

Solution. The power of the engine: $P=F v=F_{\text {conp }} v_{0}=\alpha v_{0}^{2}$, that is, the coefficient of resistance to the car's movement: $\alpha=\frac{P}{v_{0}^{2}}$ (3 points). The projection of the second law of Newton, for a small time interval, when moving with the engine off: $m \frac{\Delta v}{\Delta t}=-F_{\text {conp }}=-\alpha v$ (3 points). Considering that: $v \Delta t=\Delta s$, we get $m \Delta v=-\alpha \Delta s$ (3 points). Transitioning to final increments, we get: $m\left(0-v_{0}\right)=-\alpha s=-\frac{P}{v_{0}} s(3$ points $)$.

Final answer: $s=\frac{m v_{0}^{3}}{P}=\frac{1500 \cdot 20^{3}}{50000}=240$ meters (3 points).