5. The car engine operates with a power of $P=60 \kappa B m$. Determine the speed $v_{0}$ at which the car was traveling, if after the engine is turned off, it stops after $s=450$ m. The resistance force to the car's motion is proportional to its speed. The mass of the car is $m=1000$ kg. (15 points)
Answer: $30 \mathrm{m} / \mathrm{s}$

Solution. The power of the engine: $P=F v=F_{\text {comp }} v_{0}=\alpha v_{0}^{2}$, that is, the resistance coefficient of 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: $m \frac{\Delta v}{\Delta t}=-F_{\text {comp }}=-\alpha v$ (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: $v_{0}=\sqrt[3]{\frac{s P}{m}}=\sqrt[3]{\frac{450 \cdot 60000}{1000}}=30 \mathrm{m} / \mathrm{s}$ (3 points).