# Solution:

Convert all speeds to SI units:

$36 \kappa m / h = 10 m / s$

$54 \kappa m / h = 15 m / s$

$72 \kappa m / h = 20 m / s$

The acceleration of a car of this make:

$a=\frac{v^{2}}{2 S}=\frac{20^{2}}{2 \cdot 40}=5 \mathrm{M} / s^{2}$

Distance traveled by the first car before stopping:

$S_{1}=\frac{v_{2}^{2}}{2 a}=\frac{10^{2}}{2 \cdot 5}=10 \mathrm{M}$

Distance traveled by the second car before stopping:

$S_{2}=v_{2} \cdot 2+\frac{v_{2}^{2}}{2 a}=15 \cdot 2+\frac{15^{2}}{2 \cdot 5}=30+22.5=52.5 \mathrm{M}$

Therefore, the distance between the cars should be no less than:

$l=S_{2}-S_{1}=52.5-10=42.5$ meters

(3 points)
# Problem №1 (15 points)

Two identical cars are driving in the same direction. The speed of one is $36 \kappa \mu / h$, and the other is catching up at a speed of $54 \mathrm{km} / h$. It is known that the reaction time of the driver of the rear car to the activation of the brake lights of the front car is 2 seconds. What should be the distance between the cars so that they do not collide if the first driver decides to brake sharply? For a car of this make, the braking distance is 40 meters at a speed of $72 \kappa \mu / h$.