# Problem №1 (15 points)

A car starts moving when the traffic light turns green. As it approaches the next traffic light, the red signal lights up for it again. The graph shows the dependence of the car's speed on the distance traveled. Determine the time of the car's movement between the traffic lights.

![](https://cdn.mathpix.com/cropped/2024_05_06_9e5a3306654225a310b5g-05.jpg?height=463&width=1142&top_left_y=1082&top_left_x=137)
# Solution:

On the first segment, the dependence of speed on distance:

$v^{2} \sim S$

(3 points)

Let's find the acceleration:

$a=\frac{v^{2}}{2 S}=0.5 \, \text{m/s}^2$

The time spent on this segment:

$t_{1}=\sqrt{\frac{2 S}{a}}=\sqrt{\frac{2 \cdot 900}{0.5}}=60 \, \text{s}$

On the second segment, the motion is uniform and the time is:
$t_{2}=\frac{S}{v}=\frac{200}{30} \approx 6.7 \, \text{s}$

The time of motion on the last segment is equal to the time of motion on the first. (3 points)

Final result:

$t=2 t_{1}+t_{2}=2 \cdot 60+6.7=126.7$ seconds

## Problem №2 (10 points)

2 moles of molecular oxygen are in a vertical vessel with smooth walls, which is closed by a weightless piston. The temperature of the oxygen is $T=300 \, \text{K}$. During slow heating, the volume of the gas increased threefold, and $40 \%$ of the molecules dissociated into atoms. Determine the work done by the gas in this process.

## Solution:

Since $40 \%$ of the molecules dissociated into atoms, the amount of substance increased by 1.4 times:

$v_{2}=1.4 \cdot v_{1}=2.8$ moles

(3 points)

The process is isobaric, so from the Mendeleev-Clapeyron equation $p V=v R T$ it follows that:

$T_{2}=\frac{30}{14} T_{1}=\frac{30 \cdot 300}{14} \approx 643 \, \text{K}$

The work done by the gas in this process:

$A=p \Delta V=v_{2} R T_{2}-v_{1} R T_{1}=(2.8 \cdot 8.31 \cdot 643)-(2 \cdot 8.31 \cdot 300) \approx 9975$ J