# Problem No. 6 (10 points)

A person started walking slowly along a straight line with a speed of $v_{1}=0.5 \mathrm{~m} / \mathrm{c}$. At time $t_{1}$, he turned strictly to the right and walked with a speed twice as high. After another time interval $t_{1}$, he turned strictly to the right again, and his speed became four times the initial speed. After the last right turn, he ran with a speed of $5 v_{1}$ and returned to the initial point of his trajectory. Determine his average speed over the entire path.

Answer: $1.1 \mathrm{~m} / \mathrm{c}$

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# Solution and Evaluation Criteria:

From the problem statement, we can conclude that the person's trajectory is a rectangle.

(2 points)

![](https://cdn.mathpix.com/cropped/2024_05_06_3c4b08bfdadd8858901bg-5.jpg?height=369&width=651&top_left_y=1089&top_left_x=728)

Moreover, \( S_{2} = 2 S_{1} \)

(2 points)

Average speed:

\[ v_{C P} = \frac{S_{1} + S_{2} + S_{3} + S_{4}}{t_{1} + t_{2} + t_{3} + t_{4}} = \frac{6 S_{1}}{t_{1} + t_{1} + \frac{S_{1}}{4 v_{1}} + \frac{2 S_{1}}{5 v_{1}}} = \frac{6 S_{1}}{t_{1} + t_{1} + \frac{t_{1}}{4} + \frac{2 t_{1}}{5}} = \frac{6 S_{1}}{\frac{53}{20} t_{1}} = \frac{120}{53} v_{1} \approx 1.1 \, \text{m/s} \] (6 points)

## Problem 7 (10 points)

A person holds one end of a light board, while the other end rests on a cylinder lying on the ground. The length of the board \( L = 10 \, \text{m} \). How far will the cylinder roll by the time the person reaches it? The radius of the wheel is much smaller than the length of the board. There is no slipping.

![](https://cdn.mathpix.com/cropped/2024_05_06_3c4b08bfdadd8858901bg-5.jpg?height=186&width=1020&top_left_y=2414&top_left_x=541)

Answer: \( 10 \, \text{m} \)

## Solution and Evaluation Criteria:

The speed of the person is equal to the speed of the top point of the cylinder, which is twice the speed of the center of the cylinder.

Thus, the distance the person travels will be twice the distance the cylinder rolls.

![](https://cdn.mathpix.com/cropped/2024_05_06_3c4b08bfdadd8858901bg-6.jpg?height=66&width=774&top_left_y=601&top_left_x=181)

We get: \( S_{\text{cylinder}} = 10 \) meters