I'll analyze the uploaded image and solve the problems one by one. Let’s start with the tasks:

---

1. По какой формуле определяют проекцию вектора ускорения?

We need to identify which formula represents the projection of acceleration vector:

• Option a_x = (v_x - v_0x) / t is correct because acceleration is the change in velocity over time.

Answer: г) a_x = (v_x - v_0x) / t

---

2. В каких единицах СИ измеряют ускорение?

Acceleration is measured as the change in velocity over time, so its SI unit is meters per second squared (м/с²).

Answer: в) м/с²

---

3. На рисунке изображен график зависимости скорости v_x автомобиля от времени t. В каком интервале времени модуль ускорения будет максимальным?

Acceleration is the slope of the velocity-time graph. From the given graph:

• From 0 to 10 s, the slope is steepest, which corresponds to the maximum acceleration.

Answer: а) от 0 до 10 с

---

4. На каком из графиков представлено движение тела, имеющего наибольшее ускорение?

Acceleration is determined by the steepness of the velocity-time curve. The steeper the curve, the higher the acceleration:

• The 2nd graph (б) has the steepest slope.

Answer: б) 2

---

5. По графику зависимости скорости от времени определите проекцию ускорения тела.

To find the acceleration projection, calculate the slope for each segment of the given graph:

• Segment 1 (0–2 s):
  $a = \frac{\Delta v}{\Delta t} = \frac{4 - 0}{2 - 0} = 2 \, \text{м/с²}$

• Segment 2 (2–4 s):
  $a = \frac{\Delta v}{\Delta t} = \frac{0 - 4}{4 - 2} = -2 \, \text{м/с²}$

Answer: Acceleration is $+2 \, \text{м/с²}$ for the first segment and $-2 \, \text{м/с²}$ for the second.

---

6. Торможение электропоезда метро начинается на расстоянии 250 м от станции. Каково должно быть ускорение поезда, чтобы остановиться?

We use the equation of motion:

$v^2 = v_0^2 + 2aS$

Given:

• $v = 0$ (final velocity),
• $v_0 = 54 \, \text{км/ч} = 15 \, \text{м/с}$,
• $S = 250 \, \text{м}$.

Rearranging for $a$:

$a = \frac{v^2 - v_0^2}{2S}$

Substitute the values:

$a = \frac{0 - 15^2}{2 \cdot 250} = \frac{-225}{500} = -0.45 \, \text{м/с²}$

Answer: $a = -0.45 \, \text{м/с²}$

---

7. Движение тела описывается уравнением $x = -5 + 2t - 2t^2 + 3t^3$. Определите проекцию скорости тела через 3 секунды.

The velocity is the derivative of position with respect to time:

$v(t) = \frac{dx}{dt} = \frac{d}{dt}(-5 + 2t - 2t^2 + 3t^3)$

$v(t) = 2 - 4t + 9t^2$

Substitute $t = 3$:

$v(3) = 2 - 4(3) + 9(3^2)$

$v(3) = 2 - 12 + 81 = 71 \, \text{м/с}$

Answer: $v(3) = 71 \, \text{м/с}$

---

Let me know if you'd like further clarification or a breakdown of any step! Below are related questions to deepen your understanding:

---

Related Questions:

1. How does the slope of a velocity-time graph determine acceleration?
2. Why is the unit of acceleration $\text{м/с²}$ in the SI system?
3. Derive the formula $v^2 = v_0^2 + 2aS$ used for solving uniformly accelerated motion problems.
4. Can acceleration be negative, and what does it signify in terms of motion?
5. How would the position equation in question 7 change if acceleration were constant?

Tip:

When solving motion problems, always carefully determine which kinematic equation applies to avoid unnecessary calculations.