Question

Точка движется из состояния покоя и за время t = 20 с ее скорость увеличивается до v = 24 м/с. Определить пройденный точ­кой путь и ее полное ускорение в конце 20 с, считая движение равно­ускоренным по дуге окружности радиуса r = 600 м.

Answer 1

V = v0+a*t
24=0+a*20
a = 1.2
S=a*t²/2
S=240 м
При движении по кругу необходимо учитывать два ускорения
ar=v²/R
ar=0.96
полное ускорение =sqrt(a²+ar²)
ответ=1.54 м/с²

Answer 2

А=v^2/r. a=24^2/600=0,96 м/с^2
s=vt. s=24×20=480м

Answer 3

The question you asked is about the motion of a point along a circular arc with constant acceleration. Here is a detailed answer:

- To find the distance traveled by the point, we need to find the angular displacement of the point on the circle. Since the point starts from rest and has a constant acceleration, we can use the formula:

$$v^2 = v_0^2 + 2a_s s$$

where $v$ is the final speed, $v_0$ is the initial speed, $a_s$ is the tangential acceleration, and $s$ is the distance traveled along the arc.

- Plugging in the given values, we get:

$$24^2 = 0^2 + 2a_s (20)$$

Solving for $a_s$, we get:

$$a_s = \frac{24^2}{40} = 14.4 \text{ m/s}^2$$

- To find the angular displacement, we use the formula:

$$s = r \theta$$

where $r$ is the radius of the circle and $\theta$ is the angular displacement in radians.

- Plugging in the values, we get:

$$20 = 600 \theta$$

Solving for $\theta$, we get:

$$\theta = \frac{20}{600} = 0.0333 \text{ rad}$$

- To find the distance traveled by the point, we multiply the angular displacement by the radius:

$$s = r \theta = 600 \times 0.0333 = 19.98 \text{ m}$$

- To find the total acceleration of the point, we need to find the normal (or centripetal) acceleration of the point, which is given by the formula:

$$a_n = \frac{v^2}{r}$$

where $v$ is the speed and $r$ is the radius of the circle.

- Plugging in the values, we get:

$$a_n = \frac{24^2}{600} = 0.96 \text{ m/s}^2$$

- The total acceleration of the point is the vector sum of the tangential and normal accelerations, which are perpendicular to each other. Therefore, we can use the Pythagorean theorem to find the magnitude of the total acceleration:

$$a = \sqrt{a_s^2 + a_n^2}$$

- Plugging in the values, we get:

$$a = \sqrt{14.4^2 + 0.96^2} = 14.42 \text{ m/s}^2$$

- The direction of the total acceleration is given by the angle between the tangential and normal accelerations, which can be found by using the formula:

$$\tan \alpha = \frac{a_n}{a_s}$$

where $\alpha$ is the angle.

- Plugging in the values, we get:

$$\tan \alpha = \frac{0.96}{14.4} = 0.0667$$

Solving for $\alpha$, we get:

$$\alpha = \tan^{-1} (0.0667) = 3.82^\circ$$

- The direction of the total acceleration is toward the center of the circle and slightly ahead of the tangential direction.

- Therefore, the final answer is:

The point travels 19.98 m along the circular arc and has a total acceleration of 14.42 m/s^2 at an angle of 3.82 degrees toward the center of the circle and slightly ahead of the tangential direction.

I hope this answer helps you understand the problem better. If you want to learn more about the motion of a point along a circular arc, you can check out these web sources:

- [Касательное и нормальное ускорения точки в теоретической механике](https://www.evkova.org/kasatelnoe-i-normalnoe-uskoreniya-tochki-v-teoreticheskoj-mehanike) - [(PDF) Сборник задач по физике. 7-9 классы](https://www.academia.edu/18114569/%D0%A1%D0%B1%D0%BE%D1%80%D0%BD%D0%B8%D0%BA_%D0%B7%D0%B0%D0%B4%D0%B0%D1%87_%D0%BF%D0%BE_%D1%84%D0%B8%D0%B7%D0%B8%D0%BA%D0%B5_7_9_%D0%BA%D0%BB%D0%B0%D1%81%D1%81%D1%8B)