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Теория механики. По заданным уровнениям движения точки М, определить уровнение траектории, координаты точки, скорость точки {X=3t^2-2 {Y=4t. t=1/2с Смотреть ответ dnepr1 dnepr1 Дано параметрическое  уравнениям движения точки М: {x = 3t^2 - 2 {y = 4t. t = y/4 подставим в первое уравнение: x = (3y^2/16) - 2  или x = (3/16)y^2 - 2. Отсюда получаем каноническое уравнение параболы: x = 2*(3/32)y^2 - 2. При t = (1/2) с точка М имеет координаты: у = 4*(1/2) = 2, х = ((3/16)*2²) - 2 = (3/4) - 2 = -5/4 = -1,25. Находим скорость точки М при t = (1/2) с (в этой точке у = 2) y' = (3/16)*2y = (3/16) *2*2 = 3/4 = 0,75 м/с. Во вложениях даны график  и параметры параболы (на первом рисунке оси повёрнуты влево на 90 градусов). Новые вопросы в Другие предметы Угадайте вес мамы по её описанию (рост 170 см, 35 лет): 1. Пухлые руки, ноги 2. Попа 3. 3-й размер груди 4. Щёки 5. Животик Napisz, jakie sa zalety wysytania korespondenci droga elektroniczna, a jakie - komunikowania sie w tradycyjny sposób спасибо за внимание))) Угадайте вес мамы по её описанию (рост 170 см, 35 лет): 1. Пухлые руки, ноги 2. Попа 3. 3-й размер груди 4. Щёки 5. Животик Угадайте вес мамы по её описанию (рост 170 см, 35 лет): 1. Пухлые руки, ноги 2. Попа 3. 3-й размер груди 4. Щёки 5. Животик Предыдущий Следующий Вместе мы знаем больше Этот сайт использует cookies Политика Cookies . Вы можете указать условия хранения и доступ к cookies в своем браузере. 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Answer 1

Ваш запрос кажется немного непонятным, и, кажется, есть путаница в уравнениях и решении. Давайте попробуем пересчитать.

У вас даны параметрические уравнения движения точки М:

$x = 3t^2 - 2$ $y = 4t$

Вам нужно определить уравнение траектории, координаты точки и скорость точки при $t = \frac{1}{2}$ секунды.

Давайте подставим $t = \frac{1}{2}$ в данные уравнения:

$x = 3\left(\frac{1}{2}\right)^2 - 2 = \frac{3}{4} - 2 = -\frac{5}{4}$

$y = 4 \times \frac{1}{2} = 2$

Таким образом, координаты точки M при $t = \frac{1}{2}$ секунды: $x = -\frac{5}{4}$, $y = 2$.

Теперь найдем скорость. Для этого найдем производные $x$ и $y$ по $t$:

$\frac{dx}{dt} = 6t$ $\frac{dy}{dt} = 4$

Подставим $t = \frac{1}{2}$:

$\frac{dx}{dt} \bigg|_{t=1/2} = 6 \times \frac{1}{2} = 3$ $\frac{dy}{dt} \bigg|_{t=1/2} = 4$

Таким образом, скорость точки M при $t = \frac{1}{2}$ секунды: $V_x = 3 \, \text{м/с}$, $V_y = 4 \, \text{м/с}$.

Теперь можно записать уравнение траектории. Исходя из данных, траектория является параболой. Таким образом, уравнение траектории:

$y = ax^2 + b$

Подставим координаты точки M, чтобы найти коэффициенты $a$ и $b$:

$2 = a\left(-\frac{5}{4}\right)^2 + b$

$2 = \frac{25}{16}a + b$

Также у нас есть уравнение движения по x:

$-\frac{5}{4} = 3\left(\frac{1}{2}\right)^2 - 2$

$-\frac{5}{4} = \frac{3}{4} - 2$

$a = \frac{3}{16}$

Теперь мы можем подставить $a$ в уравнение с $b$:

$2 = \frac{25}{16} \times \frac{3}{16} + b$

$2 = \frac{75}{256} + b$

$b = \frac{437}{256}$

Таким образом, уравнение траектории:

$y = \frac{3}{16}x^2 + \frac{437}{256}$