Question

Точка движется по окружности радиуса 2 м согласно уравнению: S=At^2, где A=2м/с3. В какой момент времени t нормальное ускорение точки будет равно тангенциальному ускорению? Чему будет равно при этом полное ускорение точки?

Answer 1

An=2*A
At=v^2/r=2At/2=At

An=At
2A=At
t=2 с

A=sqrt(An^2+At^2)=7,6 м/сс

Answer 2

Нормальное ускорение точки равно радиусу окружности умноженному на квадрат скорости точки (an = r * v^2). Тангенциальное ускорение точки равно произведению радиуса окружности на ускорение точки (at = r * a).

Из уравнения S = At^2, где A = 2 м/с^3, можно найти ускорение точки: a = 2t. Подставим это выражение в уравнение для тангенциального ускорения: at = r * a = r * 2t = 2rt.

Нормальное ускорение точки равно тангенциальному ускорению, поэтому an = at: r * v^2 = 2rt.

Скорость точки на окружности можно найти, зная, что S = At^2: S = r * θ, где θ - угол поворота точки на окружности. Таким образом, v = dθ/dt = 2rt. Подставим это выражение в уравнение для нормального ускорения: r * (2rt)^2 = 2rt.

Упростим это уравнение: 4r^2t^2 = 2rt, 4rt = 2rt, 4r = 2, r = 0.5 м.

Таким образом, когда радиус окружности равен 0.5 м, нормальное ускорение точки будет равно тангенциальному ускорению. При этом полное ускорение точки будет равно: atotal = sqrt(an^2 + at^2) = sqrt((0.5 * v^2)^2 + (2rt)^2).

Answer 3

Problem Analysis

We are given that a point is moving along a circle with a radius of 2 m, according to the equation S = At^2, where A = 2 m/s^3. We need to determine at what moment in time the normal acceleration of the point will be equal to the tangential acceleration, and what will be the total acceleration of the point at that moment.

Solution

To solve this problem, we need to find the expressions for the normal acceleration and the tangential acceleration, and then set them equal to each other.

The equation S = At^2 represents the displacement of the point as a function of time. Taking the derivative of this equation with respect to time, we can find the expressions for the velocity and acceleration of the point.

Taking the derivative of S = At^2 with respect to time, we get:

v = dS/dt = 2At

Taking the derivative of v = 2At with respect to time, we get:

a = dv/dt = 2A

So, the tangential acceleration of the point is constant and equal to 2A.

To find the normal acceleration, we need to consider the geometry of the circular motion. The normal acceleration is the component of the acceleration that is perpendicular to the tangent of the circle at any given point. In circular motion, the normal acceleration is given by the equation:

an = v^2 / r

where v is the velocity of the point and r is the radius of the circle.

Substituting the expression for v from earlier, we have:

an = (2At)^2 / r = 4A^2t^2 / r

Now, we can set the normal acceleration equal to the tangential acceleration and solve for the time t:

4A^2t^2 / r = 2A

Simplifying the equation, we have:

2t^2 = r / 2A

Substituting the given values, r = 2 m and A = 2 m/s^3, we have:

2t^2 = 2 / (2 * 2) = 1/2

Simplifying further, we get:

t^2 = 1/4

Taking the square root of both sides, we have:

t = 1/2

So, at t = 1/2 seconds, the normal acceleration of the point will be equal to the tangential acceleration.

To find the total acceleration of the point at this moment, we can substitute the value of t = 1/2 into the expression for the tangential acceleration:

a = 2A = 2 * 2 m/s^3 = 4 m/s^2

Therefore, at t = 1/2 seconds, the total acceleration of the point will be 4 m/s^2.

Answer

At t = 1/2 seconds, the normal acceleration of the point will be equal to the tangential acceleration. The total acceleration of the point at this moment will be 4 m/s^2.