Question

Радиус-вектор материальной точки изменяется со временем по закону r=i*coswt+j*sinwt , где i, j – орты осей X и Y, (w – постоянная величина) . Определить: 1) модуль скорости; 2) модуль ускорения.

Answer 1

Скорость - первая производная по времени радиус-вектора, найдём её:
$v=\frac{dr(\tau)}{d\tau}=\frac{d(i*cos\omega \tau+j*sin\omega \tau)}{d\tau}=j\omega cos\omega \tau-i\omega sin\omega \tau$
Ускорение - производная скорости по времени, найдём её:
$|a|=\frac{dv}{d\tau}=\frac{d(j\omega cos\omega \tau-i\omega sin\omega \tau)}{d\tau}=\\|-(i\omega ^2cos\omega \tau+j\omega ^2sin\omega \tau)|=i\omega ^2cos\omega \tau+j\omega ^2sin\omega \tau$

Answer 2

Radius-vector of a Material Point

The radius-vector of a material point is a vector that describes the position of the point in space. In this case, the radius-vector of the material point is given by the equation:

r = i * cos(wt) + j * sin(wt)

Where: - r is the radius-vector of the material point, - i and j are the unit vectors along the X and Y axes respectively, - w is a constant value, - t is the time.

Now, let's determine the requested quantities.

1) Module of Velocity

The velocity of a material point is the derivative of its position with respect to time. To find the module of velocity, we need to differentiate the given equation with respect to time.

Differentiating the equation r = i * cos(wt) + j * sin(wt) with respect to time, we get:

v = -w * i * sin(wt) + w * j * cos(wt)

The module of velocity (|v|) is the magnitude of the velocity vector. In this case, it can be calculated as:

|v| = sqrt((-w * sin(wt))^2 + (w * cos(wt))^2)

Simplifying the expression, we have:

|v| = sqrt(w^2 * sin^2(wt) + w^2 * cos^2(wt))

Since sin^2(x) + cos^2(x) = 1, we can simplify further:

|v| = sqrt(w^2 * (sin^2(wt) + cos^2(wt)))

|v| = sqrt(w^2)

Therefore, the module of velocity is |v| = |w|.

2) Module of Acceleration

The acceleration of a material point is the derivative of its velocity with respect to time. To find the module of acceleration, we need to differentiate the velocity equation obtained earlier with respect to time.

Differentiating the equation v = -w * i * sin(wt) + w * j * cos(wt) with respect to time, we get:

a = -w^2 * i * cos(wt) - w^2 * j * sin(wt)

The module of acceleration (|a|) is the magnitude of the acceleration vector. In this case, it can be calculated as:

|a| = sqrt((-w^2 * cos(wt))^2 + (-w^2 * sin(wt))^2)

Simplifying the expression, we have:

|a| = sqrt(w^4 * cos^2(wt) + w^4 * sin^2(wt))

Since cos^2(x) + sin^2(x) = 1, we can simplify further:

|a| = sqrt(w^4 * (cos^2(wt) + sin^2(wt)))

|a| = sqrt(w^4)

Therefore, the module of acceleration is |a| = |w^2|.

To summarize: 1) The module of velocity is |v| = |w|. 2) The module of acceleration is |a| = |w^2|.

Please let me know if you need any further clarification or assistance!