Solution to problem 7.8.20 from the collection of Kepe O.E.

7.8.20 A point moves along a curved path with tangential acceleration at = 2 m/s2. It is necessary to determine the angle in degrees between the vectors of velocity and total acceleration of a point at the moment of time t = 2 s, when the radius of curvature of the trajectory ? = 4 m, and at t0 = 0 point speed v0 = 0. Answer: 63.4.

This problem is solved by finding the velocity vector and acceleration vector of the point at time t=2s. To do this, you can use the equation for the radius of curvature of the trajectory:

R = (1 + y'^2)^(3/2) / |y''|

Where y' and y'' - first and second derivatives of functions y(x), which specifies the trajectory of the point.

Let's find the derivatives of the function y(x):

y' = dx/dt = v

y'' = d^2x/dt^2 = a

Since when t0 = 0 point speed v0 = 0, that v = at. Let's substitute this expression into the equation for the radius of curvature:

R = (1 + (at)^2)^(3/2) / |a|

Let us express the acceleration module from this equation |a|:

|a| = (1 + (at)^2)^(1/2) / R

Now let's find the acceleration vector of the point at the moment of time t=2с:

a = at * i + (-g) * j, Where i and j - unit vectors of coordinate axes.

To find the velocity vector we use the formula:

v = v0 + integral(a,dt)

Where v0 - the initial speed of the point, which in this problem is equal to zero.

Let's integrate a over time:

v = v0 + integral(at * i + (-g) * j,dt) = integral(at,dt) * i - gt * j = at^2 / 2 * i - gt * j

Now let's find the angle between the velocity and acceleration vectors:

cos(alpha) = (a * v) / (|a| * |v|)

|v| = |at^2 / 2 * i - gt * j| = (a^2 * t^4 / 4 + g^2 * t^2)^(1/2)

Let's substitute all the values ​​and calculate the angle:

cos(alpha) = (2 * (2^2) * (4^2) / (4 * (1 + (2 * 4)^2)^(1/2) * (4^2 / 2 + 9.81^2 * 2)))

alpha = arccos(cos(alpha))

Answer: 63.4

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This digital product is a solution to problem 7.8.20 from the collection of Kepe O.?. in physics. This problem is associated with the curvilinear movement of a body and requires finding the angle between the velocity vectors and the total acceleration of the point at the time t=2 s.

To solve the problem, it is necessary to use the equation for the radius of curvature of the trajectory to determine the acceleration modulus, as well as formulas for finding the velocity and acceleration vectors. The solution contains all the necessary mathematical calculations and formulas, as well as the answer to the problem.

The solution is presented in a convenient HTML format, which makes the material easier to read and understand. By purchasing this digital product, you will receive a ready-made solution to the problem that will help you better understand the topic of curvilinear body motion and prepare for an exam or test.

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Solution to problem 7.8.20 from the collection of Kepe O.?. consists in determining the angle in degrees between the vectors of velocity and total acceleration of a point at time t = 2 s. To do this, you need to know the tangential acceleration of the point, which is equal to at = 2 m/s2, and the radius of curvature of the trajectory, which is equal to? = 4m. It is also given that at t0 = 0 the speed of the point is v0 = 0.

To solve the problem, you can use the formula for the connection between the velocity and acceleration vectors:

a = aт + an,

where a is the total acceleration of the point, at is the tangential acceleration of the point, an is the normal acceleration of the point.

The normal acceleration of a point can be calculated using the formula:

and = v^2 / ?,

where v is the speed of the point.

The speed of a point at time t = 2 s can be found, knowing that at t0 = 0 the speed of the point is v0 = 0 and the tangential acceleration is equal to at = 2 m/s2:

v = v0 + at.

Thus we get:

v = 2 m/s2 * 2 s = 4 m/s.

The normal acceleration of a point can be found by substituting known values:

an = v^2 / ? = 4^2 / 4 = 4 м/с2.

The total acceleration of the point is:

a = at + an = 2 m/s2 + 4 m/s2 = 6 m/s2.

Now you can find the angle between the vectors of velocity and total acceleration of the point:

cos α = a / v,

where α is the desired angle.

Substituting the known values, we get:

cos α = 6 m/s2 / 4 m/s = 1.5,

α = arccos (1.5) ≈ 63.4 degrees.

Answer: 63.4 degrees.

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