The particle moves in such a way that the coordinates depend on time

Solution tasks 10803

Given: Equations of particle motion: x = (0.4t + 1) m, y = 0.3t m Time t = 1 s

Find: Angle between radius vector and particle speed at time t = 1 s

Solution: 1. Find the speed of the particle at time t = 1 s: x' = 0.4 m/s y' = 0.3 m/s Therefore, the speed of the particle at time t = 1 s is equal to: √(x' ² + y'²) = √((0.4)² + (0.3)²) ≈ 0.5 m/s

2. Let us find the radius vector of the particle at time t = 1 s: x = (0.4 1 + 1) m = 1.4 m y = 0.3 1 m = 0.3 m Therefore, the radius vector particle at time t = 1 s is equal to: √(x² + y²) = √((1.4)² + (0.3)²) ≈ 1.41 m

3. Find the angle between the radius vector and the particle speed: cos α = (x x' + y y') / (√(x² + y²) √(x'² + y'²)) Substitute the values: cos α = (1.4 0.4 + 0.3 0.3) / (1.41 0.5) ≈ 0.89 Therefore, the angle between the radius vector and the particle speed at time t = 1 s is equal to: α ≈ acos(0.89) ≈ 29.2°

Answer: The angle between the radius vector and the particle velocity at time t = 1 s is approximately 29.2°.

Product description

Product name: Particle in motion

Description: We present to your attention a unique digital product - “Particle in Motion”. This product will allow you to better understand the laws of particle motion and learn how to solve problems related to the movement of bodies.

Product features: Detailed description of the movement of a particle, given by the equations x=(0.4t+1) m, y=0.3t m Colorful graphs demonstrating the movement of a particle Detailed solution to the problem of determining the angle between the radius vector and the speed of a particle at a certain point in time A brief record of the conditions, formulas and laws used in solving the problem

Product advantages: Convenient format for presenting material Clear explanation of complex topics Detailed solutions to problems based on real examples

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Product description "Particle in motion":

Item name: Particle in motion Description: This digital product is designed to study the laws of particle motion. In it you will find a detailed description of the movement of a particle with coordinates specified by the functions x=(0.4t+1) m, y=0.3t m, as well as colorful graphs demonstrating the movement of the particle. A special feature of the product is a detailed solution to the problem of determining the angle between the radius vector and the velocity of a particle at time t=1 s, with a brief description of the conditions, formulas and laws used in the solution. The advantages of the product are a convenient format for presenting the material, a clear explanation of complex topics and detailed solutions to problems based on real examples.

When you purchase Particles in Motion, you will receive reliable and secure access to the digital product, the ability to study the material at any convenient time, and support and assistance if you have questions about the material. Don't miss the opportunity to deepen your knowledge in the field of physics and purchase a unique digital product "Particle in Motion".

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This product is a solution to problem No. 10803, which involves finding the angle between the radius vector and the speed of a particle at a time of 1 second.

The condition of the problem states that the particle moves in such a way that its coordinates depend on time as follows: x = (0.4t + 1) m, y = 0.3t m.

To solve the problem, you need to use the formula to calculate speed:

v = (dx/dt)i + (dy/dt)j,

where i and j are unit vectors along the coordinate axes OX and OY, respectively.

Differentiating the expressions for x and y with respect to time, we obtain:

dx/dt = 0,4 м/c dy/dt = 0,3 м/c

Thus, the speed of the particle at time t = 1 second is:

v = (0,4 м/c)i + (0,3 м/c)j

Next, to find the angle between the radius vector and the speed, you need to use the formula for the scalar product of vectors:

a * b = |a| * |b| * cos(theta),

where a and b are vectors, |a| and |b| are their lengths, and theta is the angle between them.

In this case, the radius vector of the particle can be expressed as follows:

r = xi + yj,

where x and y are the coordinates of the particle at time t = 1 second.

Substituting the coordinate values, we get:

r = (0.4 m + 1) i + (0.3 m) j

The length of the radius vector will be equal to:

|r| = sqrt((0.4 m + 1)^2 + (0.3 m)^2) ≈ 1.118 m

Now we can calculate the scalar product of the vectors r and v:

r * v = (0.4 m + 1) * 0.4 m/c + 0.3 m * 0.3 m/c ≈ 0.46 m^2/c^2

It is also necessary to calculate the lengths of the vectors r and v:

|r| ≈ 1,118 м |v| ≈ 0,5 м/c

Substituting all the values ​​into the formula for the scalar product of vectors, we get:

0.46 m^2/s^2 = 1.118 m * 0.5 m/s * cos(theta)

Where do we get it from:

cos(theta) ≈ 0.823

And finally, the angle between the radius vector and the particle speed at time t = 1 second is equal to:

theta ≈ arccos(0.823) ≈ 34.1 degrees

Therefore, the answer to the problem is 34.1 degrees (rounded to one decimal place).

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