Solution to problem 7.4.12 from the collection of Kepe O.E.

7.4.12 The position of a point on the plane is specified by its radius vector r=0.3t²i+0,1t³j. It is necessary to determine the acceleration modulus of a point at an instant of time t = 2 s. (Answer 1.34)

In this problem, it is required to determine the acceleration modulus of a point specified by the radius vector on the plane at the moment of time t = 2 s. To solve it is necessary to find the antiderivative of the second derivative of the radius vector with respect to time and substitute the time value into it t = 2 s. After this, you should calculate the magnitude of the resulting acceleration vector, which will be equal to 1,34.

Solution to problem 7.4.12 from the collection of Kepe O.?.

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This product is a solution to problem 7.4.12 from the collection of Kepe O.?. mathematics. The task is to determine the acceleration modulus of a point specified by the radius vector on the plane at time t = 2 s. To solve the problem, it is necessary to find the antiderivative of the second derivative of the radius vector with respect to time and substitute the time value t = 2 s into it. After this, you should calculate the magnitude of the resulting acceleration vector, which is 1.34.

The solution to the problem is presented in digital format and can be purchased from a digital goods store. The solution is designed in a beautiful HTML format, which makes the material easy to read and understand. By purchasing this product, you receive a ready-made solution to the problem, which can be used to prepare for exams, independently study the material, or as a material for teaching mathematics. The approach used to solve the problem is simple and clear, which allows you to quickly understand the essence of the problem and get the correct answer.

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Solution to problem 7.4.12 from the collection of Kepe O.?. is associated with determining the acceleration modulus of a point on the plane at time t = 2 s. The position of a point on the plane is determined by its radius vector r = 0.3t^2 i + 0.1t^3 j.

To determine the acceleration of a point, it is necessary to calculate its time derivative twice. First, let's find the derivative of the radius vector r with respect to time t:

r' = (d/dt)(0.3t^2 i + 0.1t^3 j) = 0.6ti + 0.3t^2 j

Then we find the derivative of acceleration with respect to time:

r'' = (d/dt)(0.6ti + 0.3t^2 j) = 0.6i + 0.6tj

To find the acceleration modulus of a point at time t = 2 s, substitute t = 2 into the resulting expression and find its length:

|r''| = sqrt((0,6)^2 + (0,6*2)^2) = 1,34

Thus, the acceleration modulus of the point at time t = 2 s is 1.34.

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