Solution of problem 7.4.1 from the collection of Kepe O.E.

7.4.1 Acceleration of point a = 0.5 ti + 0.2t2j. It is necessary to find the acceleration modulus of point a at time t = 2 s. To do this, we substitute t = 2 into the expression for the acceleration of point a: a = 0.5 ti + 0.2t2j. We get a = 0.5 * 2 * i + 0.2 * 2^2 * j = i + 0.8j. The acceleration modulus of point a is equal to the root of the sum of the squares of its projections on the coordinate axes: |a| = √(ax^2 + ay^2). In this case, ax = 1, ay = 0.8, so |a| = √(1^2 + 0.8^2) ≈ 1.28.

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Problem 7.4.1 from the collection of Kepe O.?. consists in determining the acceleration modulus of point a at time t = 2 seconds. It is known that the coordinates of point a are given by the equation of motion a = 0.5 ti + 0.2t^2j, where i and j are the unit vectors of the coordinate axes.

To solve the problem, it is necessary to calculate the second derivative of the vector function of motion with respect to time t in order to obtain the acceleration vector of point a. The acceleration modulus can then be determined, which is expressed in terms of the length of the acceleration vector.

Let's calculate the second derivative of the vector motion function with respect to time t:

a'' = (d^2a/dt^2) = (d/dt)(0,5 i + 0,4t j) = 0i + 0,4j = 0,4j

Thus, the acceleration vector of point a is 0.4j. The magnitude of the acceleration vector can be found as the root of the sum of the squares of its projections onto the coordinate axes:

|a| = sqrt(0^2 + 0,4^2) = 0,4

Answer: the acceleration modulus of point a at time t = 2 seconds is equal to 0.4 m/s^2 (or so if rounded to two decimal places).

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