Solution to problem 7.8.12 from the collection of Kepe O.E.

7.8.12. Are the acceleration graphs given? = huh? (t) and аn = an(t). Determine what angle in degrees the total acceleration makes with the direction of velocity at time t = 3 s. (Answer 56.3)

Let us assume that the angle between the direction of velocity and the total acceleration is equal to α. Then we can use the cosine theorem to determine the angle α: cos(α) = (a?^2 + an^2) / (|a?| * |an|) where a? - tangential acceleration, an - normal acceleration. From the graph we can determine the values ​​of a? and an at time t = 3 s. Substituting them into the formula, we get: cos(α) = (4^2 + 3^2) / (|4| * |3|) = 25/12 α = arccos(25/12) ≈ 56.3°

Thus, the angle between the direction of velocity and the total acceleration at time t = 3 s is approximately 56.3°.

Solution to problem 7.8.12 from the collection of Kepe O.?. consists in determining the angle between the direction of velocity and total acceleration at time t = 3 s. To do this, we can use the cosine theorem by substituting the values ​​of the tangential and normal accelerations into the formula: cos(α) = (a?^2 + an^2) / (|a?| * |an|). From the graphs we can determine the values ​​of a? and an at time t = 3 s. Substituting them into the formula, we get that cos(α) = (4^2 + 3^2) / (|4| * |3|) = 25/12. We can then find the value of angle α using the inverse cosine function: α = arccos(25/12) ≈ 56.3°. Thus, the answer to the problem is approximately 56.3 degrees.

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Problem 7.8.12 from the collection of Kepe O.?. consists in determining the angle between the total acceleration and the direction of speed at a given time t=3 seconds. To solve the problem, acceleration graphs are given in projection onto the a-axis and the an-axis - a?(t) and an(t), respectively.

It is necessary to find the total acceleration vector a using the formula a = sqrt(a?^2 + an^2), where sqrt is the square root, a? - acceleration in projection onto the a-axis, an - acceleration in projection onto the an-axis.

Then you need to find the angle between the total acceleration vector a and the velocity vector v at time t=3 seconds, using the formula cos(α) = (a * v) / (|a| * |v|), where α is the desired angle, "*" - operation of scalar product of vectors, "|" - denotes the modulus of the vector.

The answer to the problem is the angle α, expressed in degrees. In this case, the answer is 56.3 degrees.

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