Solution K1-28 (Figure K1.2 condition 8 S.M. Targ 1989)

Solution to problem K1-28 (Figure K1.2, condition 8, S.M. Targ, 1989)

Under number K1 there are two tasks: K1a and K1b, which need to be solved.

Problem K1a: Point B moves in the xy plane (Fig. K1.0 - K 1.9, Table K1; the trajectory of the point in the figures is shown conditionally). The law of motion of a point is given by the equations: x = f1(t), y = f2(t), where x and y are expressed in centimeters, t in seconds. It is necessary to find the equation of the trajectory of the point, determine the speed and acceleration of the point for the moment t1 = 1 s, as well as its tangential and normal accelerations and the radius of curvature at the corresponding point of the trajectory.

The dependence x = f1(t) is indicated directly in the figures, and the dependence y = f2(t) is given in the table. K1 (for Fig. 0-2 in column 2, for Fig. 3-6 in column 3, for Fig. 7-9 in column 4). The figure number is selected according to the penultimate digit of the code, and the condition number in the table. K1 - according to the last one.

Problem K1b: A point moves along a circular arc of radius R = 2 m according to the law s = f(t), given in the table. K1 in column 5 (s - in meters, t - in seconds), where s = AM is the distance of a point from some origin A, measured along the arc of a circle. It is necessary to determine the speed and acceleration of the point at time t1 = 1 s. In the figure, it is necessary to depict vectors v and a, assuming that the point at this moment is in position M, and the positive direction of reference s is from A to M.

Solution K1-28 (Figure K1.2, condition 8, S.M. Targ, 1989)

The solution to problem K1-28 is a complex solution to two problems: K1a and K1b. In problem K1a, it is necessary to determine the equation of the trajectory of a point, the speed and acceleration of the point for the moment of time t1 = 1 s, as well as its tangential and normal accelerations and the radius of curvature at the corresponding point of the trajectory. The dependence x = f1(t) is indicated directly in the figures, and the dependence y = f2(t) is given in the table. K1 (for Fig. 0-2 in column 2, for Fig. 3-6 in column 3, for Fig. 7-9 in column 4).

Problem K1b is to determine the speed and acceleration of a point at time t1 = 1 s, when the point moves along a circular arc of radius R = 2 m according to the law s = f(t), given in table. K1 in column 5 (s - in meters, t - in seconds).

Solution K1-28 is presented in the form of visual graphs and tables, which makes the material easier to understand. The solution was made in accordance with the conditions of the problem, taking into account all the necessary formulas and solution methods. The K1-28 solution is a digital product and is sold in the digital goods store at an affordable price.

Solution K1-28 is a complex solution to two problems K1a and K1b, described in the textbook by S.M. Targa "Physics Problem Book" 1989 edition.

Problem K1a is to determine the equation of the trajectory of a point, the velocity and acceleration of the point at time t1 = 1 s, as well as its tangential and normal accelerations and the radius of curvature at the corresponding point of the trajectory. To do this, it is necessary to use the dependencies x = f1(t) and y = f2(t), presented in the figures and in table K1.

Problem K1b is to determine the speed and acceleration of a point at time t1 = 1 s, when the point moves along a circular arc of radius R = 2 m according to the law s = f(t), given in table K1.

Solution K1-28 is presented in the form of visual graphs and tables, which makes the material easier to understand. The solution was made in accordance with the conditions of the problem, taking into account all the necessary formulas and solution methods. The K1-28 solution is a digital product and is sold in the digital goods store at an affordable price.

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K1-28 is the solution to problem number 8 in condition 2 of Chapter 1 of the textbook “Problems in Physics” by S.M. Targa, published in 1989. Solution K1-28 is the answer to this problem, which is probably related to physics. A more detailed description of the product is impossible without indicating the task itself and its conditions. If you have additional information, please clarify it and I will try to help you in more detail.

Solution K1-28 consists of two problems: K1a and K1b. In problem K1a, it is necessary to find the equation for the trajectory of point B moving in the xy plane according to the law x = f1(t), y = f2(t), where t is time, x and y are expressed in centimeters. For the moment of time t1 = 1 s, it is necessary to find the speed and acceleration of the point, tangential and normal accelerations and the radius of curvature at the corresponding point of the trajectory. The dependence y = f2(t) is given in table. K1, and the dependence x = f1(t) is indicated in the figures. The figure number is selected according to the penultimate digit of the code, and the condition number in the table. K1 – according to the last one.

In problem K1b, a point moves along a circular arc of radius R = 2 m according to the law s = f(t), where s is the distance of the point from the origin A, measured along the circular arc, and t is time. It is necessary to determine the speed and acceleration of the point at time t1 = 1 s. In the figure, it is necessary to depict the velocity and acceleration vectors, assuming that the point at this moment is in position M, and the positive direction of reference s is from A to M.

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