Solution K1-63 (Figure K1.6 condition 3 S.M. Targ 1989)

The solution to problem K1-63 (Figure K1.6 condition 3 S.M. Targ 1989) consists of two parts: K1a and K1b.

Task K1a

Let's imagine that point B is moving in the xy plane, and its law of motion is given by the equations: x = f1(t), y = f2(t), where x and y are expressed in centimeters, and t in seconds. We need to find the equation for the trajectory of a point, as well as the speed and acceleration of this point at time t1 = 1 s. In addition, we need to determine the tangent and normal acceleration of the point and the radius of curvature at the corresponding point on 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.

Task K1b

Let us assume that 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), where s = AM is the distance of a point from some origin A, measured along the arc of a circle. We need to determine the speed and acceleration of the point at time t1 = 1 s. We also need to depict vectors v and a in the figure, assuming that the point at this moment is in position M, and the positive direction of reference s is from A to M.

The digital goods store presents a unique digital product - “Solution K1-63 (Figure K1.6 condition 3 S.M. Targ 1989).” This product is a solution to problem K1-63 from the textbook by S.M. Targa, published in 1989. The solution includes two parts: K1a and K1b, which describe the movement of a point in the xy plane and along a circular arc of radius R = 2 m, respectively.

The solution to problem K1-63 provides detailed calculations and graphic illustrations. Each step of the solution is accompanied by explanations and formulas, which makes it easy to understand and reproduce the solution to the problem.

The solution is designed in a beautiful html format, which makes it more convenient to read and understand. Graphic illustrations are made in the form of drawings, which are numbered and easily linked to the text of the decision.

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The solution to K1-63 is a set of problems consisting of two problems: K1a and K1b. In problem K1a, it is required to find the equation for the trajectory of point B moving in the xy plane according to the given laws of motion x = f1(t) and y = f2(t). For the moment of time t1 = 1 s, it is necessary to find the speed and acceleration of the point, 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 table K1.

In problem K1b, a 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 is the distance of the point from some origin A, measured along the arc of the circle). It is necessary to determine the speed and acceleration of the point at time t1 = 1 s. It is also required to depict vectors v and a in the figure, 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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