Solution K1-55 (Figure K1.5 condition 5 S.M. Targ 1989)

Solution of problem K1-55 (Figure K1.5 condition 5 S.M. Targ 1989)

In problems numbered K1, it is necessary to solve two problems: K1a and K1b.

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 point's trajectory. For the moment of time t1 = 1 s, it is necessary to determine 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 the table. K1 (for Fig. 0-2 in column 2, for Fig. 3-6 in column 3, for Fig. 7-9 in column

Solution K1-55 (Figure K1.5 condition 5 S.M. Targ 1989)

In this product you will find a solution to problem K1-55 from the textbook “Problems in General Physics” by S.M. Targa, published in 1989. Figure K1.5 condition 5 is part of two problems: K1a and K1b, which you have to solve.

Problem K1a

In this problem, 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 point's trajectory; for the moment of time t1 = 1 s, determine 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 the table. K1 (for Fig. 0-2 in column 2, for Fig. 3-6 in column 3, for Fig. 7-9 in column 4). As in tasks C1 - C4, 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

In this problem, 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 - 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 K1-55 (Figure K1.5 condition 5 S.M. Targ 1989) contains two problems: K1a and K1b.

Problem K1a is to determine the trajectory of point B moving in the xy plane according to the law x = f1(t), y = f2(t), where x and y are expressed in centimeters, t in seconds. 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. The figure number is selected according to the penultimate digit of the code, and the condition number in table K1 is selected according to the last one.

Problem K1b is to determine the speed and acceleration of a point moving along an arc of a circle 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 an arc of a circle. It is also necessary to depict in the figure the velocity and acceleration vectors of the point at time t1 = 1 s, assuming that the point at this moment is in position M, and the positive direction of reference s is from A to M.

Solving problems K1a and K1b from the textbook “Problems in General Physics” by S.M. Targa, published in 1989, is represented in this product.

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Solution K1-55 is a set of two problems K1a and K1b that need to be solved.

Problem K1a requires finding the equation for the trajectory of point B moving in the xy plane. The law of motion of a point is given by the equations x = f1(t) and y = f2(t), where x and y are expressed in centimeters, t - in seconds. 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 the table. K1.

Problem K1b requires finding the speed and acceleration of a point moving 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). For the moment of time t1 = 1 s, it is also necessary 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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