Question

-- Дене координаталары xo = -3 м және у, = 1 м болатын бастапқы нүктеден орын ауыстыру векторының х осіндегі проекциясы 5,2 м, ал у осіндегі проекциясы 3 м болатындай жол жүрдi. Дененің соңғы орнының коорди- наталарын табыңдар. Орын ауыстыру векторын сызыңдар. Оның модулі кандай?

Answer 1

Problem Analysis

We are given the initial position of a point D with coordinates (x0, y0) = (-3 m, 1 m) and the projection of the displacement vector from the initial point to the final point on the x-axis is 5.2 m. We are also given that the projection of the displacement vector on the y-axis is 3 m. We need to find the coordinates of the final point and the displacement vector.

Solution

To find the coordinates of the final point, we can use the given information about the projections of the displacement vector on the x-axis and y-axis. Let's denote the coordinates of the final point as (x, y).

From the given information, we know that the projection of the displacement vector on the x-axis is 5.2 m. This means that the x-coordinate of the final point is 5.2 m.

Similarly, the projection of the displacement vector on the y-axis is 3 m. This means that the y-coordinate of the final point is 3 m.

Therefore, the coordinates of the final point are (x, y) = (5.2 m, 3 m).

To find the displacement vector, we can subtract the initial position vector from the final position vector. The initial position vector is given by (x0, y0) = (-3 m, 1 m), and the final position vector is given by (x, y) = (5.2 m, 3 m).

The displacement vector can be calculated as follows:

Displacement vector = Final position vector - Initial position vector

Displacement vector = (x, y) - (x0, y0)

Displacement vector = (5.2 m, 3 m) - (-3 m, 1 m)

Displacement vector = (5.2 m - (-3 m), 3 m - 1 m)

Displacement vector = (8.2 m, 2 m)

Therefore, the displacement vector is (8.2 m, 2 m).

The magnitude of the displacement vector can be calculated using the Pythagorean theorem:

Magnitude of displacement vector = sqrt((x_displacement)^2 + (y_displacement)^2)

Magnitude of displacement vector = sqrt((8.2 m)^2 + (2 m)^2)

Magnitude of displacement vector = sqrt(67.24 m^2 + 4 m^2)

Magnitude of displacement vector = sqrt(71.24 m^2)

Magnitude of displacement vector ≈ 8.44 m

Therefore, the magnitude of the displacement vector is approximately 8.44 m.

Conclusion

The coordinates of the final point are (5.2 m, 3 m), and the displacement vector is (8.2 m, 2 m). The magnitude of the displacement vector is approximately 8.44 m.