Question

Тело переместилось из точки с координатами x = 3 , y =4 см в точку с координатами x =7 , y =5 . Определите модуль вектора перемещения и его проекции на оси координат . Помогите пожалуйста!!!

Answer 1

Вектор перемещения находим так: r=√(7-3)^2+(5-4)^2=√17
Проекция на ось x: r(x)=7-3=4
Проекция на ось y: r(y)=5-4=1

Answer 2

Calculation of Displacement Vector and its Projections

To determine the displacement vector and its projections on the coordinate axes, we need to calculate the difference between the final and initial positions.

Given: - Initial position: (x₁, y₁) = (3 cm, 4 cm) - Final position: (x₂, y₂) = (7 cm, 5 cm)

The displacement vector, Δr, can be calculated as the difference between the final and initial positions:

Δr = r₂ - r₁

where r₁ represents the initial position and r₂ represents the final position.

Let's calculate the displacement vector:

Δr = (x₂ - x₁, y₂ - y₁)

Δr = (7 cm - 3 cm, 5 cm - 4 cm)

Δr = (4 cm, 1 cm)

Therefore, the displacement vector is Δr = (4 cm, 1 cm).

To find the projections of the displacement vector on the coordinate axes, we can use the following formulas:

- The projection of Δr on the x-axis, Δx, is given by Δx = Δr · i, where i is the unit vector along the x-axis. - The projection of Δr on the y-axis, Δy, is given by Δy = Δr · j, where j is the unit vector along the y-axis.

Let's calculate the projections:

Δx = Δr · i = Δr · (1, 0) = 4 cm · 1 = 4 cm

Δy = Δr · j = Δr · (0, 1) = 1 cm · 1 = 1 cm

Therefore, the projection of the displacement vector on the x-axis is Δx = 4 cm and the projection on the y-axis is Δy = 1 cm.

In summary: - The magnitude of the displacement vector is |Δr| = √(Δx² + Δy²) = √(4² + 1²) = √17 cm. - The projection of the displacement vector on the x-axis is Δx = 4 cm. - The projection of the displacement vector on the y-axis is Δy = 1 cm.

I hope this helps! Let me know if you have any further questions.