232k views
3 votes
The initial position vector of a particle on the x,y - plane is (5,2) and its final position vector is (-2,-3) If distance is measured antimatter. a) what is the displacement of the displacement particle? b) what is the the magnitude and direction?

1 Answer

4 votes

Step-by-step explanation:

To find the displacement of the particle, we need to subtract the initial position vector from the final position vector.

Given:

Initial position vector = (5, 2)

Final position vector = (-2, -3)

a) Displacement vector = Final position vector - Initial position vector

Displacement vector = (-2, -3) - (5, 2)

Displacement vector = (-2 - 5, -3 - 2)

Displacement vector = (-7, -5)

Therefore, the displacement of the particle is (-7, -5).

b) To find the magnitude and direction of the displacement, we can use the Pythagorean theorem and trigonometry.

Magnitude of the displacement vector (|d|) can be calculated using the formula:

|d| = √(dx^2 + dy^2)

Substituting the values:

|d| = √((-7)^2 + (-5)^2)

|d| = √(49 + 25)

|d| = √74

The direction of the displacement vector (θ) can be calculated using the formula:

θ = tan^(-1)(dy / dx)

Substituting the values:

θ = tan^(-1)(-5 / -7)

θ = tan^(-1)(5/7)

Therefore, the magnitude of the displacement is approximately √74 and the direction is approximately tan^(-1)(5/7).

User Jens Meder
by
7.7k points