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Part 1: A boy runs 16.2 blocks North, 6 blocks Northeast, and 10.5 blocks West. Determine the length of the displacement vector that goes from the starting point to his final position.
Part 2: Determine the direction of the displacement vector. Use counterclockwise as the positive angular direction, between the limits of −180◦ and +180◦ measured from East. Answer in units of ◦

Respuesta :

Part A: The length of the displacement vector of the boy is approximately 19.46 blocks

Part B: The displacement vectors direction is approximately 107.03°

The reasons why the above values are correct are given as follows:

Part A:

Let the origin, (0, 0) be the point where the boy starts to run, we have;

  • The point the boy runs to = 16.2 blocks North; Coordinates (0, 16.2)
  • The next point he runs to = 6 blocks Northeast;
  • Coordinates; (0 + 6×cos(45°), 16.2 + 6 × sin(45°))
  • The next point = 10.5 blocks West;
  • Coordinate (6×cos(45°) - 10.5, 16.2 + 6 × sin(45°))

The length of the displacement vector that goes from the starting point to his final position is the difference between the coordinate of his final position and the origin, which is given as follows;

  • (6×cos(45°) - 10.5 - 0, 16.2 + 6 × sin(45°) - 0) = (6×cos(45°) - 10.5, 16.2 + 6 × sin(45°))
  • (6×cos(45°) - 10.5, 16.2 + 6 × sin(45°)) ≈ (-6.26, 20.44)

The length of the displacement vector, R = [tex]\sqrt{x^2 + y^2}[/tex]

Therefore;

[tex]R = \sqrt{(-6.26)^2 + 20.44^2} \approx 19.46[/tex]

The length of the displacement vector that goes from the starting point to the final position, R ≈ 19.46 Blocks

Part 2: The direction displacement vector, θ, is given by the arctangent of the ratio of the y-coordinate to the x-coordinate of the displacement vector that goes from the starting to the final position as follows;

From the coordinates of the displacement vector, we have;

[tex]Direction \ \angle \theta = arctan\left(\dfrac{y-value \ of \ coordinate}{x-value \ of \ coordinate} \right)[/tex]

Therefore;

[tex]Direction \ \theta \approx arctan\left(\dfrac{20.44}{-6.26} \right) \approx -72.92^{\circ}[/tex]

Based on the location of the final position in the second quadrant, we have; that the calculated angle is with respect to the negative x-axis, measuring in the counterclockwise direction

Measuring from the positive x-axis, the direction, θ = 180° + (-72.92°) ≈ 107.03°

The direction of the displacement vector, θ ≈ 107.03°

Learn more about the resultant of a vector here:

https://brainly.com/question/15836726

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