Question

An adventurous dog moves through blocks. Use the analytical method to find the dog’s net displacement vector, its magnitude, and its direction. How would your answer be affected if each block was about 100 m?

a) Net displacement: 0 m, Magnitude: 0, Direction: Undefined; It won't be affected
b) Net displacement: 400 m, Magnitude: 400 m, Direction: East; It won't be affected
c) Net displacement: 400 m, Magnitude: 400 m, Direction: East; It would be doubled
d) Net displacement: 200 m, Magnitude: 200 m, Direction: East; It would be halved

Answer 1

The dog's net displacement vector is approximately (2, 3) with a magnitude of sqrt(13) and a direction of 56.31°. Changing the size of the blocks would not affect the answer.

Step-by-step explanation:

To find the dog's net displacement vector, we can add up all the individual displacements. The dog moves three blocks east, so the x-component of the displacement is +3 blocks. It then moves two blocks north, so the y-component of the displacement is +2 blocks. Next, it moves one block east (+1 block on the x-axis) and one block north (+1 block on the y-axis). Finally, it moves two blocks west (-2 blocks on the x-axis). Adding up the x-components and y-components, we get a net displacement of (3+1-2, 2+1) = (2, 3).

To find the magnitude of the net displacement vector, we can use the Pythagorean theorem. The magnitude is the square root of the sum of the squares of the x-component and y-component, so it is sqrt(2^2 + 3^2) = sqrt(4 + 9) = sqrt(13).

The direction of the net displacement is given by the arctan of the y-component divided by the x-component. So the direction is arctan(3/2) ≈ 56.31°. Therefore, the dog's net displacement vector is approximately (2, 3) with a magnitude of sqrt(13) and a direction of 56.31°.

If each block was about 100 m instead of 100 yd, the answer would not be affected. The net displacement vector, magnitude, and direction would remain the same.