Question

PLEASE HELP ME IM ON A TIMER

Answer 1

Answer:

Displacement: 6.71 m, Direction: 63.4 degrees north of east

Explanation:

In the attached image we can aprecciate each one of the movements of the parade. Let's say that the parade started from the origin (point (0,0)) then it moves to the east 4 blocks it means now the parade is located at point (4,0).

Then the parade went to the south three blocks, so it moves to the coordinate (4,-3). After this the parade went to the west one block so the new coordinate point is (3, -3).

And finally the movement of the 0 parade was 9 blocks to the north. It means the final point is now (0,9) - (3,-3) = (3,6)

And the displacement will be defined by the folliwing vector operation:

$A (0,0) = oi + 0j\\F (3,6) = 3i + 6 j\\Displacement vector = (3-0)i + (6-0)j = 3i + 6j$

We know that the magnitude of the displacement vector is defined by the phytagoras theorem

$Displacement = \sqrt{(3)^2+ (6)^2} \\Displacement = 6.70$

And the angle will be defined by:

tan(beta)=3/6

beta = tan^-1(6/3)

beta = 63.43°