Question

vector question please help!!!!!!

Answer 1

A disoriented physics professor drives 3.24km north, then 4.79km west, and then 1.66km south. what is the direction of the resultant vector (Degrees west of north?)

Answer 2

let north be y and east be x

his position vector is r = (-4.79, 3.24 - 1.66)

if you graph it, you can see where he ended up

the direction is @ where tan@ = (opposite/adjacent) = (-4.79/1.58)

tan @ = -3.03 so @ = -71.7 degrees. If you plot this, you will
see this is 71.7 o west of north

Answer 3

wait why did you minus 1.66??

Answer 4

Subtracted the distance he went south from the distance he had gone north to get his net north-south displacement.

Answer 5

ok, do you know for a problem like this: A plane leaves Seattle, flies 84.0mi at 19.0∘ north of east, and then changes direction to 49.0∘ south of east. After flying at 118mi in this new direction, the pilot must make an emergency landing on a field. The Seattle airport facility dispatches a rescue crew. How far should the crew fly to go directly to the field? Use components to solve this problem.

do i first solvefor the components? so vectorA = 84.0 times cos19.0 ?? and the same for y and thenadd those four values ??

Answer 6

Yes, find the components, then add them to find the net vector.
It will be much clearer if you plot it, even approximately, on graph paper.
The first leg will thus have both x and y (east and north) components
x1 = 84 cos(19) and y1= 84 sin(19)
x2, y2 = etc.

eventually
r position vector = (sum of x coordinates, sum of y coordinates)
the distance from Seattle will be sqrt(r^2)
where r = (x,y) and r^2 = x^2 + y^2

Answer 7

I keep trying to graph it but I dont really understand the North south east directions. Is it like this..?and so when i plug in the 118 mi is it negative?