Ajogger runs 20 mi west and then 6.0 mi north. find the magnitude and direction of the resultant displacement.

The magnitude of the resultant displacement is 21 mi and its direction is 16.7° north of west

Explanation:

Hi there!

Please see the figure for a better understanding of the problem. The total displacement vector will be the sum of both displacements:

The vector for the first displacement is:

First displacement = (20 mi, 0)

The second displacement:

Second displacement = (0, 6.0 mi)

The resultant displacement will be:

R = (20 mi, 0) + (0, 6.0 mi) = (20 mi + 0, 0 + 6.0 mi) = (20 mi, 6.0 mi)

The magnitude of this vector will be:

The magnitude of the vector displacement is 21 mi.

To find the direction of the vector R, we have to apply trigonometry:

In a right triangle the following trigonometric rule applies:

cos θ = adjacent side to the angle/ hypotenuse

In this case:

cos θ = 20 mi / magnitude of R

θ = 16.7°

The direction of the vector is 16.7° north of west.

okay i have so many questions. but the big one here is: um wt..f is this mess? ?