Answer:
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:
[tex]|R| = \sqrt{(20 mi)^{2} + (6.0 mi)^{2}} = 21 mi[/tex]
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.
