Answer:
The displacement of the dog which is the shortest distance between the two points is 5 meters 53.13° North of East
Step-by-step explanation:
The given parameters are;
The dog's displacement due East = 3 meters
The dog's displacement due North = 4 meters
The displacement, D motion of the dog can be written in vector form as follows;
D = 3i + 4j
The magnitude of the displacement is given as follows;
[tex]\left | D\right | = \sqrt{(3 \ meters)^2 + (4 \ meters)^2} = \sqrt{25 \ meters^2} =5 \ meters[/tex]
The displacement is the shortest distance between two points
The direction is given tan⁻¹(4/3) = 53.13° North of East
The displacement of the dog which is the shortest distance between the two points = 5 meters 53.13° North of East.
