Answer:
The students displacement is 3·√3 blocks, 45° North of West
Explanation:
From the question, we have;
The distance North in which the student walks = 3.0 blocks
The next walking distance West by the student = 3.0 blocks
The displacement of the student = The shortest distance between the student start and final point
By representing the motion of the student in vector format, we have;
[tex]\mathbf{d} = 3\cdot \hat{i} + 3\cdot \hat{{j}}[/tex]
Therefore the magnitude of d is given as follows;
[tex]\left | d\right | = \sqrt{3^2 + 3^2} = 3\cdot \sqrt{3} \ blocks[/tex]
The direction of the student = tan⁻¹(3/3) = 45° West of North
The students displacement = 3·√3 blocks, 45° North of West.
