Answer:
The magnitude of the boy's displacement is 24.698 paces which is in the direction of approximately 58.24° North of East
Explanation:
The given parameters are;
The distance West the boy on a treasure hunt walks = 12 paces
The distance North the boy on a treasure hunt then walks = 21 paces
The distance East the boy on a treasure hunt then walks = 25 paces
Writing the displacement in vector format, gives;
d = -12·i + 21·j + 25·i
Which gives
d = 13·i + 21·j
The magnitude of the boy's resultant displacement is given as follows;
[tex]\left | \mathbf{d} \right | = \sqrt{d_x^2 + d_y^2}[/tex]
Where;
[tex]d_x = 13, \ d_y = 21[/tex]
[tex]\left | \mathbf{d} \right | = \sqrt{13^2 + 21^2} = \sqrt{610} \approx 24.698 \ paces[/tex]
The magnitude of the boy's displacement is 24.698 paces
The direction of the displacement is tan⁻¹(21/13) ≈ 58.24° North of East.
