Question

A student walks 5.00 m at an angle of 53.0° then walks another 8.00 m at angle of 130°. what is the magnitude of the student's final displacement?

Answer 1

This question can easily be answered using a scientific calculator via complex mode using the form r∠α wherein r is the magnitude and α is the angle. So, incorporating vector addition through scientific calculator:

5∠53° + 8∠130 = 10.34∠101.9°

Therefore, the student's displacement is 10.34 m, 101.9°. In other words, the student is heading 78.1° (180 - 101.9 = 78.1)  south of east.

Answer 2

Answer:

10.35 m

Explanation:

d1 = 5 m at angle 53 degree

d1 = 5 (Cos 53 i + Sin 53 j) = 3 i + 4 j

d2 = 8 m at an angle 130 degree

d2 = 8 (Cos 130 i + Sin 130 j) = - 5.14 i + 6.13 j

Total displacement = d = (3 - 5.14) i + (4 + 6.13) j = - 2.14 i + 10.13 j

magnitude of displacement = $\sqrt{(-2.14)^{2}+(10.13)^{2}}$ = 10.35 m