In component form, the displacement vectors become
• 350 m [S] ==> (0, -350) m
• 400 m [E 20° N] ==> (400 cos(20°), 400 sin(20°)) m
(which I interpret to mean 20° north of east]
• 550 m [N 10° W] ==> (550 cos(100°), 550 sin(100°)) m
Then the student's total displacement is the sum of these:
(0 + 400 cos(20°) + 550 cos(100°), -350 + 400 sin(20°) + 550 sin(100°)) m
≈ (280.371, 328.452) m
which leaves the student a distance of about 431.8 m from their starting point in a direction of around arctan(328.452/280.371) ≈ 50° from the horizontal, i.e. approximately 431.8 m [E 50° N].
Answer: B to C
Explanation: The line is curving inwards, practically calculating the stance that it had went down. If it went straight across, it stayed the same till a specific point, furthermore calculating the bent line bending upwards is actually a partial-raise, conclude points B to C is most likely an un-even balance, meaning it had went down; or decreasing. B to C is the decreasing segment of this equation/problem (question).
The answer is D.<span>longitudinal</span>
