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:
The time needed is 
Explanation:
From the question we are told that
The magnitude of the stimulated acceleration due gravity is 
The diameter of the spaceship is 
Generally the force acting on the spaceship is

Given that the spaceship is rotating it implies that the force experienced by the occupant is a centripetal force so

Thus

=> 
Generally the speed of this spaceship is mathematically represented as

=> ![v^2 = [\frac{2\pi}{T}] ^2](https://tex.z-dn.net/?f=v%5E2%20%20%3D%20%20%20%5B%5Cfrac%7B2%5Cpi%7D%7BT%7D%5D%20%5E2)
=> 
=> 
=> 
substituting values


2 min = 120 sec
120/15 = 8
The black horse represents 8 seconds.