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Answer:
86.51° North of West or 273.49°
Explanation:
Let V' = velocity of boat relative to the earth, v = velocity of boat relative to water and V = velocity of water.
Now, by vector addition V' = V + v'.
Since v' = 6.10 m/s in the north direction, v' = (6.10 m/s)j and V = 100 m/s in the east direction, V = (100 m/s)i. So that
V' = V + v'
V' = (100 m/s)i + (6.10 m/s)j
So, we find the direction,Ф the boat must steer to from the components of V'.
So tanФ = 6.10 m/s ÷ 100 m/s
tanФ = 0.061
Ф = tan⁻¹(0.061) = 3.49°
So, the angle from the north is thus 90° - 3.49° = 86.51° North of West or 270° + 3.49° = 273.49°
A thing the question doesn't tell us: What direction is Mr. B walking ?
-- If Passenger B is walking West, in the same direction as the moving walkway, then the stationary observer sees him passing by toward the West at 3.5 km/hr.
-- If Passenger B, for some strange reason, is walking BACKWARDS, heading east on the moving walkway, then the stationary observer sees him passing by toward the EAST, at 0.5 Km/hr.
-- We're wondering why the question ever introduced us to Passenger-A. He never figures into the question in any way. He has no effect on Passenger-B, and we never hear from him again.