Answer:
The back end of the vessel will pass the pier at 4.83 m/s
Explanation:
This is purely a kinetics question (assuming we're ignoring drag and other forces) so the weight of the boat doesn't matter. We're given:
Δd = 315.5 m
vi = 2.10 m/s
a = 0.03 m/s^2
vf = ?
The kinetics equation that incorporates all these variables is:
vf^2 = vi^2 + 2aΔd
vf = √(2.1^2 + 2(0.03)(315.5))
vf = 4.83 m/s