Answer:
6. 25 m/s
Explanation:
The problem can be modeled using Pythagoras :
a² + b² = c²
Where c = hypotenus ;, b = 6
a² + 6² = c² - - - - - (1)
Diffentiating implicity :
2a da/dt = 2c dc/dt
To lowest term
a da/dt = c dc/dt - - - - (2)
a = 8 = distance from the dock
dc/dt = pull rate = 5 m/s
From (1)
8² + 6² = c²
64 + 36 = c²
100 = c²
c = 10
Substituting into (2)
a da/dt = c dc/dt
8 da/dt = 10 (5)
8 da/dt = 50
da/dt = 50/ 8
da/dt = 6.25 m/s
Hence, speed of boat approaching the dock = 6.25 m/s