A boat is pulled into a dock by a rope attached to the bow (front end) of the boat and passing through a pulley on the dock that is 6 m higher than the bow of the boat. If the rope is pulled in at a rate of 5 m/s, at what speed is the boat approaching the dock when it is 8 m from the dock?

6. 25 m/s

Step-by-step 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

answer rather than having to add and subtract the measurements of the semi-circle, notice that the semicircle to the left of ad is actually identical in area to the blank semicircle-shaped space to the left of bc. so if we transferred the red semicircle into the blank space, we would end up with a square of side length 24 cm. the area of a square is s^2 = 24^2 = 676 cm^2.

u also answered this so i think u would know but i am here to

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