The surface of the dock is 6 feet above the water. If you pull the rope in at a rate of 2 ft/sec, how quickly is the boat approaching the dock at the moment when there is 10 feet of rope still left to pull in

The boat is approaching the dock at a rate of 2.5 ft/s.

Explanation:

Let the rope length be 'l' at any time 't', the distance of boat from dock be 'b' at any time 't'.

Given:

The height of dock above water (h) = 6 feet

Rate of pull of rope or rate of change of rope is,

As clear from the question, the height is fixed and only the length 'l' and distance 'b' varies with time 't'.

Now, the above situation represents a right angled triangle as shown below.

Using Pythagoras Theorem, we have:

Now, differentiating the above equation with time 't', we get:

Now, the distance 'b' can be calculated using 'l=10 ft' in equation (1). This gives,

Now, substituting all the given values in equation (2) and solve for . This gives,

Therefore, the boat is approaching the dock at a rate of 2.5 ft/s.

acceleration = (change in speed) / (time for the change)

change in speed = (20-120)= -100 m/s.

1.5 min = 90 sec

acceleration = -100/90 = -1.11 m/s^2.

we don't care about the 150kg or the 2,500 meters. they don't make any difference.

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