# Answer to Question #17238 in Calculus for hsd

Question #17238

A boat is pulled into a dock by a rope attached to the bow ("front") of the boat and passing through a pulley on the dock that is 1m higher than the bow of the boat. If the rope is pulled in at a rate of 1m/s, how fast is the boat approaching the dock when it is 8m from the dock?

=..........m/s

=..........m/s

Expert's answer

We need to find the horizontal component of the end of the rope. The length of a rope is

L = √((8[m])² + (1[m])²) = √65[m].

Therefore,

V = 1[m/s]*cosα = 1[m/s]*8[m]/√65[m] = 8/√65[m/s] ≈ 0.992[m/s],

where α is an angle between the rope and the water.

L = √((8[m])² + (1[m])²) = √65[m].

Therefore,

V = 1[m/s]*cosα = 1[m/s]*8[m]/√65[m] = 8/√65[m/s] ≈ 0.992[m/s],

where α is an angle between the rope and the water.

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