A rope is 1 m higher than the boat. If the rope is pulled in at a rate of 1 m/s, how fast is the boat appro?


can in terms of simple ratio.

said can form triangle, chose write out point a, doesn't matter whatever put

point
l 1m
l
dock-------------------------boat
............... 7m
*(sorry, periods don't mean when answer spaces automatically deleted disregard periods)

need complete triangle first , find distance boat point a:

use formula: a^2 + b^2 = c^2

c = square root( 7^2 + 1^2) = square root(50) = 7.07106781187

now, given rate rope being pulled @ 1m/s.

can ratios of hypotnuse ratio of lower side

set velocity of boat v

so: v / (7 m) = (1 m/s) / (7.07106781187 m)

solve v: v = [(1 m/s)(7 m)] / (7.07106781187 m)
= 0.989949493661

velocity of boat 0.9899 m/s

a boat pulled dock rope attached bow of boat , passing through pulley on dock 1 m higher bow of boat. if rope pulled in @ rate of 1 m/s, how fast boat approaching dock when 7 m dock?


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