SOLUTION: avigation. In 3 sec, Penny walks 24 ft, to the bow (front) of a tugboat. The boat is cruising at a rate of 5 ft/sec. What is Penny’s rate of travel with respect to land?

Algebra ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> SOLUTION: avigation. In 3 sec, Penny walks 24 ft, to the bow (front) of a tugboat. The boat is cruising at a rate of 5 ft/sec. What is Penny’s rate of travel with respect to land?       Log On


   

Question 364475: avigation. In 3 sec, Penny walks 24 ft, to the bow (front) of a tugboat. The boat is cruising at a rate of 5 ft/sec. What is Penny’s rate of travel with respect to land?



Answer by Alan3354(69443) About Me  (Show Source):
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In 3 sec, Penny walks 24 ft, to the bow (front) of a tugboat. The boat is cruising at a rate of 5 ft/sec. What is Penny’s rate of travel with respect to land?
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24 ft/3 sec = 8 ft/sec walking
If she goes the same direction as the boat, the speeds are added.
8 + 5 = 13 ft/sec wrt the land.