SOLUTION: A boardwalk is parallel to and 210 ft inland from a straight shoreline. A sandy beach lies between the boardwalk and the shoreline. A man is standing on the boardwalk, exactly 750

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Question 486052: A boardwalk is parallel to and 210 ft inland from a straight shoreline. A sandy beach lies between the boardwalk and the shoreline. A man is standing on the boardwalk, exactly 750 ft across the sand from his beach umbrella, which is right at the shoreline. The man walks 4ft/s on the boardwalk and 2 ft/s on the sand. How far should he walk on the boardwalk before veering off onto the sand if he wishes to reach his umbrella in exactly 4 min 45 s?
Answer by ankor@dixie-net.com(22740) (Show Source):
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: A boardwalk is parallel to and 210 ft inland from a straight shoreline.
A sandy beach lies between the boardwalk and the shoreline.
A man is standing on the boardwalk, exactly 750 ft across the sand from his beach umbrella, which is right at the shoreline.
The man walks 4ft/s on the boardwalk and 2 ft/s on the sand.
How far should he walk on the boardwalk before veering off onto the sand if he wishes to reach his umbrella in exactly 4 min 45 s?
:
Change 4 min 45 sec to 285 sec
:
Find the distance from the man to a point (p) on the boardwalk directly opposite the umbrella.
:
p =
p = 720 ft
;
Let x = the dist the man will walk on the boardwalk, before onto the sand
The distance walked on the sand will be the hypotenuse of a triangle with the legs of 210' and (720-x) or
:
Write the time equation
b.w. time + sand walk time = 285 sec
+ = 285
:
Get rid of the denominators, mult by 4, results
x + = 1140
:
After a lot math obtained a quadratic equation of:
3s^2 - 3480x + 950400 = 0
Two solutions
x = 720
x = 440 ft. the reasonable solution is the distance he walked on the boardwalk