(3) A boardwalk is parallel to and 210 feet inland from a straight shore line. There is a sandy beach between the boardwalk and the shore line. A man is standing on the boardwalk, exactly 750 feet across the sand from his beach umbrella, which is right at the water-line. The man walks 4 feet/sec on the boardwalk and 2 feet/sec on the sand. How far should he go on the boardwalk before veering off into the sand if he wishes to reach his umbrella in exactly 4 minutes and 45 seconds?
SOLUTION:
Note 4 min, 45 sec = 285 sec
Let x = time travelled on the boardwalk
y = time travelled on the sand
l = length of boardwalk
7502 = 2102 + l2
562500 = 44100 + l2
l2 = 518400
l = 720 feet
x + y = 285 sec ==> y = 285 - x
(2y)2 = 2102 + (720 - 4x)2
4y2 = 44100 + 518400 - 57600x + 16x2
y2 = 140625 - 1440x + 4x2
(285 - x)2 = 140625 - 1440x + 4x2
81225 - 570x + x2 = 140625 - 1440x + 4x2
3x2 - 870x - 59400 = 0
x2 - 290x + 19800 = 0
using the quadratic formula, we get x = 180 sec or x = 110 sec
If x = 180 sec then y = 285 - x = 285 - 180 = 105 sec
Then the distance walked on the boardwalk = 4x = (4 ft/sec)(180 sec) = 720 feet
and the distance walked on the sand = 2y = (2 ft/sec)(105 sec) = 210 feet
(In this case, he walks the entire length of the boardwalk before stepping onto the sand.)
If x = 110 sec then y = 285 - x = 285 - 110 = 175 sec
Then the distance walked on the boardwalk = 4x = (4 ft/sec)(110 sec) = 440 feet
and the distance walked on the sand = 2y = (2 ft/sec)(175 sec) = 350 feet