Asked by John
d=0.767. There are two parallel sidewalks that are separated by the park that is 600 feet wide. You are standing on the south sidewalk facing east. On the north sidewalk, 2000 feet ahead of you, is a bus stop. You can walk on the sidewalk at 5.5+d feet per sec. You can walk through the park at 3.5+d feet per sec. What is the fastest you can walk to the bus stop?
Answers
Answered by
Steve
If you walk directly to a point x feet from the bus stop, your walking distance is
park: √(600^2+(2000-x)^2)
sidewalk: x
so, your time taken (distance/speed) is
t(x,d) = √(600^2+(2000-x)^2)/(3.5+d) + x/(5.5+d)
t(x) = √(600^2+(2000-x)^2)/4.267 + x/7.267
so, just find dt/dx and set it to zero.
Then evaluate t for that value of x.
park: √(600^2+(2000-x)^2)
sidewalk: x
so, your time taken (distance/speed) is
t(x,d) = √(600^2+(2000-x)^2)/(3.5+d) + x/(5.5+d)
t(x) = √(600^2+(2000-x)^2)/4.267 + x/7.267
so, just find dt/dx and set it to zero.
Then evaluate t for that value of x.
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