Another Fibonacci-related race? Seriously?

Calculus Level 5

The city of Mathville is once again having a race. This time, however, competitiors will start at the city centre and will run 1km east, 1km north, 1 2 \dfrac{1}{2} km east, 1 3 \dfrac{1}{3} km north, 1 5 \dfrac{1}{5} km east 1 8 \dfrac{1}{8} km north ad infinitum.

If the distance displaced by the competitors can be represented as k k metres , what is ⌊ k ⌋ \lfloor{k}\rfloor ?


The answer is 2384.

Sharky Kesa by Sharky Kesa , 19, United Kingdom

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1 solution

Aaaaa Bbbbb
May 10, 2015

d i s t a n c e = D i s t a n c e E 2 + D i s t a n c e N 2 = 2384 distance = \sqrt{Distance_{E}^2+Distance_{N}^2}=\boxed{2384}

0 Helpful 0 Interesting 0 Brilliant 0 Confused

How do you determine the distance of each?

Sharky Kesa - 6 years, 1 month ago

wow!!!!but would like if some one explained a bit.....

rajdeep brahma - 3 years ago

Please elaborate your answer.

Murugesh M - 6 years, 1 month ago

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