Jump to content
BrainDen.com - Brain Teasers
  • 0

Chase on an endless road, continued


Rainman
 Share

Question

This is a sequel to the puzzle:

You have just managed to catch up to your friend in a race, on a road which goes on forever in both directions. Your friend is a bad loser, and full of adrenaline from the race. "Try and catch me this time", your friend says, and knocks you out cold. When you recover, you have no idea how much time has passed.

  • You move at constant speed x>1 m/s and your friend moves at constant speed 1 m/s.
  • Visibility is limited so you will not be able to see your friend from a distance.
  • You must not stray from the road, same goes for your friend.

Can you catch up to your friend? Does the answer depend on your speed x, and if so, what is the minimum x for which you can catch up? What is your strategy?

Link to comment
Share on other sites

4 answers to this question

Recommended Posts

  • 0

You just run down the path in one direction for a time.


If you haven't caught up to him, you turn and run back to your starting point, and then some more to compensate for the distance you may have lost, and then some more to see if you can catch up to him.
If you still haven't caught up to him, you turn and run back to where you left off in the opposite direction, compensate for lost distance, and then run for a little bit longer.
You repeat this until you've caught up to him.

Demo
You run down the path in one direction for T seconds.
If you haven't caught up to your friend, you turn and run in the other direction for T seconds to return to the starting point, and then for 2T/(x-1) more seconds to compensate for lost distance, and then for T more seconds to see if you can catch up to him.
If you still haven't caught up to your friend, you turn and run again for 2T+2T/(x-1) seconds back to where you left off, and then for 2(2T+2T(x-1))/(x-1) seconds to compensate for lost distance, and then T more seconds to see if you can catch up to him.

And then repeat, turning and running for T(n) seconds if you have turned for the nth time.

T(n+1) = T(n)*(x+1)/(x-1) + T(0)

Edited by gavinksong
Link to comment
Share on other sites

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

Guest
Answer this question...

×   Pasted as rich text.   Paste as plain text instead

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

Loading...
 Share

  • Recently Browsing   0 members

    • No registered users viewing this page.
×
×
  • Create New...