Minus' Solution to

Love Bugs
Posted August 18 - 24, 1997

Redbug has 34 cm to travel.
Bluebug has 16 cm to travel.
So Bluebug's speed is 16/34 of Redbug, so Bluebug travels at 16N/34 cm/second.

Redbug travels 34 cm in less than 5 seconds. She travels at N cm/second. Thus, N = 6.8 cm/second, or rounded to the nearest centimeter, 7 cm/second.

Although initially confused by the question, Dean Vendramin [nyr@sk.sympatico.ca] still came through with this week's starter dish. We may have to add Dean to the International Cooking-for-Sharks Team. Note: Cooking-FOR-sharks, NOT Cooking-OF-sharks!

This was confusing, but let's see if I understand things right.

By remembering that all sides of a cube are equal and using the info provided, I came up with the following :

Now the Red Bug traveled 34 cm (4 + 16 + 14). The Blue Bug traveled 16 cm (14 +2).

Now an Interger is from the set of numbers ...-3,-2,-1,0,1,2,3...

Thus if the Red Bug went N cm/sec, the Blue Bug went at a rate of N/2 cm/sec (Rationale - Using the distance traveled info I came up with this result 34/16 ~ 2.13 or 2. Using the definition of an interger and the distance travel, the Blue Bug had to be going at about the half the speed of the Red Bug (maybe it had a sore leg or four))

As for the minimum speed of the Red Bug one must divide 34cm by 5 sec and get the rate of 6.8 cm/sec.

This question was 'bugging' me, I hope I understood your question right!

Bluebug, the least devoted of the two since it only has to travel 16 cm, has a speed of (16/34)N cm/s or (.47)N cm/s while the Redbug was speeding along at (1)N cm/s.

Since Redbug, which was one speedy and devoted bug, takes less than 5 seconds to make to the rendezvous point at Point X, she could not be moving at a speed less than or equal to 6.8 cm/s while Bluebug trudged along at only 3.2 cm/s

Redbug however must have been enormously love crazed as she did not notice that she could have made it to the rendezvous by traveling in a straight line 18.9 cm long in less than 3 seconds, a short time at a speed of 6.8 cm/s.

The bluebug could not have taken a similar shortcut unless the cube were floating in space (in which case he would have had to have had a pretty strong grip to stay attached). If this were the case though, Bluebug could have traveled a slightly shorter distance of 14.1 cm in only 4.4 s.

All in all in seems to me that these two bugs must be seriously in love if they are so dazed that they're crawling all over a seemingly boring cube just to meet their date when they could have taken much faster and more direct routes and saved even more time at the end of their date for the good night kiss which will most likely leave them both so dazed that they travel over almost every single centimeter of the cube before they both make it back home.

Step One: The first thing we did was find out how far away the bugs were from point X. Red bug was 34 cm away from point X and Blue bug was 16 cm away from point X.

Step Two: Next we found out how many times faster Red bug travelled compared to Blue bug by dividing the Red bugs distance (34 cm) by the Blue bugs distance (16 cm). The answer we got is 2.125.

Step Three: We had to try and find out what N was equal to. To do this we divided the Red bugs distance (34 cm) by 5 cm per second. The answer we got is 6.8 cm per second.

Step Four: The answer is approximately 6.8 cm/sec because the time the Red bug took was less than 5 seconds.

To Minus,

If Redbug travelled at N cm/second and n was an integer, BlueBug would travel at N/2.125 cm/second. And if RedBug travelled the distance at under 5 seconds her minimum speed would be 6.8 cm/second!