Today I read an interesting thread on vozforums which discusses about a problem in Math and Physics. I would like to copy it into my blog to demonstrate the power of math and physics which helps us solving complex problems in our daily lives. There is no source code in this blog. Just math and physics. Are you ready for brainstorming? ^^.
– There are 200 peoples standing at point A.
– Walking from point A to point B takes 25 minutes.
– Driving with a car (4 seats) from point A to point B takes 5 minutes.
– Peoples can walk from point A to point B.
Questions: Which is the shortest time to take all of peoples from point A to point B? The time for getting in/out of car will be skipped.
… thinking …
… thinking …
… thinking …
– It’s time to break your brain. ^^. I just write out the idea, you must understand it. And at least you must know this formel. Distance = Time*Velocity (of course it’s only correct for moving with constant velocity)
– Call V is the velocity of a people then 5V is the velocity of the car and nominization for the distance from A to B is 25V.
– First, let the car take 4 peoples and the others walk to point B.
– After a time of T, 4 peoples get out. That means they have traveled a distance of 5VT and the others (196 peoples) have traveled a distance of VT. Now we have two groups and the distance between them is always 4VT.
– After let 4 peoples out, the car comes back to take the other 4 peoples from second group. Time to come back to the group is (5TV – TV)/(5V + 1V) = 4TV/6V = 2/3*T.
– Then the car drives to the first group (4 peoples), the timespan for catching first 4 peoples is 4VT/(5V – 1V) = T.
– So the time for coming back and catching the first group is T + 2/3*T = 5/3*T. We have 196 peoples and a car can carry 4 peoples therefore we need 49 times of driving back and driving forward so that the second group can catch the first one. Therefore after a time span of T + 49*5/3*T the second group will catch the first one and 200 peoples will be at same position and the distance they have traveled 5VT + 49*5/3*VT
– From the formel of distance above, we can see that the distance increase proportionally with the time. Therefore the shortest time unit T from A to B should be the variable of this equation 5TV + 49*5/3*T*V = 25V –> T = 15/52 Mins
– Replace this time unit T into the formel for completely shortest time T + 49*5/3*T we can calculate out that the shortest time from A to B should be 23,84 minutes. That’s fininshed. Are you still alive? ^^.