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DIsTance problem

  1. May 28, 2008 #1
    I waNT to know if my method works. lets say you have the red car at point A that goes 12 miles per hour and you have the blue car at point B that 8 miles per hour.the distance between them is 6 miles. if the blue and the red car are in motion on the same plane of course, how long will it take the red car to catch up to the blue care?,...

    Now i havent done these distance problems for a while and i just decided to see if i can approach it with a crude or unconventional way of solving the problem. so i want to know if my method is on crack or another way to say it, is it plain insane. thank u

    so here is what i did...i imediately moved point A to point B then i minus the distance that was between them.12-6=6 the 8 is over by 2..so i think the car has not caught up yet. then i....12/60=.2 of a mile per min(60)...2x5=1....1.mile= 5min..16x5=80min.i get the 16 from the original 8milesx2(2sets)so 80min=1.20min..
    solution 1.20min..

    so what do u think?lol
     
  2. jcsd
  3. May 29, 2008 #2

    HallsofIvy

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    I hope you mean "in the same direction". If they are only going "on the same plane", they might never meet!

    You think the car has not caught up yet? the red car has only moved to where the blue car was orginally, and of course the blue car has moved. I can make no sense out of "12- 6= 6 the 8 is over by 2".

    But the red car can't "immediately" move to point B. Since B is 6 miles from A and the red car is going 12 miles per hour, it will take the red car 12/6= 1/2 hour= 30 minutes just to get to B- and the blue car is now 4 miles ahead of that. That correct answer is larger than 30 minutes so, no, "1.20 min" is certainly not the solution! (Seriously, did "1.20 minutes" to catch up to a car 6 miles ahead even seem reasonable to you?)

    You could use that method if you are careful about numbers (and don't just do random things like "12- 6= 6 the 8 is over by 2"). It takes the red car 1/2 hour to go the 6 miles from A to B- and in that 1/2 hour, the blue car moves 4 miles. Now it take the red car 4/12= 1/3 hour to go that 4 miles- and the blue car has moved another 8/3 miles ahead. It takes the red car (8/3)/12= 2/9 hour to move that 8/3 miles and the blue car has moved another 8*2/9= 16/9 miles ahead. Keep going that way and you can add the times required:
    1/2+ 1/3+ 2/9+ ...

    Much better is this: the red car is moving 12- 8= 4 mph faster than the blue car. At that (relative) rate, it will take the red car 6 mi/4 mph = 3/2= 1.5 hours to catch the blue car.
     
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