For debjit

Discussion in 'Math' started by studiot, Jul 4, 2012.

  1. studiot

    Thread Starter AAC Fanatic!

    Nov 9, 2007
    5,005
    513
    Here is a nice little problem that was posted on another forum.

    enjoy


     
  2. WBahn

    Moderator

    Mar 31, 2012
    17,715
    4,787
    Just as the clock struck a new hour -- assuming the cities are in the same time zone.


    How fast was person B walking if the bridge is 1/4 mile wide?
     
  3. studiot

    Thread Starter AAC Fanatic!

    Nov 9, 2007
    5,005
    513
    Why do you want to know that? Person B walks just fast enough to cross the bridge in 5 minutes. That is calculable from the information given.
     
  4. WBahn

    Moderator

    Mar 31, 2012
    17,715
    4,787
    I already know it. I was throwing it out as a confirming question.

    I don't want to just give the answer and spoil other people's fun, particularly if you were posing it specifically to one person as a challenge of some kind.

    So by answering my question, and coming up with a nice round exact answer, it provides strong evidence that I actually did solve the problem. Of course, my answer involving the clock bells also does this -- the two combined make it pretty much a sure thing.
     
  5. studiot

    Thread Starter AAC Fanatic!

    Nov 9, 2007
    5,005
    513
    Yes I think you have it.

    I understand the question was posed at a conference for high school physics teachers.
     
  6. WBahn

    Moderator

    Mar 31, 2012
    17,715
    4,787
    It's a nice question, with a lot of interesting variants. Not too hard to set up correctly and not too hard to solve once you do. But fairly easy to mess up if you don't think through it with some care.
     
  7. MrChips

    Moderator

    Oct 2, 2009
    12,413
    3,353
    Very intriguing. I don't know if there is quick solution but it took me awhile (about an hour) doing it graphically.

    (The length of the bridge is irrelevant).

    Here is another question: At what time do they meet?
     
    Last edited: Jul 4, 2012
  8. WBahn

    Moderator

    Mar 31, 2012
    17,715
    4,787
    2 min 44 & 4/9 seconds after they reached the bridge.

    I solved in about 10 minutes algebraically, but part of that is probably because I caught what I thought was an inconsistency in the problem, namely that one person left the bridge a minute after the other person did, but yet it stated that they were moving at a constant speed for the entire trip. So it very quickly donned on me that the difference had to be related to how long it took each to cross the bridge. So I first solved for how long that was (which is two lines of algebra). With that, the two simultaneous equations can be written down pretty much by inspection and then five lines of algebra and then plugging about eight numbers into a calculator.
     
  9. debjit625

    Well-Known Member

    Apr 17, 2010
    790
    186
    Sorry for late response as last weak I was very busy with my own studies.
    The question is very interesting and thanks to studiot .Obviously it was for me as I think I am only the Debjit in AAC,so I am going to answer it I guess. I will try to explain it very well as many students may have problem solving this.

    Here is the picture ,which describes almost every thing...
    [​IMG]

    As you can look at the picture what I have did, their is two town 'X' and 'Z' in between their is a bridge, now the total time taken by person 'A' to walk from town 'X' to 'Z' is t_A which is given its from 10:18 am to 1:30 pm i.e.. 192 min,the total time taken by person 'B' to walk from town 'Z' to 'X' is t_B ' which is also given 9:00 am to 11:40 am i.e.. 160 min.

     t_A = 192

     t_B = 160

    Now as speed/distance and time are involved we could write it like this...
    Now both covers the same distance 'D' at their respective speed S_A is the speed for person 'A' and S_B is the speed for person 'B'.

    S_A = {D \over t_A}

    S_A = {D \over 192}

    D = S_A * 192
    And

    S_B = {D \over t_B}

    S_B = {D \over 160}

    D = S_B * 160

    or

    S_B * 160 =  S_A * 192

    {S_B \over S_A}=  {160 \over 192} = {6 \over 5}

    The ratio describes that the speed of person 'B' is 6/5 times faster than the speed of person 'A'.
    As in question and by looking at the picture it says person 'B' took t_q time to cross the bridge and person 'A' took t_y time to cross the bridge or its also given person 'A' took 1 min extra to cross the bridge so we can write

     t_y = t_q + 1

    Now in the whole journey both the person maintained same speed so at the bridge also they maintained the same speed,let be the distance of the bridge be D_B ',then we can write

    S_A = {D_B \over t_q + 1}

    D_B = S_A*(t_q + 1)

    and
    S_B = {D_B \over t_q}

    D_B = S_B*t_q
    or

    {S_B\over S_A} = {(t_q + 1) \over t_q}

    {6\over 5} = {(t_q + 1) \over t_q}

    6t_q = 5t_q + 5

    t_q = 5 min

    then

    t_y = 6 min

    Now in the picture its clear the total time (which is given) taken by each person is divided into three parts,for person 'A' its  t_x , t_y , t_z and for person 'B' its  t_p , t_q , t_r , from these information we can write some equations...
    For person 'A'

     t_x + t_y + t_z = 192

     t_x + t_z = 186

    For person 'B'

     t_p + t_q + t_r = 160

     t_p + t_r = 155 ------ Eq 1

    Now above both the equation describes the same thing so its means we have our first equation but we need to find some other equation form the same event.

    Well we can say this as per the question, that the real clock time taken by person 'A' to walk from town 'X' to the one end of the bridge is the sum of the time when he started (10:18 am) and  t_x ,and it is equal to the real clock time taken by person 'B' to walk from town 'Z' to the other end of the bridge is also the sum of the time when he started (9:00 am) and  t_r .We can write it like this...

     10:18 am + t_x = 9:00 am + t_r

    Converting the real clock time to min for calculation

     618 + t_x = 540 + t_r

     t_x - t_r= -78

    We can write t_x as

     {6 \over 5} t_p - t_r = -78

     6t_p - 5t_r = -390 ----- Eq 2

    So we got 2 equations ,now we can solve it

     t_p + t_r = 155 ------ Eq 1

     6t_p - 5t_r = -390 ----- Eq 2
    Multiply equation 1 with 6

     6t_p + 6t_r = 930 ------ Eq 1

     6t_p - 5t_r = -390 ----- Eq 2

    Subtracting eq 2 from eq 1
     11t_r = 1320

     t_r = 120 min

    And
     t_x - t_r= -78

     t_x = 42 min

    So person B started at 9:00 am and after 120 min he reached the bridge i.e.. 11:00 am and person A started at 10:18 am and 42 min he reached at bridge i.e.. 11:00 am.

    Good Luck
     
    • dia.jpg
      dia.jpg
      File size:
      17.6 KB
      Views:
      53
  10. debjit625

    Well-Known Member

    Apr 17, 2010
    790
    186
    Sorry its not the correct answer... it will be 11:00 am
     
  11. debjit625

    Well-Known Member

    Apr 17, 2010
    790
    186
    By the way studiot where you got this question from...
     
  12. WBahn

    Moderator

    Mar 31, 2012
    17,715
    4,787
    Different question. I got the 11:00am, but didn't want to post it and spoil your fun outright. The indication that I actually did get the right answer is that I said that they got there just as the clock struck a new hour. The only new hour at which they were both on the road was 11:00am.

    The answer in your quote was for a follow-up question by MrChips, namely when do they actually meet. So this answer is relative to the undisclosed time at which they both arrive at the bridge.

    Here's how I did it:

    Preliminaries:

    The total distance traveled by each is D.
    Person A takes a total of 3h 12m = 192min
    Person B takes a total of 2h 40m = 160min
    Person A leaves their side 78min after Person B leaves theirs.

    Thus
    Person A is walking at a speed of Va = D/192min
    Person B is walling at a speed of Vb = D/160min

    If the bridge has a length B, then the time it takes each person to walk this is:
    Person A: Ta = B/Va
    Person B: Tb = B/Vb

    Thus, Ta = Tb(Vb/Va) = Tb((D/160min)/(D/192min) = (6/5)Tb

    It is given that Person A takes 1min longer, hence:

    Tb = Ta - 1min = (6/5)Tb - 1min
    Tb = 5min

    Now for the main event:

    The distance away from City X is X and, for each person this is given by:

    Xa(t) = Va*(t-78min)
    Xb(t) = D-Vb*t

    where 't' is the time since 9:00am.

    If we were looking for the time at which they meet, we would be looking for the time such that Xa(t) = Xb(t). However, because we are looking for the time at which they get to their respective sides of the bridge, we are looking for t such that Xa(t) puts Person A at their side of the bridge and Xa(t+5min) puts Person B at that same side (Person A's side).

    Thus
    Xa(t) = Xb(t+5min)
    (D/192min)*(t-78min) = D-(D/160min)*(t+5min)
    (1/192min)*(t-78min) = 1-(1/160min)*(t+5min)
    t-78min = 192min - (192/160min)*(t+5min)
    t = 270min - (6/5)*(t+5min) = 264min - (6/5)t
    (11/5)t = 120min

    Thus, the time at which they both arrive at their ends of the bridge is

    T = 9:00am + 120min = 11:00am
     
  13. debjit625

    Well-Known Member

    Apr 17, 2010
    790
    186
    My bad I didn't noticed...

    Well the answer is already given,but its not explained I will do that...
    The process will be the same...here is another diagram
    [​IMG]
    Let 'M' be the place where they meet or cross each other,and t_x be the time taken by person 'A' to walk from town 'X' to 'M' and t_y be the time taken by person 'A' to walk from 'M' to town 'Z'.
    On the other hand t_q be the time taken by person 'B' to walk from town 'Z' to 'M' and t_q be the time taken by person 'B' to walk from 'M' to town 'X'.
    Many values we already have from earlier solution...like

     t_A = 192 min

     t_B = 160 min

    and the speed of person 'B' is 6/5 times faster than the speed of person 'A'.
    So we can write

     t_x + t_y = 192 --- Eq 1.1

     t_p + t_q = 160 --- Eq 1.2

    We can say this as per the question, that the real clock time taken by person 'A' to walk from town 'X' to 'M' is the sum of the time when he started (10:18 am) and t_x,and it is equal to the real clock time taken by person 'B' to walk from town 'Z' to 'M' which is the sum of the time when he started (9:00 am) and t_q.We can write it like this...

    10:18 am + t_x = 9:00 am + t_q

    Converting the real clock time to min for calculation

    618 + t_x = 540 + t_q

    t_x - t_q = -78

    {6 \over 5}t_p - t_q = -78

    6t_p - 5t_q = -390---- Eq 1.3

    Multiply 6 with Eq 1.2

     6t_p + 6t_q = 960 --- Eq 1.2

    Subtract Eq 1.3 from Eq 1.2

     11tq = 1350

     tq = 122.{\overline{72}}

    In hr/min/sec format t_q is approx 2hrs , 02 mins and 44 sec

    So it will be around 11:02:44 am when they both meet.

    Good Luck
     
    Last edited: Jul 9, 2012
  14. studiot

    Thread Starter AAC Fanatic!

    Nov 9, 2007
    5,005
    513
    You might like to have another go?

    A leaves the bridge at 1106 and B at 1105

    Glad you liked it by the way.
     
  15. MrChips

    Moderator

    Oct 2, 2009
    12,413
    3,353
    On the original question, I did mine graphically.



    [​IMG]


    The time scale is referenced to 9:00am. Hence B starts out at time 0.
    A starts out 78 min later.

    They both travel the same distance D. The actual distance is irrelevant. So I use arbitrary units where D = 160. This simplifies the math.

    B reaches his destination 160 min later (red line).

    while A reaches his destination at time 270 min (blue line).

    B takes 160 min while A takes 192 min.

    The speed of A = D/192
    The speed of B = D/160
    The ratio of A's speed to B's speed is 5:6

    If A took 1 more minute to cross the bridge, then A took 6 minutes to cross the bridge while B took only 5 minutes.

    So what we do is make A start off 6 minutes earlier, that is we shift A's line early by 6 minutes (purple line).

    Where the red and purple lines intersect is the time when they both reach the bridge at the same time. This is solved by straight forward geometry.
     
  16. WBahn

    Moderator

    Mar 31, 2012
    17,715
    4,787
    Just a boo-boo, since the time in minutes is correct. But it does underscore the utility of always asking if the answer makes sense, something that I, too, could have done better on a couple of posts in the last few days.
     
  17. debjit625

    Well-Known Member

    Apr 17, 2010
    790
    186
    Sorry it will be 2 min rather 22min i.e.. 11:02:44 am ,I was in hurry my calculations are correct.
     
Loading...