SPEED question

[XIPP]

Relations between South and North Korea have been tense lately. In this backdrop, a North Korean troop 5 meters long starts marching. A soldier at the end of the file steps out and starts marching forward at a higher speed. On reaching the head of the column, he immediately turns around and marches back at the same speed. As soon as he reaches the end of the file, the troop marching and it is found that the troop has moved exactly 5 metres. Which of the following could be the distance travelled by the soldier?

This question is from here and the link also possesses answer, which I can't understand. how did we get to know (5/x+y) + (5/x–y) = 5/y etc.? after calculation how we got x/y = (1 +- 2^1/2)/1? If anyone knows these answers please share.

Thanks.

[Bankelal]

Hi XIPP,

Here's how we reached this equation:  (5/x+y) + (5/x–y) = 5/y 

Let the speed of soldier and troop be x,y m/s resp.

thus, speed of soldier wrt troop(Relative speed of soldier when he is traversing from start of file to end)=x-y m/s (direction of both soldier and troop are same)

        speed of soldier wrt troop(Relative speed of soldier when he is traversing from end of file to start)=x- (-y)=x+y m/s 

(direction of both soldier and troop are opposite)

Now total total time in which soldier returned to his original position = time in which troop moved 5 m.

during both journeys of soldier he traversed 5 m. wrt troop, thus, we considered his relative speed wrt troop.

(since time=t1+t2; t1=5/(x-y); t2=5/(x+y); time of troop= 5/y)

And we finally have 

5/(x-y)+ 5/(x+y)=5/y.

On Sat, Nov 30, 2013 at 7:13 PM, XIPP [via Discussion forum] <[hidden email]> wrote:

Relations between South and North Korea have been tense lately. In this backdrop, a North Korean troop 5 meters long starts marching. A soldier at the end of the file steps out and starts marching forward at a higher speed. On reaching the head of the column, he immediately turns around and marches back at the same speed. As soon as he reaches the end of the file, the troop marching and it is found that the troop has moved exactly 5 metres. Which of the following could be the distance travelled by the soldier?

This question is from here and the link also possesses answer, which I can't understand. how did we get to know (5/x+y) + (5/x–y) = 5/y etc.? after calculation how we got x/y = (1 +- 2^1/2)/1? If anyone knows these answers please share.

Thanks.

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NAML

[Bankelal]

Sorry missed complete sol:

distance covered by soldier = x*t,

distance covered by troop= y*t = 5 m,

And we have obtained 

 x/y = (1 + 2^1/2)/1, (As speed cannot be negative)

or  (x*t)/(y*t) = (1 + 2^1/2)/1,

or Ds/5=(1 + 2^1/2)/1

 

or Ds=5*(1 + 2^1/2)

On Sat, Nov 30, 2013 at 7:13 PM, XIPP [via Discussion forum] <[hidden email]> wrote:

Relations between South and North Korea have been tense lately. In this backdrop, a North Korean troop 5 meters long starts marching. A soldier at the end of the file steps out and starts marching forward at a higher speed. On reaching the head of the column, he immediately turns around and marches back at the same speed. As soon as he reaches the end of the file, the troop marching and it is found that the troop has moved exactly 5 metres. Which of the following could be the distance travelled by the soldier?

This question is from here and the link also possesses answer, which I can't understand. how did we get to know (5/x+y) + (5/x–y) = 5/y etc.? after calculation how we got x/y = (1 +- 2^1/2)/1? If anyone knows these answers please share.

Thanks.

---

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NAML

[XIPP]

Thank you sir. How did you derive x/y ? I am stuck here:
see this  and I can't get  x/y = (1 +- 2^1/2)/1. Where am I wrong?

[Bankelal]

Hi XIPP,

My name is Aditya! (Well i prefer Adi :) )

For our earlier eqn:

5/(x-y)+ 5/(x+y)=5/y.

Take 5/x common from both sides, and we have

1/(1 - y/x) + 1/(1 + y/x) = 1/(y/x)

Take y/x = z.(for simplification purpose only)

=> 1/(1-z) + 1/(1+z) = 1/z   

=> 2/(1-z^2)=1/z

or z^2 -2z -1 = 0

On solving z, 

z = (2 +-8^(1/2))/2 = (1+-2^1/2)/1 =x/y

And that's last missing piece of our puzzle!

On Sun, Dec 1, 2013 at 10:57 AM, XIPP [via Discussion forum] <[hidden email]> wrote:

Thank you sir. How did you derive x/y ? I am stuck here:
see this  and I can't get  x/y = (1 +- 2^1/2)/1. Where am I wrong?

---

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NAML

[XIPP]

Thank you so much.

By the way, are you preparing for DSE/ISI masters in Econ., or else?

[Bankelal]

No need to mention that since we both are from same troop marching towards DSE/lSI entrance.

On Mon, Dec 2, 2013 at 12:54 PM, XIPP [via Discussion forum] <[hidden email]> wrote:

Thank you so much.

By the way, are you preparing for DSE/ISI masters in Econ., or else?

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NAML