Four persons K, L, M, and N are initially at the four corners of a square of side 'd'. Each person now moves with a uniform speed of 'v' in such a way that K always moves directly towards L, L directly towards M, M directly towards N and N directly towards K. At what time will the four persons meet?
6 Answers
This is an excellent question!
Intuitively, it may seem like the four people will never meet, because the person they are moving towards is always moving away from them. The key is that while the person they are moving towards is moving, they are not actually moving away from the person chasing them. In fact, the person moving away will always move perpendicular to the line between them and the person chasing them, so they never get further away from their pursuer. On the other hand, the person chasing them is always moving towards them at speed v. Therefore, the distance between these two people decreases at speed v, and the time it takes for them to meet is [math]\frac{d}{v}.[/math].
For the more mathematically inclined, this may not seem like much of a proof, so here is a way to show this with a little bit of calculus:
First, let's look at what would happen if all of the people were to move in their initial direction (set only by the initial positions of K,L,M,N) for a time [math]t = \Delta t[/math].
![]()
It is not hard to show that the red shape must also be a square. This is not surprising but important for our calculation. Let's take the square at an arbitrary time t, with side lengths [math]x(t)[/math]. Notice that we have a right triangle with side lengths [math]x(t)-v\Delta t[/math], [math]v\Delta t[/math], and [math]x(t+\Delta t)[/math]. Using the pythagorean theorem, we obtain the equation:
[math]x(t+\Delta t)^2 = (x(t)-v\Delta t)^2 + (v\Delta t)^2[/math]
Expanding the squared equations we get:
[math]x(t+\Delta t)^2 = x(t)^2-2x(t)v\Delta t + 2(v\Delta t)^2[/math]
Now, let [math]\alpha(t) = x(t)^2[/math]:
[math]\alpha(t+\Delta t) = \alpha(t)-2\sqrt{\alpha(t)}v\Delta t + 2(v\Delta t)^2[/math]
More algebra, move [math]\alpha(t)[/math] over, divide by [math]\Delta t[/math]
[math]\frac{\alpha(t+\Delta t)-\alpha(t)}{\Delta t} = -2\sqrt{\alpha(t)}v+ 2v^2\Delta t[/math]
Now notice if we let [math]\Delta t[/math] go to 0 (which represents continuous motion), the left side is just a derivative:
[math]\frac{d\alpha(t)}{dt} = -2\sqrt{\alpha(t)}v[/math]
Now this is just a differential equation that is easy to solve with separation of variables:
[math]\frac{d\alpha}{\sqrt{\alpha(t)}} = -2v dt[/math]
Note, our integral over [math]\alpha(t)[/math] will go from [math]d^2[/math] to [math]x(t)^2[/math].
[math]\int_{d^2}^{x(t)^2}\frac{d\alpha(t)}{\sqrt{\alpha(t)}} = \int_{0}^{t}-2v dt [/math]
[math]\big(2\sqrt{\alpha(t)}\big)_{d^2}^{x(t)^2} = -2vt[/math]
[math]|x|-|d| = -vt[/math]
Now we can solve for t when [math]x=0[/math] (d is assumed to be positive, so we drop the absolute value):
[math]t=\frac{d}{v}[/math]
Which is what we got with the first method!
Intuitively, it may seem like the four people will never meet, because the person they are moving towards is always moving away from them. The key is that while the person they are moving towards is moving, they are not actually moving away from the person chasing them. In fact, the person moving away will always move perpendicular to the line between them and the person chasing them, so they never get further away from their pursuer. On the other hand, the person chasing them is always moving towards them at speed v. Therefore, the distance between these two people decreases at speed v, and the time it takes for them to meet is [math]\frac{d}{v}.[/math].
For the more mathematically inclined, this may not seem like much of a proof, so here is a way to show this with a little bit of calculus:
First, let's look at what would happen if all of the people were to move in their initial direction (set only by the initial positions of K,L,M,N) for a time [math]t = \Delta t[/math].
It is not hard to show that the red shape must also be a square. This is not surprising but important for our calculation. Let's take the square at an arbitrary time t, with side lengths [math]x(t)[/math]. Notice that we have a right triangle with side lengths [math]x(t)-v\Delta t[/math], [math]v\Delta t[/math], and [math]x(t+\Delta t)[/math]. Using the pythagorean theorem, we obtain the equation:
[math]x(t+\Delta t)^2 = (x(t)-v\Delta t)^2 + (v\Delta t)^2[/math]
Expanding the squared equations we get:
[math]x(t+\Delta t)^2 = x(t)^2-2x(t)v\Delta t + 2(v\Delta t)^2[/math]
Now, let [math]\alpha(t) = x(t)^2[/math]:
[math]\alpha(t+\Delta t) = \alpha(t)-2\sqrt{\alpha(t)}v\Delta t + 2(v\Delta t)^2[/math]
More algebra, move [math]\alpha(t)[/math] over, divide by [math]\Delta t[/math]
[math]\frac{\alpha(t+\Delta t)-\alpha(t)}{\Delta t} = -2\sqrt{\alpha(t)}v+ 2v^2\Delta t[/math]
Now notice if we let [math]\Delta t[/math] go to 0 (which represents continuous motion), the left side is just a derivative:
[math]\frac{d\alpha(t)}{dt} = -2\sqrt{\alpha(t)}v[/math]
Now this is just a differential equation that is easy to solve with separation of variables:
[math]\frac{d\alpha}{\sqrt{\alpha(t)}} = -2v dt[/math]
Note, our integral over [math]\alpha(t)[/math] will go from [math]d^2[/math] to [math]x(t)^2[/math].
[math]\int_{d^2}^{x(t)^2}\frac{d\alpha(t)}{\sqrt{\alpha(t)}} = \int_{0}^{t}-2v dt [/math]
[math]\big(2\sqrt{\alpha(t)}\big)_{d^2}^{x(t)^2} = -2vt[/math]
[math]|x|-|d| = -vt[/math]
Now we can solve for t when [math]x=0[/math] (d is assumed to be positive, so we drop the absolute value):
[math]t=\frac{d}{v}[/math]
Which is what we got with the first method!
This question can be answered in quite elementary terms, there is basically no need in any advanced math.
Since by symmetry at every moment each person is in a corner of some square, and they meet in the center of the square, you can merely decompose the velocity vector of any of them (pointing out along the side of some square) in two components (one of them is the projection on the diagonal of the square, and the other is its orthogonal complement).
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Then each of persons should cover half of length of the diagonal of square, i.e. [math]\frac{d}{\sqrt{2}}[/math] by speed [math]\frac{v}{\sqrt{2}}[/math]. The elapsed time is then obviously [math]\frac{\frac{d}{\sqrt{2}}}{\frac{v}{\sqrt{2}}} = \frac{d}{v}. [/math]
Roughly speaking, you can just ignore the part of velocity vector which is responsible for rotation of the picture thus taking into account only the projection of the velocity vector onto the diagonal of the square.
Since by symmetry at every moment each person is in a corner of some square, and they meet in the center of the square, you can merely decompose the velocity vector of any of them (pointing out along the side of some square) in two components (one of them is the projection on the diagonal of the square, and the other is its orthogonal complement).
Then each of persons should cover half of length of the diagonal of square, i.e. [math]\frac{d}{\sqrt{2}}[/math] by speed [math]\frac{v}{\sqrt{2}}[/math]. The elapsed time is then obviously [math]\frac{\frac{d}{\sqrt{2}}}{\frac{v}{\sqrt{2}}} = \frac{d}{v}. [/math]
Roughly speaking, you can just ignore the part of velocity vector which is responsible for rotation of the picture thus taking into account only the projection of the velocity vector onto the diagonal of the square.
Well answered by Kyle Lund, but for a TL/DR answer:
Say you are K. You are the centre of your universe. At any point in time, N is moving towards you with a speed "v". This translates to N's relative velocity to you being "v"- as N is always moving towards you and the relative directional vector is preserved (since N's velocity is perpendicular to yours). Hence you and N meet at time T + d/v. Having successfully stalked L, and so on, you all meet at the same time.
This question was easy due to the sameness of speed- as Kyle Lund shows in his calculus method, you keep getting that square, maintaining perpendicularity of velocity. Were the speeds different, the resultant relative velocity would have to be factored in.
Say you are K. You are the centre of your universe. At any point in time, N is moving towards you with a speed "v". This translates to N's relative velocity to you being "v"- as N is always moving towards you and the relative directional vector is preserved (since N's velocity is perpendicular to yours). Hence you and N meet at time T + d/v. Having successfully stalked L, and so on, you all meet at the same time.
This question was easy due to the sameness of speed- as Kyle Lund shows in his calculus method, you keep getting that square, maintaining perpendicularity of velocity. Were the speeds different, the resultant relative velocity would have to be factored in.
CAN ANYONE COUNTER THIS SOLUTION :-
Consider a small time interval ∆t at the beginning of journey . In this time interval,all of them travel a distance v∆t in the direction of person they are facing.(See figure) Through this process, all persons travel along the four different quadrants of the circles with one end at their starting position and another at the center of square O.The radius of each arc is d/2 and length of arc is 1/4(2π (d/2)> = πd/4. Thus the time taken by each person moving with speed v to reach O along he quadrant arc is t= πd/(4v) >> Different answer from all other answers available out there.
