Four persons K, L, M, and N are initially at the four corners of a square of side d. Each persons now moves with a uniform speed v in such a way that K always moves directly towards L, L directly towards M, M directly towards N and N directly toward K. The four persons will meet at a time ..........??
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This is an excellent question!
Note - The diagram is for 1st Method...
1st 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).
Then each of persons should cover half of length of the diagonal of square, i.e. d/√2 by speed v/√2. The elapsed time is then obviously d/√2/v/√2=d/v.
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.
2nd 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 dv.
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 t=Δt.
It is not hard to show that the small 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 x(t). Notice that we have a right triangle with side lengths x(t)−vΔt, vΔt, and x(t+Δt). Using the pythagorean theorem, we obtain the equation:
x(t+Δt)2=(x(t)−vΔt)^2+(vΔt)^2
Expanding the squared equations we get:
x(t+Δt)^2=x(t)^2−2x(t)vΔt+2(vΔt)^2
Now, let α(t)=x(t)^2:
α(t+Δt)=α(t)−2α(t)√vΔt+2(vΔt)^2
More algebra, move α(t) over, divide by Δt
α(t+Δt)−α(t)/Δt=−2α(t)√v+2v^2Δt
Now notice if we let Δt go to 0 (which represents continuous motion), the left side is just a derivative:
dα(t)/dt=−2α(t)√v
Now this is just a differential equation that is easy to solve with separation of variables:
dα/√α(t)=−2vdt
Note, our integral over α(t) will go from d^2to x(t)^2.
∫d^2to x(t)^2[dα(t)/√α(t)]=∫0tot(−2vdt)
(2√α(t))x(t)^2/d2=−2vt
|x|−|d|=−vt
Now we can solve for t when x=0 (d is assumed to be positive, so we drop the absolute value):
t=d/v
Which is what we got with the first method!
Note - The diagram is for 1st Method...
1st 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).
Then each of persons should cover half of length of the diagonal of square, i.e. d/√2 by speed v/√2. The elapsed time is then obviously d/√2/v/√2=d/v.
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.
2nd 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 dv.
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 t=Δt.
It is not hard to show that the small 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 x(t). Notice that we have a right triangle with side lengths x(t)−vΔt, vΔt, and x(t+Δt). Using the pythagorean theorem, we obtain the equation:
x(t+Δt)2=(x(t)−vΔt)^2+(vΔt)^2
Expanding the squared equations we get:
x(t+Δt)^2=x(t)^2−2x(t)vΔt+2(vΔt)^2
Now, let α(t)=x(t)^2:
α(t+Δt)=α(t)−2α(t)√vΔt+2(vΔt)^2
More algebra, move α(t) over, divide by Δt
α(t+Δt)−α(t)/Δt=−2α(t)√v+2v^2Δt
Now notice if we let Δt go to 0 (which represents continuous motion), the left side is just a derivative:
dα(t)/dt=−2α(t)√v
Now this is just a differential equation that is easy to solve with separation of variables:
dα/√α(t)=−2vdt
Note, our integral over α(t) will go from d^2to x(t)^2.
∫d^2to x(t)^2[dα(t)/√α(t)]=∫0tot(−2vdt)
(2√α(t))x(t)^2/d2=−2vt
|x|−|d|=−vt
Now we can solve for t when x=0 (d is assumed to be positive, so we drop the absolute value):
t=d/v
Which is what we got with the first method!
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