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A cockroach is crawling along the walls inside a cubical room that has an edge length of 3 m. If the cockroach starts from the back lower left hand corner of the cube and finishes at the front upper right hand corner, what is the magnitude of the displacement of the cockroach?

Simple question, but I don't understand the answer they provide. In my mind the cockroach has a displacement of 3m. Can someone verify or point my intuition in the right direction.

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4 Answers 4

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Displacemtns are vector quantities which are determined on the basis of the final and initial position of the object. WHen you have a cube

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Let's say the insect traveled form one of the blue balls to another blue ball. Let's represent each of them by vectors from a reference origin $O$. Then the initial position vector $r_0$ and final position vector $r_1$.

Hence your displacements would be $r_1 - r_0$. which in magnitude will be $|r_1-r_0|$ i.e. the distance between the two blue balls.

One can use Pythagoras theorem to get the desired displacement in magnitude and direction of the vector would be from initial positron to final position.

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  • $\begingroup$ the cockroach has to go along the walls, unless it's one of those Indian flying cockroaches, or a pregnant female, which I don't think was the intended meaning. $\endgroup$
    – Ron Maimon
    Apr 15, 2012 at 6:25
  • $\begingroup$ Sorry, question asked for displacement. $\endgroup$
    – Ron Maimon
    Apr 15, 2012 at 6:44
  • $\begingroup$ @RonMaimon: duh..it's the american ones which fly...en.wikipedia.org/wiki/….. :P :P $\endgroup$ Apr 15, 2012 at 12:23
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Remember that the displacement is the the vector that goes from the starting point to the finishing point. In your case, it is the diagonal that bonds the back lower left corner and the front upper right corner.

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  • $\begingroup$ Except the cockroach can't fly. $\endgroup$
    – Ron Maimon
    Apr 15, 2012 at 6:25
  • $\begingroup$ @Ron-Maimon It has nothing to do with it's ability to fly. The displacement IS DEFINED in that way. Maybe you are thinking in DISTANCE which a totally different measure in physics. Read more here: physicsclassroom.com/class/1dkin/u1l1c.cfm $\endgroup$
    – Néstor
    Apr 15, 2012 at 6:29
  • $\begingroup$ @RonMaimon, I see from your answer that you actually confused distance with displacement, which was what the OP asked for :P. $\endgroup$
    – Néstor
    Apr 15, 2012 at 6:36
  • $\begingroup$ Yes, I misinterpreted it as distance. My bad. No need for refs, you are right. $\endgroup$
    – Ron Maimon
    Apr 15, 2012 at 6:43
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The cube is three dimensional so you could use the Pythagorean theorem for many dimensions $c^2 = x^2 + y^2 + z^2 $. Here $x$, $y$ and $z$ are the distances traveled in each of the directions, and $c$ is the shortest distance, which is diagonally from corner to corner (also the displacement).

$x = y = z = 3 $

$c = \sqrt ( 3^2 + 3^2 + 3^2 )$

$c = \sqrt 27 $

$c = 3 \sqrt 3 $

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The cockroach first crawls along the floor to the halfway point of one of the two edges of the floor which are not attached to the starting point. This initial distance is

$$ S_1 = \sqrt{3^2+(1.5)^2} = {3\sqrt{5}\over 2} $$

Then the cockroach crawls up the wall to the corner, from the midpoint of the edge to the endpoint. This distance is equal to the initial distance.

So the total distance travelled is $3\sqrt{5}$. To prove that this is the minimum distance path along the crawling surface, fold out the two crawling walls, and draw the straight line between the initial and final points--- this reproduces the path above.

EDIT: except the question asked for displacement, not distance, and this is a trivial application of the pythagorean theorem, and the answer is $3\sqrt{3}$.

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  • $\begingroup$ I call it the "trithagorean theorem", but that's just me :) $\endgroup$ Apr 15, 2012 at 8:24
  • $\begingroup$ I am very appreciative of the quick response, but I am still not seeing the whole picture. Using Pythagoras theorem, it appears as though the displacement will be 3 + 3√2. The diagonal resultant distance between two corners + the distance from the bottom of that corner to the top right. @Ron - I am not clear on where the value 1.5 comes into the distance calculation, perhaps this is where I am missing the full picture. $\endgroup$
    – Kurt
    Apr 15, 2012 at 19:10
  • $\begingroup$ Ok, I am clear on arriving at 3√5/2. When the two panels of the cube are folded out we have a base length of 6m and a height of 3m. Then D^2 = 3^2 + 6^2, which leaves us with 3√5. Do we divide by 2 because the displacement is equal to the area under the line, which creates a triangle? Leaving me with 3√5/2, the total distance travelled, but the question is asking for displacement. However, I don't see movement in any negative vector quantity that would change the displacement, in this case, from varying to the distance travelled. $\endgroup$
    – Kurt
    Apr 15, 2012 at 19:27
  • $\begingroup$ @Kurt: Yes--- the issue with the divided by 2 is that each half-path on the individual face is this big--- the total distance travelled is $3\sqrt{5}$ as you say. The addition of displacements is vectorial, component by component, and the two individual displacements are perpendicular. I left this answer up, even though the original interpretation was totally wrong, because I imagined you were confused the same way I was. It's a trick question, and the only way to get it wrong is to misinterpret it as asking something nontrivial. I don't like these kinds of questions. They penalize thinking. $\endgroup$
    – Ron Maimon
    Apr 15, 2012 at 20:06

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