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A rabbit trying to escape a fox runs north for 8.0m, darts northwest for 1.0m, then drops 1.0m down a hole into its burrow. What is the magnitude of the net displacement of the rabbit?

So I drew two triangles and tried to solve it using pythagorean theorem but I got stuck. I had a small triangle with two sides that are 1 and one unknown side (not sure how to find this side).

Help please?

Lil
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    Why not just literally add the three vectors together (this will give you the net displacement vector) and then find the norm of the sum? The only one that requires any figuring is the second vector and it's still straightforward. – rnrstopstraffic Jul 23 '17 at 07:47
  • Change your title –  Aug 25 '17 at 07:06

3 Answers3

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You may represent the displacement as a vector $\left(\begin{matrix}x\\y\\z\end{matrix}\right)$. $x$ is the distance to the East, $y$ the distance to the North, and $z$ the distance up into the sky. I'll write it inline like $(x,y,z)$ because it's more compact.

  1. Initially, the rabbit and fox are in the same place. The displacement is $(0,0,0).$
  2. The rabbit runs north for $8 \mathrm m$. That's $8$ units in the $y$ direction. New displacement is $(0,8,0).$
  3. The rabbit darts North-West $1\mathrm m$. That's $(-1/ \sqrt 2, 1 / \sqrt 2, 0)$ (use pytharogas to work out East and North displacements of this move separately). New displacement is $(0,8,0) + (-1/\sqrt 2, 1/\sqrt 2,0) = (-1/\sqrt 2, 8 + 1/\sqrt 2, 0)$.
  4. The rabbit burrows $1 \mathrm m$ down, decreasing its $z$ coordinate by $1$. Final displacement is $$(-1/\sqrt 2, \; \; 8 + 1/\sqrt 2, \; \; -1).$$

Fortunately, Pythagorean's theorem generalises to any dimension. So to find the distance from the rabbit to the fox, we take the root of the sum of the squares of the displacements in each orthogonal direction ($x,y$ and $z$):

$$\begin{align}\text{Distance} &= \sqrt{\left( - \frac 1 {\sqrt 2}\right)^2 + \left(8+ \frac 1 {\sqrt 2}\right)^2 + (-1)^2}\\ &=\sqrt{\frac 1 2 + \left(64 + \frac 1 2 + \frac{16}{\sqrt 2}\right)+1}\\ &=\sqrt{66 + \frac{16}{\sqrt 2}}\\ &=\sqrt{66 + 8 \sqrt 2}\\ &\approx 8.79\mathrm m\end{align}$$

Myridium
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The rabbit trying to escape a fox runs north for $8.0m$, darts northwest for $1.0m$, $\ldots$

Let's give this part a treatment of complex numbers. The situation will be similar to rotating the point $i$ by an angle $\frac{π}{4}$, and then shifting the origin to the point $-8i$. The new position of the rabbit is given by the complex number

$$ z = ie^{\frac{iπ}{4}} + 8i = -\frac{1}{\sqrt{2}} + \left(8+\frac{1}{\sqrt{2}}\right)i. $$

$\ldots$ then drops $1.0m$ down a hole into its burrow.

Let's revert back to vectors for this purpose. The situation then is similar to adding a component of $-\hat{k}$ to the vector $-\frac{1}{\sqrt{2}}\hat{i} +\left(8+\frac{1}{\sqrt{2}}\right)\hat{j}$. The final position is

$$ \vec{v} = -\frac{1}{\sqrt{2}}\hat{i} +\left(8+\frac{1}{\sqrt{2}}\right)\hat{j} -\hat{k}. $$

The magnitude of displacement is given by $|\vec{v}|$, where

$$ |\vec{v}| = \sqrt{v_x^2+v_y^2+v_z^2} \\ = \sqrt{\left(\frac{1}{\sqrt{2}}\right)^2 + \left(8+\frac{1}{\sqrt{2}}\right)^2 + 1} \\ = \sqrt{66 +8\sqrt{2}}. $$

Nilabro Saha
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Hint: It is easier to find the horizontal displacement first. What is the long side of the triangle at ground level? You will have to use the law of cosines as the two sides are not perpendicular. Then note that the vector down the hole is perpendicular to this one, so you have another right triangle.

Your approach was fine, too. You had a $1-1-\sqrt 2$ right triangle from the northwest and down, but then assessing the angle with the north is difficult.

Another approach, I suspect the intended one, is just to find the final coordinates of the rabbit in $\Bbb R^3$. Then you have perpendicular sides and can use Pythagoras.

Ross Millikan
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  • any way you can provide a picture of your diagram? – Lil Oct 15 '14 at 23:37
  • I have a small triangle with 2 sides of 1m and then the smallest side is part of another triangle. For the law of cosines would i use the two 1m sides and the 90 degree angle? – Lil Oct 15 '14 at 23:38
  • No, the triangle along the ground has sides of $8,1$. The triangle with two $1$ sides is the NW run and the drop down the hole. They are perpendicular, so the diagonal is $\sqrt 2$. I don't see an easy way to get the angle between the $8$ and $\sqrt 2$ for use in the law of cosines. That is why I suggested finding the long side of the $8,1$ triangle, because you have that angle. This long side is still perpendicular to the $1$ down. – Ross Millikan Oct 16 '14 at 00:09