Question

A room has dimensions ${}^{\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{m}\mathbf{}\left(\mathrm{height}\right)\mathbf{}\mathbf{\times }\mathbf{3}\mathbf{.}\mathbf{70}\mathbf{\times }\mathbf{4}\mathbf{.}\mathbf{30}\mathbf{}\mathbf{m}}$. A fly starting at one corner flies around, ending up at the diagonally opposite corner. (a) What is the magnitude of its displacement? (b) Could the length of its path be less than this magnitude? (c) Greater? (d) Equal? (e) Choose a suitable coordinate system and express the components of the displacement vector in that system in unit-vector notation. (f) If the fly walks, what is the length of the shortest path? (Hint: This can be answered without calculus. The room is like a box. Unfold its walls to flatten them into a plane.)

Short answer

1. The magnitude of the displacement of the fly is${}^{\mathbf{6}\mathbf{.}\mathbf{42}\mathbf{}\mathbf{m}}$
   

2. The length of its path can’t be less than the displacement.

3. The path length can be greater than the displacement.

4. The path length can be equal to the displacement.

5. The displacement vector is${}^{\mathbf{(}\mathbf{3}\mathbf{.}\mathbf{7}\mathbf{}\mathbf{m}\mathbf{)}\mathbf{}\hat{\mathbf{i}}\mathbf{}\mathbf{(}\mathbf{4}\mathbf{.}\mathbf{3}\mathbf{)}\mathbf{}\hat{\mathbf{j}}\mathbf{}\mathbf{+}\mathbf{(}\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{)}\mathbf{}\hat{\mathbf{k}}}\mathbf{}\mathbf{}$.

6. The length of the shortest path if the fly walks along the sides of the room is${}^{\mathbf{7}\mathbf{.}\mathbf{96}\mathbf{}\mathbf{m}}$

Step-by-step solution

Given data

The dimensions of the room are${}^{{}^{\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{m}\mathbf{}\mathbf{\left(}\mathbf{height}\mathbf{\right)}\mathbf{}\mathbf{\times }\mathbf{3}\mathbf{.}\mathbf{70}\mathbf{\times }\mathbf{4}\mathbf{.}\mathbf{30}\mathbf{}\mathbf{m}}}$

Understanding the displacement

Displacement may be defined as the shortest distance between two points.

The magnitude of the displacement ${}^{\mathbf{d}}$can be expressed in three dimension as:

$\mathit{d}\mathbf{=}\sqrt{{\left(\Delta x\right)}^{\mathbf{2}}\mathbf{+}{\left(\Delta y\right)}^{\mathbf{2}}\mathbf{+}{\left(\Delta z\right)}^{\mathbf{2}}}$ … (i)

Here,${}^{\mathbf{\Delta }\mathbf{x}}$,${}^{\mathbf{\Delta }\mathbf{y}}$, and ${}^{\mathbf{\Delta }\mathbf{z}}$can be consider as width, length, and height respectively.

(a) Determination of the magnitude of displacement

If a fly is start moving from one corner (the ceiling) and the opposite corner is on the floor, the magnitude of the displacement can be calculated by equation (i) as:

$$\begin{gathered} \mathbf{d}\mathbf{=}\sqrt{{\mathbf{(}\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}\mathbf{)}}^{\mathbf{2}}\mathbf{+}{\mathbf{(}\mathbf{3}\mathbf{.}\mathbf{7}\mathbf{}\mathbf{m}\mathbf{)}}^{\mathbf{2}}\mathbf{+}{\mathbf{(}\mathbf{4}\mathbf{.}\mathbf{3}\mathbf{}\mathbf{m}\mathbf{)}}^{\mathbf{2}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{417}\mathbf{}\mathbf{m} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{\approx }\mathbf{6}\mathbf{.}\mathbf{42}\mathbf{}\mathbf{m} \end{gathered}$$

Thus, the magnitude of the displacement is${}^{\mathbf{6}\mathbf{.}\mathbf{42}\mathbf{}\mathbf{m}}$ .

(b) To find out if the length of path can be less than the magnitude

Since, the straight line gives the shortest distance between two points, length of the path can’t be less than the above displacement.

If the fly crawls along the edges of the room, the path length would

$\mathbf{3}\mathbf{+}\mathbf{3}\mathbf{.}\mathbf{7}\mathbf{+}\mathbf{4}\mathbf{.}\mathbf{3}\mathbf{=}\mathbf{11}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}$

Thus, the path length can’t be less than the displacement.

(c) To find out if the length of path can be greater than the magnitude

Since, the straight line gives the shortest distance between two points, length of the path can’t be less than the above displacement.

If the fly crawls along the edges of the room, the path length would

$\mathbf{3}\mathbf{+}\mathbf{3}\mathbf{.}\mathbf{7}\mathbf{+}\mathbf{4}\mathbf{.}\mathbf{3}\mathbf{=}\mathbf{11}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}$

Thus, the path length can be greater than the displacement.

(d) To find out if the length of path can be equal to the magnitude

Since, the straight line gives the shortest distance between two points, length of the path can’t be less than the above displacement.

If the fly flies along the displacement vector, then the path length can be equal to the displacement.

(e) Express the components of the displacement vector in that system in unit-vector notation

Taking 3.7 m along x-axis, 4.30 m along y-axis and 3.0 m along z axis, the displacement vector can be written as,

$\mathbf{d}\mathbf{}\mathbf{=}\left(3.7m\right)\mathbf{}\mathbf{i}\mathbf{}\left(4.3m\right)\mathbf{}\hat{\mathbf{j}}\mathbf{}\left(3.0m\right)\mathbf{}\hat{\mathbf{k}}$

If we flatten out the corners of the box, we would get the rectangles as shown in figure.

(f) Determination of the length of the shortest path

There can be three different routes to reach to the top, ${}^{{\mathbf{R}}_{\mathbf{1}\mathbf{\text{'}}}}$${}^{{\mathbf{R}}_{\mathbf{2}}}$, and ${}^{{\mathbf{R}}_{\mathbf{3}}}$, thus it can be concluded that ${}^{{\mathbf{R}}_{\mathbf{1}}}$would be smaller than ${}^{{\mathbf{R}}_{\mathbf{2}}}$, and ${}^{{\mathbf{R}}_{\mathbf{3}}}$.

So, the length of the shortest path is,

$$\begin{gathered} {\mathbf{R}}_{\mathbf{1}}\mathbf{=}\sqrt{{\left(6.7\right)}^{\mathbf{2}}\mathbf{+}{\left(4.3\right)}^{\mathbf{2}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{7}\mathbf{.}\mathbf{96}\mathbf{}\mathbf{m} \end{gathered}$$

Thus, the length of the shortest path is${}^{\mathbf{7}\mathbf{.}\mathbf{96}\mathbf{}\mathbf{m}}$ .