Magnitude of Displacement problem

[KrissyFivee]

Homework Statement

A room has dimensions 3.15 m (height) 3.85 (width) m 4.10 (length) m. A fly starting at one corner flies around, ending up at the diagonally opposite corner.

(a) What is the magnitude of its displacement? (solved and got 6.45m)
(b) Could the length of its path be less than this magnitude?
(c) Could the length of its path be greater than this magnitude?
(d) Could the length of its path be equal to this magnitude?
(e) Take xyz axes so that the x-axis is parallel to the width, the y-axis is parallel to the length, and the z axis is parallel to the height. Express the components of the displacement vector.
(f) If the fly walks rather than flies, what is the length of the shortest path it can take? (Hint: This can be answered without calculus. The room is like a box. Unfold the walls and flatten them into a plane.)

Homework Equations

|A| = sqrt(A^2+B^2+C^2)

The Attempt at a Solution

I don't even know where to begin with this problem . . .

So far all I've got is the displacement is 6.45m, which is correct.

 

[LawrenceC]

Hint for b,c, and d:
The shortest distance between two points in space is a straight line.

For part e, I think the question is looking for the i,j,k components of the displacement vector so they are talking about a straight line. From the starting position, what are the components of the displacement.

For part f, sketch the box unfolded then eyeball the shortest path on the surface of the unfolded box. Determine its length by using a method like the one you used in part a.

 

[KrissyFivee]

Thank you for the help :) I got it! I just didn't understand what the question was asking for!

 

[LawrenceC]

Good!

 

[asteriatic]

To answer the other parts of the problem, we can use the given dimensions of the room to visualize the fly's movement. Since it starts at one corner and ends at the diagonally opposite corner, we can imagine the fly flying along the diagonal of the room.

(a) The magnitude of displacement is the length of this diagonal, which can be calculated using the Pythagorean theorem: d = √(3.15^2 + 3.85^2 + 4.10^2) = 6.45 m.

(b) The length of the fly's path cannot be less than the magnitude of displacement, as it must move at least that distance to reach the opposite corner.

(c) The length of the fly's path can be greater than the magnitude of displacement if it does not fly along the diagonal, but takes a longer route around the room.

(d) The length of the fly's path can be equal to the magnitude of displacement if it flies along the diagonal, as this is the shortest distance between the two points.

(e) To express the components of the displacement vector, we can use the given axes. Since the x-axis is parallel to the width, the x component of the displacement will be equal to the width of the room, 3.85 m. Similarly, the y component will be equal to the length of the room, 4.10 m, and the z component will be equal to the height of the room, 3.15 m. Therefore, the displacement vector can be expressed as (3.85, 4.10, 3.15).

(f) If the fly walks instead of flies, the length of the shortest path it can take is along the edges of the room. This can be visualized by unfolding the walls and flattening them into a plane. The shortest path would then be the diagonal of this flattened rectangle, which can be calculated using the Pythagorean theorem: d = √(3.85^2 + 4.10^2) = 5.29 m. This is shorter than the displacement when the fly flies, as expected.