A room has dimensions 3.00 m (height) 3.70 m 4.30 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 than this magnitude? (d) Equal to this magnitude? (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 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 its walls to flatten them into a plane.)
I don't know please answer this question
(a) If one corner is on the ceiling and the opposite corner is on the floor, the straight-line distance between them is
sqrt (a^2 + b^2 + c^2) = 6.417 m
(b) How could it?
(c) Suppose it did not fly in a straight line. Then what?
(d) Can the magnitude equal the magnitude? Does 2 = 2?
(e) 2.00i + 3.70 j + 4.30 k
(f) Take the hint. There are two possible short routes to the far corner; the path length depends upon which nonadjacent wall the fly walks to first.
One route gives a minimum length of
sqrt[(4.3+3)^2 +3.7^2] = 8.18 m
and the other has a minimum length of
sqrt[(3.7+3)^2 +4.3^2] = 7.96 m
To find the answers to the questions, we can break them down step-by-step:
a) The magnitude of the fly's displacement is the distance between the starting and ending points. Since the points are diagonally opposite corners of the room, we can think of the displacement as a diagonal of a rectangular parallelepiped. Using the Pythagorean theorem, the magnitude of the diagonal can be found as:
d^2 = h^2 + l^2 + w^2
where:
- d is the magnitude of the diagonal
- h is the height of the room (3.00 m)
- l is the length of the room (4.30 m)
- w is the width of the room (3.70 m)
Plugging in the values, we get:
d^2 = (3.00 m)^2 + (4.30 m)^2 + (3.70 m)^2
Calculating this gives us:
d ≈ 6.19 m
So the magnitude of the fly's displacement is approximately 6.19 meters.
b) The length of the fly's path cannot be less than the magnitude of its displacement. The displacement represents the shortest straight-line distance between the starting and ending points, and any path shorter than this would not reach the ending point.
c) The length of the fly's path cannot be greater than the magnitude of its displacement either. Any path longer than the displacement would involve unnecessary detours and cannot be the shortest path.
d) The length of the fly's path cannot be equal to the magnitude of its displacement. If the path were equal, then the path would also be straight and would coincide with the displacement.
e) To express the components of the displacement vector in a suitable coordinate system, we can use a Cartesian coordinate system with one corner of the room as the origin (0, 0, 0). We can choose the positive x-axis to point towards the opposite corner along the length dimension, the positive y-axis to point towards the opposite corner along the width dimension, and the positive z-axis to point upwards towards the ceiling.
The coordinates of the starting point are (0, 0, 0), and the coordinates of the ending point are (l, w, h).
So the displacement vector can be expressed in unit-vector notation as:
vector D = l * i + w * j + h * k
Where i, j, and k are the unit vectors along the x, y, and z axes, respectively.
f) If the fly walks instead of flies, the length of the shortest path it can take can be found by unfolding the walls of the room into a plane, as suggested in the hint. The room can be unfolded into a rectangular shape.
In this case, the length of the shortest path is the perimeter of the rectangular shape, which can be calculated as:
shortest path length = 2(l + w)
Plugging in the values, we get:
shortest path length = 2(4.30 m + 3.70 m)
Calculating this gives us:
shortest path length = 2(8.00 m)
So the length of the shortest path the fly can take if it walks is 16.00 meters.
To answer these questions, we'll need to use the concepts of displacement and distance.
(a) The magnitude of the fly's displacement is the straight line distance between its starting and ending points. In this case, it's the distance between the opposite corners of the room. Let's call this distance d.
To find the magnitude, we can use the Pythagorean theorem since we have a right-angled triangle formed by the three dimensions of the room. The formula is:
d = √(height^2 + width^2 + length^2)
Plugging in the values, we get:
d = √(3.00^2 + 3.70^2 + 4.30^2)
≈ √(9 + 13.69 + 18.49)
≈ √(41.18)
≈ 6.42 meters
So the magnitude of the fly's displacement is approximately 6.42 meters.
(b) No, the length of the path cannot be less than the magnitude of displacement because the magnitude represents the shortest straight-line distance between two points.
(c) Yes, the length of the path could be greater than the magnitude of displacement if the fly takes a longer or more circuitous route from one corner to the other.
(d) The length of the path cannot be equal to the magnitude of displacement because the magnitude represents the shortest distance.
(e) Choosing a suitable coordinate system helps us express the components of the displacement vector. Let's assume the origin (0, 0, 0) is at the starting corner of the fly.
We can then express the position of the starting corner as (0, 0, 0) and the position of the ending corner as (3.00, 3.70, 4.30) in meters. The displacement vector is the difference between these two points:
Displacement = (3.00, 3.70, 4.30) - (0, 0, 0)
= (3.00, 3.70, 4.30) in meters
So the components of the displacement vector in unit-vector notation are (3.00 m, 3.70 m, 4.30 m).
(f) If the fly walks rather than flies, the shortest path it can take would be by walking along the walls of the room. The trick here is to unfold the walls and flatten them into a plane.
Since the room is like a box, unfolding the walls will create a rectangle with dimensions 3.00 m x 3.70 m. To find the shortest path, we simply calculate the perimeter of this rectangle:
Shortest path length = 2 * (3.00 + 3.70)
= 2 * 6.70
= 13.40 meters
So, if the fly walks, the length of the shortest path it can take is 13.40 meters.