A particle has displacement 12cm towards east and 9cm towards north and then 6cm vertically upward. The magnitude of the displacement of the particle is:
Calculating Particle Displacement Magnitude in 3D
This problem asks us to find the magnitude of the total displacement of a particle that moves sequentially in three mutually perpendicular directions: east, north, and vertically upward.
Displacement is a vector quantity, meaning it has both magnitude and direction. When a particle undergoes multiple displacements, the total displacement is the vector sum of the individual displacements. Since the movements are along perpendicular directions (east, north, upward), we can consider these directions as corresponding to the x, y, and z axes of a 3D Cartesian coordinate system.
Representing Displacements as Vectors
Let's assign the directions to axes:
- East: Positive x-axis
- North: Positive y-axis
- Vertically Upward: Positive z-axis
The given displacements are:
- Displacement 1: 12 cm towards east. Vector component in x-direction = +12 cm.
- Displacement 2: 9 cm towards north. Vector component in y-direction = +9 cm.
- Displacement 3: 6 cm vertically upward. Vector component in z-direction = +6 cm.
The total displacement vector, let's call it $\vec{d}$, can be written as:
\(\vec{d} = d_x \hat{i} + d_y \hat{j} + d_z \hat{k}\)
Where:
- \(d_x = 12\) cm (east)
- \(d_y = 9\) cm (north)
- \(d_z = 6\) cm (upward)
So, the total displacement vector is \(\vec{d} = (12 \hat{i} + 9 \hat{j} + 6 \hat{k})\) cm.
Finding the Magnitude of Total Displacement
The magnitude of a vector \(\vec{v} = a \hat{i} + b \hat{j} + c \hat{k}\) in three dimensions is given by the formula:
\(|\vec{v}| = \sqrt{a^2 + b^2 + c^2}\)
In our case, the components are \(a=12\), \(b=9\), and \(c=6\). The magnitude of the total displacement \(|\vec{d}|\) is:
\(|\vec{d}| = \sqrt{(12 \; \text{cm})^2 + (9 \; \text{cm})^2 + (6 \; \text{cm})^2}\)
Step-by-Step Magnitude Calculation
Let's calculate the squares of the components:
- \(12^2 = 12 \times 12 = 144\)
- \(9^2 = 9 \times 9 = 81\)
- \(6^2 = 6 \times 6 = 36\)
Now, add these values:
\(144 + 81 + 36 = 225 + 36 = 261\)
Finally, take the square root of the sum to find the magnitude:
\(|\vec{d}| = \sqrt{261}\)
The magnitude of the displacement is \(\sqrt{261}\) cm.
Comparing with Options
Let's look at the given options:
- Option 1: \(\sqrt {216} \;cm\)
- Option 2: \(\sqrt {131} \;cm\)
- Option 3: \(\sqrt {261} \;cm\)
- Option 4: \(\sqrt {331} cm\)
Our calculated magnitude is \(\sqrt{261}\) cm, which matches Option 3.
| Direction | Displacement (cm) | Component | Component Squared |
|---|---|---|---|
| East (x) | 12 | 12 | \(12^2 = 144\) |
| North (y) | 9 | 9 | \(9^2 = 81\) |
| Upward (z) | 6 | 6 | \(6^2 = 36\) |
| Sum of Squares: | \(144 + 81 + 36 = 261\) | ||
| Magnitude (\(\sqrt{\text{Sum}}\)): | \(\sqrt{261}\) cm | ||
Conclusion on Particle Displacement
The magnitude of the total displacement of the particle, after moving 12cm east, 9cm north, and 6cm vertically upward, is \(\sqrt{261}\) cm. This is because the displacements are in mutually perpendicular directions, allowing us to use the Pythagorean theorem extended to three dimensions.
Revision Table: Key Concepts
| Concept | Description | Relevance to Problem |
|---|---|---|
| Displacement | A vector quantity representing the change in position of an object. It is the straight-line distance and direction from the starting point to the ending point. | We need to find the magnitude of the total displacement vector. |
| Vector Addition | Combining two or more vectors to find a resultant vector. For perpendicular components, this is done using the Pythagorean theorem (or its 3D extension). | The total displacement is the vector sum of the individual displacements. |
| Vector Magnitude | The length or size of a vector. For a vector \(\vec{v} = v_x \hat{i} + v_y \hat{j} + v_z \hat{k}\), the magnitude is \(\sqrt{v_x^2 + v_y^2 + v_z^2}\). | We need to calculate the magnitude of the total displacement vector using its components. |
| Perpendicular Components | Vectors or components that are at right angles (\(90^\circ\)) to each other. Displacements along East-West, North-South, and vertical directions are typically perpendicular. | The east, north, and upward displacements are perpendicular, simplifying vector addition. |
Additional Information on Displacement and Distance
It's important to distinguish between displacement and distance. Distance is a scalar quantity representing the total length of the path traveled, while displacement is a vector quantity representing the shortest distance between the initial and final points, along with direction.
- In this problem, the distance traveled would be the sum of the magnitudes of the individual movements: \(12 \, \text{cm} + 9 \, \text{cm} + 6 \, \text{cm} = 27 \, \text{cm}\).
- The magnitude of the displacement, as calculated, is \(\sqrt{261}\) cm, which is approximately 16.16 cm.
The magnitude of the displacement is generally less than or equal to the total distance traveled. They are equal only if the motion is along a straight line in one direction.
Understanding vector components and how to calculate the magnitude of a resultant vector from perpendicular components is crucial in physics problems involving motion in 2D or 3D.
Important Questions from Position, path length and displacement
The ratio of magnitude of displacement to distance for a moving object is always.
