Velocity and acceleration of a particle are `v=(2 hati-4 hatj) m/s` and `a=(-2 hati+4 hatj) m/s^2` Which type of motion is this?

Explanation

To determine the type of motion of a particle given its velocity and acceleration vectors, we can follow these steps:

### Step 1: Identify the Given Vectors
The velocity \( \mathbf{v} \) and acceleration \( \mathbf{a} \) of the particle are given as:
- \( \mathbf{v} = (2 \hat{i} - 4 \hat{j}) \, \text{m/s} \)
- \( \mathbf{a} = (-2 \hat{i} + 4 \hat{j}) \, \text{m/s}^2 \)

### Step 2: Check for Constancy
Next, we observe that both vectors are constant. There are no time-dependent functions involved, indicating that both the velocity and acceleration do not change over time.

### Step 3: Analyze the Relationship Between Velocity and Acceleration
To understand the motion type, we need to analyze the relationship between the velocity and acceleration vectors. We can factor out a negative sign from the acceleration vector:
\[ \mathbf{a} = -2 \hat{i} + 4 \hat{j} = -1 \cdot (2 \hat{i} - 4 \hat{j}) = -\mathbf{v} \]
This shows that the acceleration vector \( \mathbf{a} \) is in the opposite direction to the velocity vector \( \mathbf{v} \).

### Step 4: Calculate Magnitudes
Now, we calculate the magnitudes of both vectors:
- Magnitude of velocity \( |\mathbf{v}| \):
\[ |\mathbf{v}| = \sqrt{(2)^2 + (-4)^2} = \sqrt{4 + 16} = \sqrt{20} \]
- Magnitude of acceleration \( |\mathbf{a}| \):
\[ |\mathbf{a}| = \sqrt{(-2)^2 + (4)^2} = \sqrt{4 + 16} = \sqrt{20} \]

### Step 5: Conclusion on Motion Type
Since the magnitudes of the velocity and acceleration are equal and the acceleration is directed opposite to the velocity, this indicates that the particle is undergoing **uniformly retarded motion**. The particle is slowing down because the acceleration is acting in the opposite direction to the velocity.

### Final Answer
The type of motion of the particle is **uniformly retarded motion**.

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