A particle `P` is at the origin starts with velocity `u=(2hati-4hatj)m//s` with constant acceleration `(3hati+5hatj)m//s^(2)`. After travelling for `2s` its distance from the origin is

Options

A. 10 m
B. 20.6 m
C. 9.8 m
D. 11.7 m

Explanation

To solve the problem step by step, we will use the equations of motion to find the displacement of the particle after 2 seconds and then calculate the distance from the origin.

### Step 1: Identify the given values
- Initial velocity, \( \mathbf{u} = 2 \hat{i} - 4 \hat{j} \, \text{m/s} \)
- Acceleration, \( \mathbf{a} = 3 \hat{i} + 5 \hat{j} \, \text{m/s}^2 \)
- Time, \( t = 2 \, \text{s} \)

### Step 2: Use the equation of motion to find displacement
The formula for displacement \( \mathbf{s} \) is given by:
\[ \mathbf{s} = \mathbf{u} t + \frac{1}{2} \mathbf{a} t^2 \]

### Step 3: Calculate the first term \( \mathbf{u} t \)
\[ \mathbf{u} t = (2 \hat{i} - 4 \hat{j}) \cdot 2 = 4 \hat{i} - 8 \hat{j} \]

### Step 4: Calculate the second term \( \frac{1}{2} \mathbf{a} t^2 \)
First, calculate \( t^2 \):
\[ t^2 = 2^2 = 4 \]
Now calculate \( \frac{1}{2} \mathbf{a} t^2 \):
\[ \frac{1}{2} \mathbf{a} t^2 = \frac{1}{2} (3 \hat{i} + 5 \hat{j}) \cdot 4 = (1.5 \cdot 4) \hat{i} + (2.5 \cdot 4) \hat{j} = 6 \hat{i} + 10 \hat{j} \]

### Step 5: Combine the two terms to find displacement
Now, add the two results:
\[ \mathbf{s} = (4 \hat{i} - 8 \hat{j}) + (6 \hat{i} + 10 \hat{j}) = (4 + 6) \hat{i} + (-8 + 10) \hat{j} = 10 \hat{i} + 2 \hat{j} \]

### Step 6: Calculate the magnitude of the displacement to find the distance from the origin
The distance \( d \) from the origin is given by the magnitude of the displacement vector:
\[ d = |\mathbf{s}| = \sqrt{(10)^2 + (2)^2} = \sqrt{100 + 4} = \sqrt{104} \]
Calculating this gives:
\[ d \approx 10.2 \, \text{m} \]

### Final Answer
The distance from the origin after 2 seconds is approximately \( 10.2 \, \text{m} \).

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