A particle has initial velocity of `(3hat(i)+4hat(j)) m//s` and a acceleration of `(4hat(i)-3hat(j)) m//s^(2)`. Its speed after one second will be equal to :-

Options

A. `0`
B. `10 m//s`
C. `5sqrt(2) m//s`
D. `25 m//s`

Explanation

To find the speed of the particle after one second, we will follow these steps:

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

### Step 2: Use the formula for final velocity
The final velocity \( \mathbf{v} \) can be calculated using the equation:
\[ \mathbf{v} = \mathbf{u} + \mathbf{a} t \]

### Step 3: Substitute the values into the equation
Substituting the values we have:
\[ \mathbf{v} = (3 \hat{i} + 4 \hat{j}) + (4 \hat{i} - 3 \hat{j}) \cdot 1 \]

### Step 4: Simplify the equation
Now, simplifying the right side:
\[ \mathbf{v} = (3 \hat{i} + 4 \hat{j}) + (4 \hat{i} - 3 \hat{j}) = (3 + 4) \hat{i} + (4 - 3) \hat{j} \]
\[ \mathbf{v} = 7 \hat{i} + 1 \hat{j} \, \text{m/s} \]

### Step 5: Calculate the magnitude of the velocity
The speed (magnitude of velocity) is given by:
\[ |\mathbf{v}| = \sqrt{(7)^2 + (1)^2} \]
\[ |\mathbf{v}| = \sqrt{49 + 1} = \sqrt{50} \]

### Step 6: Simplify the magnitude
We can simplify \( \sqrt{50} \):
\[ |\mathbf{v}| = \sqrt{25 \cdot 2} = 5\sqrt{2} \, \text{m/s} \]

### Conclusion
Thus, the speed of the particle after one second is \( 5\sqrt{2} \, \text{m/s} \).