Velocity and acceleration of a particle at some instant of time are `v = (3hati +4hatj) ms^(-1)` and `a =- (6hati +8hatj)ms^(-2)` respectively. At the same instant particle is at origin. Maximum x-coordinate of particle will be

Options

A. 1.5 m
B. 0.75m
C. 2.25m
D. 4m

Explanation

To find the maximum x-coordinate of the particle given its initial velocity and acceleration, we can follow these steps:

### Step-by-step Solution:

1. **Identify Initial Conditions:**
- The initial velocity of the particle is given as:
\[ \mathbf{v} = 3 \hat{i} + 4 \hat{j} \, \text{m/s} \]
- The acceleration of the particle is given as:
\[ \mathbf{a} = -6 \hat{i} + 8 \hat{j} \, \text{m/s}^2 \]
- The particle is initially at the origin, which means its initial position is:
\[ \mathbf{r}_0 = 0 \hat{i} + 0 \hat{j} \]

2. **Determine the Maximum x-coordinate:**
- The maximum x-coordinate occurs when the velocity in the x-direction becomes zero.
- The initial velocity in the x-direction (\(u_x\)) is 3 m/s, and the acceleration in the x-direction (\(a_x\)) is -6 m/s².

3. **Use the Third Equation of Motion:**
- The third equation of motion states:
\[ v^2 = u^2 + 2as \]
- Here, \(v\) is the final velocity (0 m/s at maximum x-coordinate), \(u\) is the initial velocity (3 m/s), \(a\) is the acceleration (-6 m/s²), and \(s\) is the displacement in the x-direction (which we want to find).

4. **Set Up the Equation:**
- Plugging in the values:
\[ 0 = (3)^2 + 2(-6)s \]
- This simplifies to:
\[ 0 = 9 - 12s \]

5. **Solve for Displacement \(s\):**
- Rearranging gives:
\[ 12s = 9 \implies s = \frac{9}{12} = \frac{3}{4} \, \text{m} \]

6. **Conclusion:**
- The maximum x-coordinate of the particle is:
\[ \text{Maximum x-coordinate} = 0.75 \, \text{m} \]