A particle starts from the origin at `t=0` with an initial velocity of `3.0hati` m/s and moves in the x-y plane with a constant cacceleration `(6.0hati+4.0hatij)m//s^(2).` The x-coordinate of the particle at the instant when its y-coordinates is 32 m is D meters. The value of D is :

Options

A. 60
B. 32
C. 40
D. 50

Answer

To solve the problem, we need to find the x-coordinate of a particle when its y-coordinate is 32 meters. The particle starts from the origin with an initial velocity and moves under constant acceleration.

### Step-by-Step Solution:

1. **Identify Initial Conditions:**
- Initial position, \( \vec{r_0} = 0 \) (origin)
- Initial velocity, \( \vec{u} = 3 \hat{i} \) m/s (only in the x-direction)
- Acceleration, \( \vec{a} = (6 \hat{i} + 4 \hat{j}) \) m/s²

2. **Determine the time taken to reach y = 32 m:**
- The equation for the y-coordinate is given by:
\[ y = u_y t + \frac{1}{2} a_y t^2 \]
- Here, \( u_y = 0 \) (initial velocity in the y-direction) and \( a_y = 4 \) m/s² (acceleration in the y-direction).
- Plugging in the values:
\[ 32 = 0 \cdot t + \frac{1}{2} \cdot 4 \cdot t^2 \]
- Simplifying this gives:
\[ 32 = 2t^2 \]
- Therefore:
\[ t^2 = 16 \implies t = 4 \text{ seconds} \]

3. **Calculate the x-coordinate at t = 4 seconds:**
- The equation for the x-coordinate is:
\[ x = u_x t + \frac{1}{2} a_x t^2 \]
- Here, \( u_x = 3 \) m/s (initial velocity in the x-direction) and \( a_x = 6 \) m/s² (acceleration in the x-direction).
- Plugging in the values:
\[ x = 3 \cdot 4 + \frac{1}{2} \cdot 6 \cdot (4^2) \]
- Simplifying this gives:
\[ x = 12 + \frac{1}{2} \cdot 6 \cdot 16 \]
\[ x = 12 + 48 \]
\[ x = 60 \text{ meters} \]

4. **Conclusion:**
- The x-coordinate \( D \) when the y-coordinate is 32 meters is \( D = 60 \) meters.

### Final Answer:
The value of \( D \) is **60 meters**.