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 will break it down into steps to find the x-coordinate (D) when the y-coordinate is 32 m.

### Step 1: Identify the given data
- Initial velocity in the x-direction, \( u_x = 3.0 \, \text{m/s} \)
- Initial velocity in the y-direction, \( u_y = 0 \, \text{m/s} \)
- Acceleration in the x-direction, \( a_x = 6.0 \, \text{m/s}^2 \)
- Acceleration in the y-direction, \( a_y = 4.0 \, \text{m/s}^2 \)
- We need to find the x-coordinate when the y-coordinate \( y = 32 \, \text{m} \).

### Step 2: Use the second equation of motion for the y-direction
The second equation of motion for the y-direction is given by:
\[ y = u_y t + \frac{1}{2} a_y t^2 \]
Substituting the known values:
\[ 32 = 0 + \frac{1}{2} (4) t^2 \]
This simplifies to:
\[ 32 = 2 t^2 \]
Dividing both sides by 2:
\[ t^2 = 16 \]
Taking the square root:
\[ t = 4 \, \text{s} \]

### Step 3: Use the second equation of motion for the x-direction
Now that we have the time \( t = 4 \, \text{s} \), we can find the x-coordinate using the second equation of motion for the x-direction:
\[ x = u_x t + \frac{1}{2} a_x t^2 \]
Substituting the known values:
\[ x = (3)(4) + \frac{1}{2} (6)(4^2) \]
Calculating each term:
\[ x = 12 + \frac{1}{2} (6)(16) \]
\[ x = 12 + 3 \times 16 \]
\[ x = 12 + 48 \]
\[ x = 60 \, \text{m} \]

### Conclusion
The x-coordinate \( D \) when the y-coordinate is 32 m is:
\[ D = 60 \, \text{m} \]