A particle starts from origin at `t=0` with a constant velocity `5hatj m//s` and moves in `x-y` plane under action of a force which produce a constant acceleration of `(3hati+2hatj)m//s^(2)` the `y`-coordinate of the particle at the instant its `x` co-ordinate is `24 m` in `m` is

Options

A. 48
B. 60
C. 12
D. 24

Answer

To solve the problem, we will break it down into steps. The particle starts from the origin with a constant velocity and is acted upon by a force that produces a constant acceleration. We need to find the y-coordinate of the particle when its x-coordinate is 24 m.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Initial velocity in the y-direction, \( u_y = 5 \, \hat{j} \, \text{m/s} \)
- Initial velocity in the x-direction, \( u_x = 0 \, \hat{i} \, \text{m/s} \)
- Acceleration in the x-direction, \( a_x = 3 \, \hat{i} \, \text{m/s}^2 \)
- Acceleration in the y-direction, \( a_y = 2 \, \hat{j} \, \text{m/s}^2 \)
- Displacement in the x-direction, \( s_x = 24 \, \text{m} \)

2. **Use the Equation of Motion in the x-direction:**
The equation of motion in the x-direction is given by:
\[ s_x = u_x t + \frac{1}{2} a_x t^2 \]
Substituting the known values:
\[ 24 = 0 \cdot t + \frac{1}{2} \cdot 3 \cdot t^2 \]
Simplifying this gives:
\[ 24 = \frac{3}{2} t^2 \]
Multiplying both sides by \( \frac{2}{3} \):
\[ t^2 = \frac{48}{3} = 16 \]
Taking the square root:
\[ t = 4 \, \text{s} \]

3. **Use the Equation of Motion in the y-direction:**
Now that we have the time \( t = 4 \, \text{s} \), we can find the y-coordinate using the equation of motion in the y-direction:
\[ s_y = u_y t + \frac{1}{2} a_y t^2 \]
Substituting the known values:
\[ s_y = 5 \cdot 4 + \frac{1}{2} \cdot 2 \cdot (4^2) \]
Calculating each term:
\[ s_y = 20 + \frac{1}{2} \cdot 2 \cdot 16 \]
\[ s_y = 20 + 16 = 36 \, \text{m} \]

4. **Final Result:**
The y-coordinate of the particle when the x-coordinate is 24 m is:
\[ \boxed{36 \, \text{m}} \]