A particle starts from origin at `t=0` with a constant velocity `5hati 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 `84 m` in `m` is

Options

A. 12 m
B. 24 m
C. 36 m
D. 48 m

Explanation

To solve the problem, we need to determine the y-coordinate of a particle at the moment its x-coordinate reaches 84 meters. The particle starts from the origin with an initial velocity and is subject to a constant acceleration.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Initial velocity in the x-direction, \( u_x = 5 \, \hat{i} \, \text{m/s} \)
- Acceleration, \( \vec{a} = 3 \, \hat{i} + 2 \, \hat{j} \, \text{m/s}^2 \)
- Displacement in the x-direction, \( s_x = 84 \, \text{m} \)

2. **Write the Equation for Displacement in the x-direction:**
The displacement in the x-direction can be expressed as:
\[ s_x = u_x t + \frac{1}{2} a_x t^2 \]
where \( a_x = 3 \, \text{m/s}^2 \).

3. **Substitute the Known Values:**
Plugging in the values, we have:
\[ 84 = 5t + \frac{1}{2} \cdot 3 \cdot t^2 \]
Simplifying this gives:
\[ 84 = 5t + \frac{3}{2} t^2 \]
Rearranging leads to:
\[ \frac{3}{2} t^2 + 5t - 84 = 0 \]
To eliminate the fraction, multiply through by 2:
\[ 3t^2 + 10t - 168 = 0 \]

4. **Solve the Quadratic Equation:**
We can use the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 3, b = 10, c = -168 \):
\[ t = \frac{-10 \pm \sqrt{10^2 - 4 \cdot 3 \cdot (-168)}}{2 \cdot 3} \]
\[ t = \frac{-10 \pm \sqrt{100 + 2016}}{6} \]
\[ t = \frac{-10 \pm \sqrt{2116}}{6} \]
\[ t = \frac{-10 \pm 46}{6} \]
This gives two possible solutions for \( t \):
\[ t = \frac{36}{6} = 6 \quad \text{(valid, since time cannot be negative)} \]
\[ t = \frac{-56}{6} \quad \text{(not valid)} \]

5. **Calculate the y-coordinate:**
Now, we need to find the y-coordinate when \( t = 6 \) seconds. The displacement in the y-direction is given by:
\[ s_y = u_{y} t + \frac{1}{2} a_y t^2 \]
Since the initial velocity in the y-direction \( u_y = 0 \):
\[ s_y = 0 + \frac{1}{2} \cdot 2 \cdot t^2 \]
\[ s_y = \frac{1}{2} \cdot 2 \cdot (6)^2 \]
\[ s_y = 36 \, \text{m} \]

### Final Answer:
The y-coordinate of the particle when its x-coordinate is 84 m is **36 m**.