A particle starts from the origin at `t= 0 s` with a velocity of `10.0 hatj m//s` and moves in the `xy`-plane with a constant acceleration of `(8hati+2hatj)m//s^(-2)`. Then `y`-coordinate of the particle in `2 sec` is

Options

A. `24 m`
B. `16 m`
C. `8 m`
D. `12 m`

Explanation

To find the y-coordinate of the particle at \( t = 2 \, \text{s} \), we can use the equations of motion in two dimensions. The motion of the particle is described by its initial velocity and constant acceleration.

### Step-by-Step Solution:

1. **Identify Initial Conditions:**
- The initial position of the particle is at the origin:
\[ \mathbf{r}_0 = 0 \hat{i} + 0 \hat{j} \, \text{m} \]
- The initial velocity is given as:
\[ \mathbf{u} = 0 \hat{i} + 10 \hat{j} \, \text{m/s} \]
- The constant acceleration is:
\[ \mathbf{a} = 8 \hat{i} + 2 \hat{j} \, \text{m/s}^2 \]

2. **Use the Equation of Motion:**
The position vector \( \mathbf{r} \) at time \( t \) can be calculated using the equation:
\[ \mathbf{r}(t) = \mathbf{r}_0 + \mathbf{u} t + \frac{1}{2} \mathbf{a} t^2 \]

3. **Substituting the Values:**
Since \( \mathbf{r}_0 = 0 \):
\[ \mathbf{r}(t) = \mathbf{u} t + \frac{1}{2} \mathbf{a} t^2 \]
Substituting \( \mathbf{u} \) and \( \mathbf{a} \):
\[ \mathbf{r}(t) = (0 \hat{i} + 10 \hat{j}) t + \frac{1}{2} (8 \hat{i} + 2 \hat{j}) t^2 \]

4. **Calculating for \( t = 2 \, \text{s} \):**
\[ \mathbf{r}(2) = (0 \hat{i} + 10 \hat{j}) \cdot 2 + \frac{1}{2} (8 \hat{i} + 2 \hat{j}) \cdot (2^2) \]
\[ = (0 \hat{i} + 20 \hat{j}) + \frac{1}{2} (8 \hat{i} + 2 \hat{j}) \cdot 4 \]
\[ = (0 \hat{i} + 20 \hat{j}) + (16 \hat{i} + 4 \hat{j}) \]
\[ = 16 \hat{i} + (20 + 4) \hat{j} \]
\[ = 16 \hat{i} + 24 \hat{j} \]

5. **Extract the y-coordinate:**
The y-coordinate of the particle at \( t = 2 \, \text{s} \) is:
\[ y = 24 \, \text{m} \]

### Final Answer:
The y-coordinate of the particle in 2 seconds is \( 24 \, \text{m} \).

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