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

Options

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

Answer

To solve the problem, we need to find the y-coordinate of a particle after 2 seconds, given its initial velocity and constant acceleration. Let's break it down step by step.

### Step 1: Identify the given values
- Initial velocity \( \mathbf{u} = 10 \hat{j} \) m/s
- Acceleration \( \mathbf{a} = 8 \hat{i} + 2 \hat{j} \) m/s²
- Time \( t = 2 \) s

### Step 2: Use the equation of motion
The equation of motion to find the displacement \( \mathbf{s} \) is given by:
\[ \mathbf{s} = \mathbf{u}t + \frac{1}{2} \mathbf{a} t^2 \]

### Step 3: Focus on the y-component
Since we are interested in the y-coordinate, we will only consider the y-components of the initial velocity and acceleration.
- The initial velocity in the y-direction: \( u_y = 10 \) m/s
- The acceleration in the y-direction: \( a_y = 2 \) m/s²

### Step 4: Substitute the values into the equation
Now we can substitute the values into the equation for the y-coordinate:
\[ s_y = u_y t + \frac{1}{2} a_y t^2 \]
Substituting the known values:
\[ s_y = (10 \, \text{m/s})(2 \, \text{s}) + \frac{1}{2} (2 \, \text{m/s}^2)(2 \, \text{s})^2 \]

### Step 5: Calculate each term
Calculating the first term:
\[ s_y = 20 \, \text{m} + \frac{1}{2} (2)(4) \]
Calculating the second term:
\[ s_y = 20 \, \text{m} + \frac{1}{2} \cdot 8 = 20 \, \text{m} + 4 \, \text{m} \]

### Step 6: Add the results
Now, adding the two results together:
\[ s_y = 20 \, \text{m} + 4 \, \text{m} = 24 \, \text{m} \]

### Final Answer
Thus, the y-coordinate of the particle after 2 seconds is:
\[ \boxed{24 \, \text{m}} \]

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