At t=0 and x=0 , an initially stationary blue car begins to accelerate at a constant rate of `2.0m//s^(2) ` in the positive direction of the x -axis At =t = 2s a red car travelling in an adjecent lane and in the same direction passes x =0 with a speed of ` 8.0 m//s ` and a constant accleration of ` 3.0 m//s^(2) ` .the time when red car passes the blue car is :

Options

A. 4.2 s
B. 3.6 s
C. 2.9 s
D. None of these

Explanation

To solve the problem of when the red car passes the blue car, we will analyze the motion of both cars using the equations of motion.

### Step-by-Step Solution:

1. **Identify the motion parameters of both cars:**
- **Blue Car:**
- Initial position, \( x_{B0} = 0 \)
- Initial velocity, \( u_B = 0 \) (stationary)
- Acceleration, \( a_B = 2.0 \, \text{m/s}^2 \)
- Time of motion, \( t_B = t \) (we will find this time)

- **Red Car:**
- Initial position, \( x_{R0} = 0 \)
- Initial velocity, \( u_R = 8.0 \, \text{m/s} \) (starts at \( t = 2 \, \text{s} \))
- Acceleration, \( a_R = 3.0 \, \text{m/s}^2 \)
- Time of motion, \( t_R = t - 2 \) (since it starts moving 2 seconds later)

2. **Write the equations of motion for both cars:**
- **Distance covered by the Blue Car:**
\[ x_B = u_B t + \frac{1}{2} a_B t^2 = 0 + \frac{1}{2} (2) t^2 = t^2 \]

- **Distance covered by the Red Car:**
\[ x_R = u_R (t - 2) + \frac{1}{2} a_R (t - 2)^2 \]
Expanding this:
\[ x_R = 8(t - 2) + \frac{1}{2} (3)(t - 2)^2 \]
\[ x_R = 8t - 16 + \frac{3}{2}(t^2 - 4t + 4) = 8t - 16 + \frac{3}{2}t^2 - 6t + 6 \]
\[ x_R = \frac{3}{2}t^2 + 2t - 10 \]

3. **Set the distances equal to find when the cars meet:**
\[ x_B = x_R \]
\[ t^2 = \frac{3}{2}t^2 + 2t - 10 \]

4. **Rearranging the equation:**
\[ t^2 - \frac{3}{2}t^2 - 2t + 10 = 0 \]
\[ -\frac{1}{2}t^2 - 2t + 10 = 0 \]
Multiplying through by -2 to eliminate the fraction:
\[ t^2 + 4t - 20 = 0 \]

5. **Use the quadratic formula to solve for \( t \):**
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1, b = 4, c = -20 \):
\[ t = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot (-20)}}{2 \cdot 1} \]
\[ t = \frac{-4 \pm \sqrt{16 + 80}}{2} \]
\[ t = \frac{-4 \pm \sqrt{96}}{2} \]
\[ t = \frac{-4 \pm 4\sqrt{6}}{2} \]
\[ t = -2 \pm 2\sqrt{6} \]

6. **Select the positive root:**
\[ t = -2 + 2\sqrt{6} \]
Approximating \( \sqrt{6} \approx 2.45 \):
\[ t \approx -2 + 4.9 \approx 2.9 \, \text{s} \]

### Conclusion:
The time when the red car passes the blue car is approximately \( t \approx 2.9 \, \text{s} \).