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. (a)4.2 s
B. (b)3.6 s
C. (c)2.9 s
D. (d)None of these

Explanation

To solve the problem of when the red car passes the blue car, we can break it down into a series of steps:

### Step-by-Step Solution:

1. **Define the Motion of the Blue Car:**
- The blue car starts from rest (initial velocity \( u_B = 0 \)) and accelerates at a constant rate \( a_B = 2.0 \, \text{m/s}^2 \).
- The position of the blue car as a function of time \( t \) can be described by the equation of motion:
\[ x_B = u_B t + \frac{1}{2} a_B t^2 \]
- Substituting the values:
\[ x_B = 0 + \frac{1}{2} \cdot 2 \cdot t^2 = t^2 \]

2. **Define the Motion of the Red Car:**
- The red car passes \( x = 0 \) at \( t = 2 \, \text{s} \) with an initial speed \( u_R = 8.0 \, \text{m/s} \) and accelerates at \( a_R = 3.0 \, \text{m/s}^2 \).
- The position of the red car as a function of time \( t \) (considering it starts moving at \( t = 2 \, \text{s} \)) is given by:
\[ x_R = u_R (t - 2) + \frac{1}{2} a_R (t - 2)^2 \]
- Substituting the values:
\[ x_R = 8(t - 2) + \frac{1}{2} \cdot 3 (t - 2)^2 \]
- Simplifying this:
\[ x_R = 8t - 16 + \frac{3}{2}(t^2 - 4t + 4) \]
\[ x_R = 8t - 16 + \frac{3}{2}t^2 - 6t + 6 \]
\[ x_R = \frac{3}{2}t^2 + 2t - 10 \]

3. **Set the Positions Equal to Find When They Meet:**
- To find the time when the red car passes the blue car, set \( x_B = x_R \):
\[ t^2 = \frac{3}{2}t^2 + 2t - 10 \]
- Rearranging gives:
\[ 0 = \frac{3}{2}t^2 - t^2 + 2t - 10 \]
\[ 0 = \frac{1}{2}t^2 + 2t - 10 \]
\[ 0 = t^2 + 4t - 20 \]

4. **Solve the Quadratic Equation:**
- Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 1 \), \( b = 4 \), and \( c = -20 \).
\[ t = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot (-20)}}{2 \cdot 1} \]
\[ t = \frac{-4 \pm \sqrt{16 + 80}}{2} \]
\[ t = \frac{-4 \pm \sqrt{96}}{2} \]
\[ t = \frac{-4 \pm 4\sqrt{6}}{2} \]
\[ t = -2 \pm 2\sqrt{6} \]

5. **Select the Positive Root:**
- Since time cannot be negative, we take:
\[ t = -2 + 2\sqrt{6} \]
- Approximating \( \sqrt{6} \approx 2.45 \):
\[ t \approx -2 + 4.9 \approx 2.9 \, \text{s} \]

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