Two bikes `A and B` start from a point. A moves with uniform speed `40 m//s and B` starts from rest with uniform acceleration `2 m//s^2`. If `B` starts at `t = 10` and `A` starts from the same point at `t = 10 s`, then the time during the journey in which `A` was ahead of `B` is :

Options

A. `20s`
B. `8s`
C. `10s`
D. `A` is never ahead of `B`

Explanation

To solve the problem, we need to analyze the motion of both bikes A and B and determine the time during which bike A was ahead of bike B.

### Step-by-Step Solution:

1. **Identify the Motion of Bike A**:
- Bike A moves with a uniform speed of \(40 \, \text{m/s}\).
- Bike A starts at \(t = 10 \, \text{s}\).

2. **Identify the Motion of Bike B**:
- Bike B starts from rest (initial velocity \(u = 0\)) and accelerates uniformly at \(2 \, \text{m/s}^2\).
- Bike B starts at \(t = 0 \, \text{s}\).

3. **Determine the Distance Covered by Each Bike**:
- For bike A, the distance covered after \(t\) seconds (starting from \(t = 10 \, \text{s}\)) is:
\[ s_A = 40(t - 10) \quad \text{(for } t \geq 10\text{)} \]
- For bike B, the distance covered after \(t\) seconds is given by the equation of motion:
\[ s_B = ut + \frac{1}{2} a t^2 = 0 + \frac{1}{2} \cdot 2 \cdot t^2 = t^2 \quad \text{(for } t \geq 0\text{)} \]

4. **Set the Distances Equal to Find When B Catches A**:
- We need to find when bike B catches up to bike A, which occurs when:
\[ s_A = s_B \]
- Thus, we set:
\[ 40(t - 10) = t^2 \]
- Rearranging gives:
\[ t^2 - 40t + 400 = 0 \]

5. **Solve the Quadratic Equation**:
- We can use the quadratic formula \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\[ a = 1, \quad b = -40, \quad c = 400 \]
- The discriminant is:
\[ b^2 - 4ac = (-40)^2 - 4 \cdot 1 \cdot 400 = 1600 - 1600 = 0 \]
- Since the discriminant is zero, there is one solution:
\[ t = \frac{40}{2} = 20 \, \text{s} \]

6. **Determine the Time During Which A Was Ahead of B**:
- Bike A starts at \(t = 10 \, \text{s}\) and bike B catches up at \(t = 20 \, \text{s}\).
- Therefore, bike A was ahead of bike B from \(t = 10 \, \text{s}\) to \(t = 20 \, \text{s}\).
- The total time during which A was ahead of B is:
\[ 20 \, \text{s} - 10 \, \text{s} = 10 \, \text{s} \]

### Final Answer:
The time during the journey in which bike A was ahead of bike B is **10 seconds**.