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. `20 s`
B. `8s`
C. `10 s`
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 is ahead of bike B.

### Step-by-Step Solution:

1. **Identify the Motion of Bike A:**
- Bike A moves with a uniform speed of \( 40 \, \text{m/s} \).
- Since it starts at \( t = 10 \, \text{s} \), the distance covered by A after \( t \) seconds (where \( t \) is the total time from the start) can be expressed as:
\[ s_A = 40(t - 10) \quad \text{for } t \geq 10 \]
- Here, \( (t - 10) \) is the time A has been moving.

2. **Identify the Motion of Bike B:**
- Bike B starts from rest with a uniform acceleration of \( 2 \, \text{m/s}^2 \) at \( t = 10 \, \text{s} \).
- The distance covered by B after \( t \) seconds can be expressed as:
\[ s_B = 0 + \frac{1}{2} \cdot 2 \cdot (t - 10)^2 = (t - 10)^2 \quad \text{for } t \geq 10 \]

3. **Set the Distances Equal to Find When B Catches A:**
- To find when B catches A, we set the distances equal:
\[ 40(t - 10) = (t - 10)^2 \]
- Rearranging gives:
\[ (t - 10)^2 - 40(t - 10) = 0 \]
- Factoring out \( (t - 10) \):
\[ (t - 10)((t - 10) - 40) = 0 \]
- This gives us two solutions:
\[ t - 10 = 0 \quad \Rightarrow \quad t = 10 \, \text{s} \quad \text{(initial condition)} \]
\[ t - 10 = 40 \quad \Rightarrow \quad t = 50 \, \text{s} \]

4. **Determine the Time A Was Ahead of B:**
- Bike A starts moving at \( t = 10 \, \text{s} \) and is ahead until \( t = 50 \, \text{s} \).
- Therefore, the time during which A was ahead of B is:
\[ 50 \, \text{s} - 10 \, \text{s} = 40 \, \text{s} \]

### Final Answer:
The time during the journey in which A was ahead of B is \( 40 \, \text{s} \).