Two cars A and B are at rest at same point initially. If A starts with uniform velocity of 40 m/sec and B starts in the same direction with constant acceleration of` 4m//s^(2) ,`then B will catch A after :

Options

A. 10 sec
B. 20 sec
C. 30 sec
D. 35 sec

Explanation

To solve the problem, we need to determine when car B, which starts from rest and accelerates, will catch up to car A, which is moving at a constant speed.

### Step-by-Step Solution:

1. **Identify the Variables:**
- For car A:
- Initial velocity, \( u_A = 40 \, \text{m/s} \)
- Acceleration, \( a_A = 0 \, \text{m/s}^2 \) (since it moves with uniform velocity)
- For car B:
- Initial velocity, \( u_B = 0 \, \text{m/s} \) (starts from rest)
- Acceleration, \( a_B = 4 \, \text{m/s}^2 \)

2. **Write the Displacement Equations:**
- The displacement of car A after time \( t \):
\[ s_A = u_A t + \frac{1}{2} a_A t^2 = 40t + 0 = 40t \]
- The displacement of car B after time \( t \):
\[ s_B = u_B t + \frac{1}{2} a_B t^2 = 0 + \frac{1}{2} \cdot 4 t^2 = 2t^2 \]

3. **Set the Displacements Equal:**
- Since car B catches car A when they have traveled the same distance:
\[ s_A = s_B \]
\[ 40t = 2t^2 \]

4. **Rearrange the Equation:**
- Rearranging gives:
\[ 2t^2 - 40t = 0 \]
- Factor out \( t \):
\[ t(2t - 40) = 0 \]

5. **Solve for \( t \):**
- This gives two solutions:
\[ t = 0 \quad \text{(initial time)} \]
\[ 2t - 40 = 0 \implies t = 20 \, \text{seconds} \]

6. **Conclusion:**
- Car B will catch car A after \( t = 20 \, \text{seconds} \).

### Final Answer:
Car B will catch car A after **20 seconds**.

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