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 of when car B catches up to car A, we can follow these steps:

### Step 1: Define the motion of car A
Car A is moving with a uniform velocity of 40 m/s. The distance covered by car A after time \( t \) can be expressed as:
\[ S_A = v_A \cdot t = 40 \cdot t \]
where \( S_A \) is the distance traveled by car A, \( v_A \) is the velocity of car A, and \( t \) is the time in seconds.

### Step 2: Define the motion of car B
Car B starts from rest and accelerates at a constant rate of 4 m/s². The distance covered by car B after time \( t \) can be expressed using the equation of motion:
\[ S_B = u_B \cdot t + \frac{1}{2} a_B t^2 \]
Since car B starts from rest, \( u_B = 0 \), so:
\[ S_B = 0 \cdot t + \frac{1}{2} \cdot 4 \cdot t^2 = 2t^2 \]
where \( S_B \) is the distance traveled by car B and \( a_B \) is the acceleration of car B.

### Step 3: Set the distances equal
Since car B catches up to car A when they have traveled the same distance, we can set the two equations equal to each other:
\[ S_A = S_B \]
Thus:
\[ 40t = 2t^2 \]

### Step 4: Rearrange the equation
Rearranging the equation gives us:
\[ 2t^2 - 40t = 0 \]
Factoring out \( t \):
\[ t(2t - 40) = 0 \]

### Step 5: Solve for \( t \)
Setting each factor to zero gives us:
1. \( t = 0 \) (the initial time when both cars are at the same point)
2. \( 2t - 40 = 0 \) which simplifies to \( t = 20 \) seconds.

### Conclusion
Thus, car B will catch car A after \( t = 20 \) seconds.

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