A car is standing 200m behind a bus , which is also at rest . The two. Start moving at the same instant but with different forward accelerations. The bus has acceleration `2 ms ^(-2)` and The car has acceleration `4 ms^(-2)` The car will catch up will the bus after time :

Options

A. `sqrt(120)s`
B. 15 s
C. `sqrt(110)s`
D. `10 sqrt(2) s`

Answer

To solve the problem, we need to find the time at which the car catches up with the bus. Let's break it down step by step.

### Step 1: Understand the Initial Conditions
- The car is initially 200 meters behind the bus.
- Both vehicles start from rest.
- The acceleration of the bus (AB) is \(2 \, \text{m/s}^2\).
- The acceleration of the car (AC) is \(4 \, \text{m/s}^2\).

### Step 2: Set Up the Equations of Motion
Using the second equation of motion, we can express the distance traveled by both the bus and the car after time \(t\).

1. **For the Bus:**
\[ S_B = U_B t + \frac{1}{2} A_B t^2 \]
Since the bus starts from rest (\(U_B = 0\)):
\[ S_B = 0 + \frac{1}{2} \cdot 2 \cdot t^2 = t^2 \quad \text{(Equation 1)} \]

2. **For the Car:**
\[ S_C = U_C t + \frac{1}{2} A_C t^2 \]
Since the car also starts from rest (\(U_C = 0\)):
\[ S_C = 0 + \frac{1}{2} \cdot 4 \cdot t^2 = 2t^2 \quad \text{(Equation 2)} \]

### Step 3: Relate the Distances
The car is initially 200 meters behind the bus, so when the car catches up, the distance traveled by the car (SC) will be equal to the distance traveled by the bus (SB) plus 200 meters:
\[ S_C = S_B + 200 \]
Substituting the equations from above:
\[ 2t^2 = t^2 + 200 \]

### Step 4: Solve for Time \(t\)
Rearranging the equation:
\[ 2t^2 - t^2 = 200 \]
\[ t^2 = 200 \]
Taking the square root of both sides:
\[ t = \sqrt{200} = \sqrt{100 \cdot 2} = 10\sqrt{2} \, \text{seconds} \]

### Conclusion
The time at which the car catches up with the bus is \(10\sqrt{2}\) seconds.

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