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 of when the car will catch up with the bus, we can use the equations of motion. Let's break it down step by step.

### Step 1: Understand the initial conditions
- The bus is at rest and has an acceleration of \( a_b = 2 \, \text{m/s}^2 \).
- The car is also at rest but has a higher acceleration of \( a_c = 4 \, \text{m/s}^2 \).
- The initial distance between the car and the bus is \( d = 200 \, \text{m} \).

### Step 2: Write the equations of motion
Since both vehicles start from rest, we can use the second equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
where \( s \) is the distance traveled, \( u \) is the initial velocity (which is 0 for both), \( a \) is the acceleration, and \( t \) is the time.

For the bus:
\[ s_b = 0 \cdot t + \frac{1}{2} \cdot 2 \cdot t^2 = t^2 \]

For the car:
\[ s_c = 0 \cdot t + \frac{1}{2} \cdot 4 \cdot t^2 = 2t^2 \]

### Step 3: Set up the equation for when the car catches up
The car will catch up with the bus when the distance traveled by the car minus the distance traveled by the bus equals the initial distance between them:
\[ s_c - s_b = 200 \]
Substituting the expressions for \( s_c \) and \( s_b \):
\[ 2t^2 - t^2 = 200 \]
This simplifies to:
\[ t^2 = 200 \]

### Step 4: Solve for time \( t \)
Taking the square root of both sides:
\[ t = \sqrt{200} = \sqrt{100 \times 2} = 10\sqrt{2} \, \text{seconds} \]

### Final Answer
The car will catch up with the bus after \( t = 10\sqrt{2} \) seconds.

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