A car is standing 800 m behind a bus, which is also at rest. The two start moving at the same instant but the different forward accelerations. The bus has acceleration `4m//s^(2)` and the car has acceleration `8m//s^(2)`. The car will catch up with the bus after a time of:

Options

A. 20s
B. 10s
C. 5s
D. 15s

Answer

To solve the problem of when the car will catch up with the bus, we can follow these steps:

### Step 1: Understand the initial conditions
- The car is initially 800 meters behind the bus.
- Both vehicles start from rest, meaning their initial velocities are 0 m/s.
- The bus has an acceleration of \( a_b = 4 \, \text{m/s}^2 \).
- The car has an acceleration of \( a_c = 8 \, \text{m/s}^2 \).

### Step 2: Write the equations for displacement
Using the equation of motion for displacement, which is given by:
\[ S = ut + \frac{1}{2} a t^2 \]
where \( S \) is the displacement, \( u \) is the initial velocity, \( a \) is the acceleration, and \( t \) is the time.

For the car:
- Initial velocity \( u_c = 0 \)
- Acceleration \( a_c = 8 \, \text{m/s}^2 \)

Thus, the displacement of the car after time \( t \) is:
\[ S_c = 0 \cdot t + \frac{1}{2} \cdot 8 \cdot t^2 = 4t^2 \]

For the bus:
- Initial velocity \( u_b = 0 \)
- Acceleration \( a_b = 4 \, \text{m/s}^2 \)

Thus, the displacement of the bus after time \( t \) is:
\[ S_b = 0 \cdot t + \frac{1}{2} \cdot 4 \cdot t^2 = 2t^2 \]

### Step 3: Set up the equation for catching up
The car will catch up with the bus when the distance traveled by the car equals the distance traveled by the bus plus the initial distance of 800 meters:
\[ S_c = S_b + 800 \]
Substituting the expressions we derived:
\[ 4t^2 = 2t^2 + 800 \]

### Step 4: Solve for time \( t \)
Rearranging the equation:
\[ 4t^2 - 2t^2 = 800 \]
\[ 2t^2 = 800 \]
\[ t^2 = 400 \]
Taking the square root of both sides:
\[ t = \sqrt{400} = 20 \, \text{seconds} \]

### Conclusion
The car will catch up with the bus after \( t = 20 \) seconds.

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