A car accelerates from rest at a constant rate of `2 m s^(-2)` for some time. The it retatds at a constant rate of `4 m s^(-2)` and comes to rest. It remains in motion for `6 s`.

Options

A. Its maximum speed is `8 m s^(-1)`
B. Its maximum speed is `6 m s^(-1)`
C. It travelled a total distance of `24 m`
D. It travelled a total distance of `18 m`

Explanation

To solve the problem step by step, we will analyze the motion of the car during its acceleration and deceleration phases.

### Step 1: Understand the motion phases
The car accelerates from rest at a constant rate of \(2 \, \text{m/s}^2\) for a time \(t_1\), then it decelerates at a constant rate of \(4 \, \text{m/s}^2\) for a time \(t_2\), and the total time of motion is given as \(6 \, \text{s}\).

### Step 2: Define the equations for each phase
1. **Acceleration Phase**:
- Initial velocity \(u = 0\)
- Acceleration \(a = 2 \, \text{m/s}^2\)
- Final velocity after acceleration \(v = u + at = 0 + 2t_1 = 2t_1\)

2. **Deceleration Phase**:
- Initial velocity \(u = v = 2t_1\)
- Final velocity \(v = 0\)
- Deceleration \(a = -4 \, \text{m/s}^2\)
- Using the equation \(v = u + at\):
\[ 0 = 2t_1 - 4t_2 \implies 4t_2 = 2t_1 \implies t_2 = \frac{t_1}{2} \]

### Step 3: Relate total time to \(t_1\) and \(t_2\)
Given that the total time is \(6 \, \text{s}\):
\[ t_1 + t_2 = 6 \implies t_1 + \frac{t_1}{2} = 6 \]
Combining terms:
\[ \frac{3t_1}{2} = 6 \implies 3t_1 = 12 \implies t_1 = 4 \, \text{s} \]
Now substituting \(t_1\) back to find \(t_2\):
\[ t_2 = \frac{t_1}{2} = \frac{4}{2} = 2 \, \text{s} \]

### Step 4: Calculate maximum speed \(v\)
Using \(t_1\) to find the maximum speed:
\[ v = 2t_1 = 2 \times 4 = 8 \, \text{m/s} \]

### Step 5: Calculate distances traveled during each phase
1. **Distance during acceleration \(d_1\)**:
Using the equation \(s = ut + \frac{1}{2}at^2\):
\[ d_1 = 0 \cdot t_1 + \frac{1}{2} \cdot 2 \cdot (4^2) = \frac{1}{2} \cdot 2 \cdot 16 = 16 \, \text{m} \]

2. **Distance during deceleration \(d_2\)**:
Using the same equation:
\[ d_2 = (2t_1)t_2 + \frac{1}{2}(-4)(t_2^2) = (8)(2) + \frac{1}{2}(-4)(2^2) = 16 - 8 = 8 \, \text{m} \]

### Step 6: Calculate total distance
The total distance \(D\) traveled by the car:
\[ D = d_1 + d_2 = 16 + 8 = 24 \, \text{m} \]

### Final Answers
- Maximum speed \(v = 8 \, \text{m/s}\)
- Total distance traveled \(D = 24 \, \text{m}\)