A car accelerates from rest at a constant rate of `2ms^(-2)` for some time. Then, it retards at a constant rete of `4ms^(-2)` and comes and rest. If the total time for which it remains in motion is 3 s, Then the total distance travelled is

Options

A. 2 m
B. 3 m
C. 4 m
D. 6 m

Answer

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

### Step 1: Define Variables
- Let \( a_1 = 2 \, \text{m/s}^2 \) (acceleration)
- Let \( a_2 = -4 \, \text{m/s}^2 \) (deceleration)
- Let \( t_1 \) = time of acceleration
- Let \( t_2 \) = time of deceleration
- We know that \( t_1 + t_2 = 3 \, \text{s} \)

### Step 2: Find Final Velocity after Acceleration
Since the car starts from rest:
- Initial velocity \( u = 0 \)
- The final velocity after acceleration \( v_1 \) can be calculated using the formula:
\[ v_1 = u + a_1 t_1 = 0 + 2 t_1 = 2 t_1 \]

### Step 3: Relate Final Velocity to Deceleration
The car comes to rest after deceleration, so:
- Final velocity \( v_2 = 0 \)
- Using the formula for deceleration:
\[ v_2 = v_1 + a_2 t_2 \]
Substituting the values:
\[ 0 = 2 t_1 - 4 t_2 \]
This simplifies to:
\[ 2 t_1 = 4 t_2 \]
or
\[ t_1 = 2 t_2 \]

### Step 4: Substitute into Total Time Equation
Now, substitute \( t_1 = 2 t_2 \) into the total time equation:
\[ t_1 + t_2 = 3 \]
\[ 2 t_2 + t_2 = 3 \]
\[ 3 t_2 = 3 \]
\[ t_2 = 1 \, \text{s} \]

### Step 5: Find \( t_1 \)
Now substitute \( t_2 \) back to find \( t_1 \):
\[ t_1 = 2 t_2 = 2 \times 1 = 2 \, \text{s} \]

### Step 6: Calculate Final Velocity \( v_1 \)
Now calculate \( v_1 \):
\[ v_1 = 2 t_1 = 2 \times 2 = 4 \, \text{m/s} \]

### Step 7: Calculate Distances
Now we can calculate the distance traveled during acceleration and deceleration:
1. **Distance during acceleration**:
\[ s_1 = u t_1 + \frac{1}{2} a_1 t_1^2 = 0 \times 2 + \frac{1}{2} \times 2 \times (2^2) = \frac{1}{2} \times 2 \times 4 = 4 \, \text{m} \]

2. **Distance during deceleration**:
\[ s_2 = v_1 t_2 + \frac{1}{2} a_2 t_2^2 = 4 \times 1 + \frac{1}{2} \times (-4) \times (1^2) = 4 - 2 = 2 \, \text{m} \]

### Step 8: Total Distance
Now, add both distances to get the total distance traveled:
\[ s_{total} = s_1 + s_2 = 4 + 2 = 6 \, \text{m} \]

### Final Answer
The total distance traveled by the car is **6 meters**.

---