A car, starting from rest, moves in a straight line with constant acceleration . ` 4m//s^(2) ` After some time, it immediately starts retarding at constant rate ` 2m//s^(2) ` until it comes to rest. If the car moved for a total time 18 seconds, the total distance covered is

Explanation

To solve the problem, we need to break it down into two parts: the acceleration phase and the deceleration phase.

### Step 1: Define the variables
- Let \( a_1 = 4 \, \text{m/s}^2 \) (acceleration)
- Let \( a_2 = -2 \, \text{m/s}^2 \) (retardation)
- Let \( t_1 \) be the time of acceleration
- Let \( t_2 \) be the time of retardation
- Total time \( t_1 + t_2 = 18 \, \text{s} \)

### Step 2: Find the relationship between \( t_1 \) and \( t_2 \)
From the equations of motion, we know that:
1. The final velocity after acceleration \( v = u + a_1 t_1 \)
Since the car starts from rest, \( u = 0 \):
\[ v = 4 t_1 \]

2. The final velocity after deceleration is 0, so:
\[ 0 = v + a_2 t_2 \implies 0 = 4 t_1 - 2 t_2 \]
Rearranging gives:
\[ 2 t_2 = 4 t_1 \implies t_2 = 2 t_1 \]

### Step 3: Substitute \( t_2 \) into the total time equation
Substituting \( t_2 = 2 t_1 \) into \( t_1 + t_2 = 18 \):
\[ t_1 + 2 t_1 = 18 \implies 3 t_1 = 18 \implies t_1 = 6 \, \text{s} \]
Then, substituting back to find \( t_2 \):
\[ t_2 = 2 t_1 = 2 \times 6 = 12 \, \text{s} \]

### Step 4: Calculate the distance during acceleration
Using the formula for distance under constant acceleration:
\[ s = ut + \frac{1}{2} a t^2 \]
Since \( u = 0 \):
\[ s_1 = 0 + \frac{1}{2} \times 4 \times (6^2) = 2 \times 36 = 72 \, \text{m} \]

### Step 5: Calculate the distance during deceleration
Using the same formula for distance during deceleration:
\[ s_2 = v t_2 + \frac{1}{2} a_2 t_2^2 \]
Where \( v = 4 t_1 = 4 \times 6 = 24 \, \text{m/s} \):
\[ s_2 = 24 \times 12 + \frac{1}{2} \times (-2) \times (12^2) \]
Calculating each term:
\[ s_2 = 288 - \frac{1}{2} \times 2 \times 144 = 288 - 144 = 144 \, \text{m} \]

### Step 6: Calculate the total distance
The total distance covered by the car is:
\[ s_{\text{total}} = s_1 + s_2 = 72 + 144 = 216 \, \text{m} \]

### Final Answer
The total distance covered by the car is **216 meters**.