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 step by step, we will break it down into parts based on the motion of the car under acceleration and then under retardation.

### Step 1: Determine the time intervals
Let \( t_1 \) be the time during which the car accelerates and \( t_2 \) be the time during which the car decelerates. According to the problem, the total time of motion is given as:
\[ t_1 + t_2 = 18 \text{ seconds} \]

### Step 2: Calculate the final velocity after acceleration
The car starts from rest, so the initial velocity \( u = 0 \). The acceleration \( a = 4 \, \text{m/s}^2 \). The final velocity \( v \) after time \( t_1 \) can be calculated using the formula:
\[ v = u + at = 0 + 4t_1 = 4t_1 \]

### Step 3: Set up the equation for deceleration
During the deceleration phase, the car decelerates at \( -2 \, \text{m/s}^2 \) until it comes to rest. Therefore, the final velocity \( v = 0 \) and the initial velocity for this phase is \( 4t_1 \). Using the equation:
\[ v = u + at \]
we can write:
\[ 0 = 4t_1 - 2t_2 \]
Rearranging gives:
\[ 2t_2 = 4t_1 \quad \Rightarrow \quad t_2 = 2t_1 \]

### Step 4: Substitute \( t_2 \) in the total time equation
Substituting \( t_2 = 2t_1 \) into the total time equation:
\[ t_1 + 2t_1 = 18 \]
This simplifies to:
\[ 3t_1 = 18 \quad \Rightarrow \quad t_1 = 6 \text{ seconds} \]
Now substituting back to find \( t_2 \):
\[ t_2 = 2t_1 = 2 \times 6 = 12 \text{ seconds} \]

### Step 5: Calculate the distance covered during acceleration
Using the formula for distance under constant acceleration:
\[ s = ut + \frac{1}{2} a t^2 \]
Substituting \( u = 0 \), \( a = 4 \, \text{m/s}^2 \), and \( t = t_1 = 6 \):
\[ s_1 = 0 \cdot 6 + \frac{1}{2} \cdot 4 \cdot (6)^2 = 0 + 2 \cdot 36 = 72 \text{ meters} \]

### Step 6: Calculate the distance covered during deceleration
Using the same distance formula for the deceleration phase:
\[ s' = ut + \frac{1}{2} a t^2 \]
Here, the initial velocity \( u = 4t_1 = 4 \times 6 = 24 \, \text{m/s} \), \( a = -2 \, \text{m/s}^2 \), and \( t = t_2 = 12 \):
\[ s_2 = 24 \cdot 12 + \frac{1}{2} \cdot (-2) \cdot (12)^2 \]
Calculating this gives:
\[ s_2 = 288 - \frac{1}{2} \cdot 2 \cdot 144 = 288 - 144 = 144 \text{ meters} \]

### Step 7: Calculate the total distance covered
The total distance covered by the car is the sum of the distances during acceleration and deceleration:
\[ s_{\text{total}} = s_1 + s_2 = 72 + 144 = 216 \text{ meters} \]

### Final Answer
The total distance covered by the car is:
\[ \boxed{216 \text{ meters}} \]