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, we will break it down into steps:

### Step 1: Identify the given data
- Initial velocity (u) = 0 m/s (the car starts from rest)
- Acceleration (a) = 2 m/s² (during the acceleration phase)
- Retardation (deceleration) = 4 m/s² (during the deceleration phase)
- Total time of motion (T) = 3 s

### Step 2: Define the time intervals for acceleration and deceleration
Let:
- \( T_1 \) = time during acceleration
- \( T_2 \) = time during deceleration

From the problem, we know:
\[ T_1 + T_2 = 3 \, \text{s} \]

### Step 3: Relate the time intervals using the final velocity
At the end of the acceleration phase, the final velocity \( V \) can be expressed as:
\[ V = u + a \cdot T_1 \]
Since \( u = 0 \):
\[ V = 2 \cdot T_1 \]

During the deceleration phase, the final velocity becomes 0:
\[ 0 = V - 4 \cdot T_2 \]
Substituting for \( V \):
\[ 0 = 2 \cdot T_1 - 4 \cdot T_2 \]
Rearranging gives:
\[ 2 \cdot T_1 = 4 \cdot T_2 \]
or
\[ T_1 = 2 \cdot T_2 \]

### Step 4: Substitute \( T_1 \) in the total time equation
Substituting \( T_1 = 2 \cdot T_2 \) into the total time equation:
\[ 2 \cdot T_2 + T_2 = 3 \]
\[ 3 \cdot T_2 = 3 \]
\[ T_2 = 1 \, \text{s} \]

### Step 5: Find \( T_1 \)
Using \( T_2 \) to find \( T_1 \):
\[ T_1 = 2 \cdot T_2 = 2 \cdot 1 = 2 \, \text{s} \]

### Step 6: Calculate the distance traveled during acceleration (S1)
Using the equation of motion:
\[ S_1 = u \cdot T_1 + \frac{1}{2} a \cdot T_1^2 \]
Substituting the values:
\[ S_1 = 0 \cdot 2 + \frac{1}{2} \cdot 2 \cdot (2^2) \]
\[ S_1 = 0 + \frac{1}{2} \cdot 2 \cdot 4 \]
\[ S_1 = 4 \, \text{m} \]

### Step 7: Calculate the distance traveled during deceleration (S2)
Using the equation of motion:
\[ S_2 = V \cdot T_2 - \frac{1}{2} \cdot 4 \cdot T_2^2 \]
Substituting for \( V \):
\[ S_2 = (2 \cdot T_1) \cdot T_2 - \frac{1}{2} \cdot 4 \cdot (1^2) \]
Substituting \( T_1 = 2 \):
\[ S_2 = (2 \cdot 2) \cdot 1 - 2 \]
\[ S_2 = 4 - 2 = 2 \, \text{m} \]

### Step 8: Calculate the total distance traveled
Total distance \( S \) is:
\[ S = S_1 + S_2 = 4 + 2 = 6 \, \text{m} \]

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

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