A car accelerates from rest at constant rate of `2ms^(-2)` for some time. Immediately after this, it retards at a constant rate of `4ms^(-2)` and comes to rest. The total time for which it remains in motion is 3 s. Taking the moment of start of motion as t = 0, answer the following questions.
What is the distance travelled by it during the time of increasing speed ?

Options

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

Answer

To solve the problem step by step, let's break it down into manageable parts.

### Given Data:
- Initial velocity, \( u = 0 \) (the car starts from rest)
- Acceleration, \( a = 2 \, \text{m/s}^2 \)
- Retardation, \( b = 4 \, \text{m/s}^2 \)
- Total time of motion, \( T = 3 \, \text{s} \)

### Step 1: Define the Time Intervals
Let \( t_1 \) be the time during which the car accelerates. Therefore, the time during which the car is decelerating is \( T - t_1 = 3 - t_1 \).

### Step 2: Find the Maximum Velocity
Using the formula for velocity under constant acceleration:
\[ v_{\text{max}} = u + a t_1 \]
Since \( u = 0 \):
\[ v_{\text{max}} = 0 + 2 t_1 = 2 t_1 \]

### Step 3: Relate Maximum Velocity to Deceleration
During the deceleration phase, the car comes to rest, so:
\[ 0 = v_{\text{max}} - b(3 - t_1) \]
Substituting \( v_{\text{max}} = 2 t_1 \):
\[ 0 = 2 t_1 - 4(3 - t_1) \]
Expanding and rearranging:
\[ 2 t_1 = 12 - 4 t_1 \]
\[ 2 t_1 + 4 t_1 = 12 \]
\[ 6 t_1 = 12 \]
\[ t_1 = 2 \, \text{s} \]

### Step 4: Calculate Maximum Velocity
Now substituting \( t_1 \) back to find \( v_{\text{max}} \):
\[ v_{\text{max}} = 2 t_1 = 2 \times 2 = 4 \, \text{m/s} \]

### Step 5: Calculate the Distance During Acceleration
The distance traveled during the acceleration phase can be calculated using the formula:
\[ d_1 = u t_1 + \frac{1}{2} a t_1^2 \]
Substituting the values:
\[ d_1 = 0 \times 2 + \frac{1}{2} \times 2 \times (2)^2 \]
\[ d_1 = 0 + \frac{1}{2} \times 2 \times 4 = 4 \, \text{m} \]

### Final Answer
The distance traveled by the car during the time of increasing speed is \( 4 \, \text{m} \).

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