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.
At t = 3 s, the speed of the car is zero. The time for which it increases its speed is

Options

A. 1 s
B. 2 s
C. 3 s
D. 1.5 s

Answer

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

### Step 1: Understand the Motion Phases
The car has two phases of motion:
1. **Acceleration Phase**: The car accelerates from rest at a constant rate of \(2 \, \text{m/s}^2\) for a time \(t_1\).
2. **Deceleration Phase**: The car then decelerates at a constant rate of \(4 \, \text{m/s}^2\) for a time \(t_2\) until it comes to rest.

### Step 2: Total Time of Motion
The total time for which the car is in motion is given as \(3 \, \text{s}\). Therefore, we can write:
\[ t_1 + t_2 = 3 \quad \text{(Equation 1)} \]

### Step 3: Calculate the Final Velocity After Acceleration
Using the first equation of motion:
\[ v = u + at \]
where:
- \(u = 0\) (initial velocity, since the car starts from rest),
- \(a = 2 \, \text{m/s}^2\),
- \(t = t_1\).

Thus, the final velocity \(v\) after the acceleration phase is:
\[ v = 0 + 2t_1 = 2t_1 \quad \text{(Equation 2)} \]

### Step 4: Calculate the Time of Deceleration
During the deceleration phase, we again use the first equation of motion:
\[ v = u + at \]
where:
- \(u = 2t_1\) (initial velocity at the start of deceleration),
- \(a = -4 \, \text{m/s}^2\) (deceleration),
- \(v = 0\) (final velocity, since the car comes to rest).

Thus, we can write:
\[ 0 = 2t_1 - 4t_2 \]
Rearranging gives:
\[ 4t_2 = 2t_1 \quad \Rightarrow \quad t_2 = \frac{t_1}{2} \quad \text{(Equation 3)} \]

### Step 5: Substitute \(t_2\) in Total Time Equation
Substituting \(t_2\) from Equation 3 into Equation 1:
\[ t_1 + \frac{t_1}{2} = 3 \]
To solve this, we can express it as:
\[ \frac{2t_1 + t_1}{2} = 3 \]
\[ \frac{3t_1}{2} = 3 \]
Multiplying both sides by 2:
\[ 3t_1 = 6 \quad \Rightarrow \quad t_1 = 2 \, \text{s} \]

### Step 6: Find \(t_2\)
Using Equation 3 to find \(t_2\):
\[ t_2 = \frac{t_1}{2} = \frac{2}{2} = 1 \, \text{s} \]

### Conclusion
The time for which the car increases its speed (the acceleration phase) is:
\[ \boxed{2 \, \text{s}} \]