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.
If the time of acceleration is `t_(1)`, then the speed of the car at `t=t_(1)` is

Options

A. `2t_(1)`
B. `4t_(1)`
C. `gt 2t_(1)`
D. `lt 2t_(1)`

Answer

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

### Step 1: Understand the motion phases
The car accelerates from rest at a constant rate of \(2 \, \text{m/s}^2\) for a time \(t_1\), and then it decelerates at a constant rate of \(4 \, \text{m/s}^2\) until it comes to rest. The total time of motion is given as \(3 \, \text{s}\).

### Step 2: Set up the equations
1. **Acceleration Phase**:
- Initial velocity \(u = 0\) (since the car starts from rest)
- Acceleration \(a = 2 \, \text{m/s}^2\)
- Time of acceleration \(t = t_1\)
- Final velocity at the end of acceleration \(v = u + at = 0 + 2t_1 = 2t_1\)

2. **Deceleration Phase**:
- The time of deceleration can be expressed as \(t_2 = 3 - t_1\) (since the total time is \(3 \, \text{s}\)).
- Deceleration \(a' = -4 \, \text{m/s}^2\)
- The initial velocity for this phase is the final velocity from the acceleration phase, which is \(v = 2t_1\).

### Step 3: Use the equation of motion for deceleration
Using the equation of motion:
\[ v_f = v_i + a' t_2 \]
Where:
- \(v_f = 0\) (final velocity when the car comes to rest)
- \(v_i = 2t_1\) (initial velocity at the start of deceleration)
- \(a' = -4 \, \text{m/s}^2\)
- \(t_2 = 3 - t_1\)

Substituting these values into the equation:
\[ 0 = 2t_1 - 4(3 - t_1) \]

### Step 4: Solve for \(t_1\)
Expanding and rearranging the equation:
\[ 0 = 2t_1 - 12 + 4t_1 \]
\[ 0 = 6t_1 - 12 \]
\[ 6t_1 = 12 \]
\[ t_1 = 2 \, \text{s} \]

### Step 5: Calculate the speed at \(t = t_1\)
Now, substituting \(t_1\) back into the equation for speed:
\[ v = 2t_1 = 2 \times 2 = 4 \, \text{m/s} \]

### Final Answer
The speed of the car at \(t = t_1\) is \(4 \, \text{m/s}\).

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