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.
Let t be a time instant greater than `t_(1)` but less than 3 s. The velocity at this time instant is

Options

A. `4(t-t_(1))`
B. `2t_(1)+4(t-t_(1))`
C. `2t_(1)-4(t-t_(1))`
D. `4t`

Answer

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

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

### Step 2: Finding Time of Acceleration
Let \(t_1\) be the time during which the car accelerates. Since the total time is \(3 \, \text{s}\), the time of deceleration will be \(3 - t_1\).

### Step 3: Finding Maximum Velocity
Using the equation of motion for the acceleration phase:
- Initial velocity \(u = 0\)
- Acceleration \(a = 2 \, \text{m/s}^2\)
- Time \(t = t_1\)

The final velocity \(v\) at time \(t_1\) can be calculated using the formula:
\[ v = u + at \]
Substituting the values:
\[ v = 0 + 2 \cdot t_1 = 2t_1 \]
Thus, the maximum velocity \(v_{\text{max}} = 2t_1\).

### Step 4: Analyzing the Deceleration Phase
During the deceleration phase, the initial velocity \(u\) is the maximum velocity \(2t_1\), and the deceleration \(a = -4 \, \text{m/s}^2\). The time for this phase is \(t - t_1\).

### Step 5: Finding Velocity at Time \(t\)
Using the equation of motion for the deceleration phase:
\[ v = u + at \]
Substituting the values:
\[ v = 2t_1 - 4(t - t_1) \]
This simplifies to:
\[ v = 2t_1 - 4t + 4t_1 = 6t_1 - 4t \]

### Step 6: Conclusion
The velocity at any time \(t\) (where \(t_1 < t < 3\)) is given by:
\[ v = 6t_1 - 4t \]