A particle is moving along `x`-axis with constant acceleration. At `t=0`, the particle is at `x=3m` and `(dx)/(dt)=+4m//s`. The maximum value of `x` co-ordiante of the particle is observed 2 seconds later. Starting from `t=0` sec, after what time, particle reaches its initial position again?

Options

A. `4s`
B. `6s`
C. `8s`
D. `12s`

Answer

To solve the problem step by step, we will analyze the motion of the particle under constant acceleration.

### Step 1: Understand the Initial Conditions
At time \( t = 0 \):
- The position of the particle is \( x_0 = 3 \, \text{m} \).
- The initial velocity is \( v_0 = \frac{dx}{dt} = 4 \, \text{m/s} \).

### Step 2: Determine the Acceleration
The problem states that the maximum value of the \( x \)-coordinate is observed 2 seconds later. This indicates that the particle comes to rest at this maximum position. We can use the equation of motion:
\[ v = u + at \]
At \( t = 2 \, \text{s} \), the final velocity \( v = 0 \):
\[ 0 = 4 + a(2) \]
From this, we can solve for acceleration \( a \):
\[ a = -\frac{4}{2} = -2 \, \text{m/s}^2 \]

### Step 3: Calculate the Maximum Position
Now, we can calculate the maximum position \( x_{max} \) using the equation of motion:
\[ s = ut + \frac{1}{2}at^2 \]
Substituting \( u = 4 \, \text{m/s} \), \( a = -2 \, \text{m/s}^2 \), and \( t = 2 \, \text{s} \):
\[ s = 4(2) + \frac{1}{2}(-2)(2^2) \]
Calculating this gives:
\[ s = 8 - 4 = 4 \, \text{m} \]
Thus, the maximum position \( x_{max} \) is:
\[ x_{max} = x_0 + s = 3 + 4 = 7 \, \text{m} \]

### Step 4: Determine the Time to Return to Initial Position
The particle will return to its initial position after reaching the maximum position. The time taken to return to the initial position can be calculated by finding the time taken to travel back from \( x_{max} = 7 \, \text{m} \) to \( x_0 = 3 \, \text{m} \).

Using the same equation of motion:
\[ s = ut + \frac{1}{2}at^2 \]
Here, the initial velocity \( u = 0 \) (at the maximum position) and the distance \( s = -4 \, \text{m} \) (since it is returning to the left):
\[ -4 = 0 + \frac{1}{2}(-2)t^2 \]
This simplifies to:
\[ -4 = -t^2 \quad \Rightarrow \quad t^2 = 4 \quad \Rightarrow \quad t = 2 \, \text{s} \]

### Step 5: Total Time Calculation
The total time taken to return to the initial position:
\[ t_{total} = t_{to\,max} + t_{to\,initial} = 2 + 2 = 4 \, \text{s} \]

### Final Answer
The particle reaches its initial position again after \( 4 \, \text{s} \).

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