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 use the equations of motion under constant acceleration.

### Step 1: Identify the initial conditions
At time \( t = 0 \):
- Position \( x_0 = 3 \, \text{m} \)
- Velocity \( u = \frac{dx}{dt} = 4 \, \text{m/s} \)

### Step 2: Determine the acceleration
The particle comes to rest after 2 seconds. Therefore, we can use the equation of motion:
\[ v = u + at \]
where:
- \( v = 0 \, \text{m/s} \) (final velocity when the particle comes to rest)
- \( u = 4 \, \text{m/s} \) (initial velocity)
- \( t = 2 \, \text{s} \) (time to come to rest)

Substituting the values:
\[ 0 = 4 + a(2) \]
Rearranging gives:
\[ a(2) = -4 \quad \Rightarrow \quad a = -2 \, \text{m/s}^2 \]

### Step 3: Calculate the distance traveled in the first 2 seconds
Using the equation:
\[ s = ut + \frac{1}{2} a t^2 \]
Substituting the known values:
\[ s = 4(2) + \frac{1}{2}(-2)(2^2) \]
Calculating:
\[ s = 8 - 4 = 4 \, \text{m} \]
Thus, the particle moves 4 m in the first 2 seconds.

### Step 4: Determine the maximum position
The maximum position \( x_{max} \) after 2 seconds is:
\[ x_{max} = x_0 + s = 3 + 4 = 7 \, \text{m} \]

### Step 5: Calculate the time to return to the initial position
Now, the particle will move back to the initial position \( x_0 = 3 \, \text{m} \) from \( x_{max} = 7 \, \text{m} \). The distance to cover is:
\[ d = x_{max} - x_0 = 7 - 3 = 4 \, \text{m} \]
Now, the particle starts from rest at \( x_{max} \) with an acceleration of \( 2 \, \text{m/s}^2 \) (the same magnitude but positive since it is now accelerating back towards the initial position).

Using the same equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
where \( u = 0 \) (initial velocity at \( x_{max} \)):
\[ 4 = 0 \cdot t + \frac{1}{2}(2)t^2 \]
This simplifies to:
\[ 4 = t^2 \]
Thus:
\[ t = 2 \, \text{s} \]

### Step 6: Total time to return to the initial position
The total time taken to return to the initial position is:
\[ t_{total} = t_1 + t_2 = 2 + 2 = 4 \, \text{s} \]

### Final Answer
The particle reaches its initial position again after **4 seconds**.

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