The position of a particle is given by
`x=2(t-t^(2))`
where `t` is expressed in seconds and `x` is in metre.
The particle

Options

A. never does to negative `x`-axis
B. never goes to positive `x`-axis
C. starts from the origin goes up to `x=1/2m` in the positive `x` -axis and then moves in opposites direction
D. has zero initial velocity

Answer

To solve the problem, we need to analyze the motion of a particle whose position is given by the equation:

\[ x = 2(t - t^2) \]

where \( t \) is in seconds and \( x \) is in meters. We will go through the steps to understand the motion of the particle.

### Step 1: Identify the Position Function
The position of the particle is given by:

\[ x(t) = 2(t - t^2) \]

This is a quadratic function of time \( t \).

### Step 2: Determine the Velocity Function
To find the velocity of the particle, we need to differentiate the position function with respect to time \( t \):

\[ v(t) = \frac{dx}{dt} = \frac{d}{dt}[2(t - t^2)] \]
\[ v(t) = 2(1 - 2t) \]

### Step 3: Find the Time When Velocity is Zero
To find when the particle stops (i.e., when the velocity is zero), we set the velocity function to zero:

\[ 2(1 - 2t) = 0 \]
\[ 1 - 2t = 0 \]
\[ 2t = 1 \]
\[ t = \frac{1}{2} \text{ seconds} \]

### Step 4: Determine the Position at \( t = 0 \) and \( t = \frac{1}{2} \)
Now, we will find the position of the particle at \( t = 0 \) and \( t = \frac{1}{2} \):

1. At \( t = 0 \):
\[ x(0) = 2(0 - 0^2) = 0 \text{ meters} \]

2. At \( t = \frac{1}{2} \):
\[ x\left(\frac{1}{2}\right) = 2\left(\frac{1}{2} - \left(\frac{1}{2}\right)^2\right) \]
\[ = 2\left(\frac{1}{2} - \frac{1}{4}\right) = 2\left(\frac{1}{4}\right) = \frac{1}{2} \text{ meters} \]

### Step 5: Determine the Position at \( t = 1 \)
Next, we find the position at \( t = 1 \):

\[ x(1) = 2(1 - 1^2) = 2(1 - 1) = 0 \text{ meters} \]

### Step 6: Analyze the Motion
- The particle starts at the origin (0 meters) at \( t = 0 \).
- It moves to \( x = \frac{1}{2} \) meters at \( t = \frac{1}{2} \) seconds, where it comes to a stop (velocity = 0).
- After \( t = \frac{1}{2} \) seconds, the particle starts moving back towards the origin and returns to \( x = 0 \) meters at \( t = 1 \) second.

### Conclusion
The particle starts from the origin, moves to a maximum position of \( \frac{1}{2} \) meters, and then returns to the origin at \( t = 1 \) second.

### Final Answer
The correct statement about the motion of the particle is that it starts from the origin, goes up to \( x = \frac{1}{2} \) meters, and then moves in the opposite direction.