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 maximum value of position co-ordinate of particle on positive `x`-axis is

Options

A. `1m`
B. `2m`
C. `1/2m`
D. `4m`

Answer

To find the maximum position coordinate of the particle given by the equation \( x = 2(t - t^2) \), we will follow these steps:

### Step 1: Differentiate the position function
We start by differentiating the position function \( x(t) = 2(t - t^2) \) with respect to time \( t \).

\[ \frac{dx}{dt} = \frac{d}{dt}[2(t - t^2)] = 2\left(\frac{d}{dt}[t] - \frac{d}{dt}[t^2]\right) \]
Using the power rule for differentiation:
\[ \frac{dx}{dt} = 2(1 - 2t) = 2 - 4t \]

### Step 2: Set the derivative to zero
To find the critical points, we set the derivative equal to zero:

\[ 2 - 4t = 0 \]

### Step 3: Solve for \( t \)
Solving the equation for \( t \):

\[ 4t = 2 \implies t = \frac{1}{2} \text{ seconds} \]

### Step 4: Verify if it is a maximum
To confirm that this critical point is indeed a maximum, we will take the second derivative of \( x(t) \).

\[ \frac{d^2x}{dt^2} = \frac{d}{dt}[2 - 4t] = -4 \]

Since the second derivative is negative (\( -4 < 0 \)), this indicates that the function has a maximum at \( t = \frac{1}{2} \) seconds.

### Step 5: Calculate the maximum position
Now we substitute \( t = \frac{1}{2} \) back into the original position equation to find the maximum position \( x \):

\[ x\left(\frac{1}{2}\right) = 2\left(\frac{1}{2} - \left(\frac{1}{2}\right)^2\right) \]
Calculating this:

\[ x\left(\frac{1}{2}\right) = 2\left(\frac{1}{2} - \frac{1}{4}\right) = 2\left(\frac{1}{4}\right) = \frac{1}{2} \text{ meters} \]

### Conclusion
The maximum value of the position coordinate of the particle on the positive x-axis is \( \frac{1}{2} \) meters.

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