A particle moves along x-axis as ` x= t (t-2)+(t-2)^(2) ` . Which of the following is true?

Options

A. The initial velocity of particle is 4
B. The accekeration of particle varies with time
C. The particle is at origin at t = 0
D. None of these

Explanation

To solve the problem, we need to analyze the motion of the particle given by the equation \( x = t(t-2) + (t-2)^2 \).

### Step 1: Simplify the equation for \( x \)

We start with the equation:
\[ x = t(t-2) + (t-2)^2 \]
Expanding both terms:
\[ x = t^2 - 2t + (t^2 - 4t + 4) \]
Combining like terms:
\[ x = t^2 - 2t + t^2 - 4t + 4 = 2t^2 - 6t + 4 \]

### Step 2: Calculate the velocity \( v \)

Velocity is the derivative of position with respect to time:
\[ v = \frac{dx}{dt} = \frac{d}{dt}(2t^2 - 6t + 4) \]
Using the power rule:
\[ v = 4t - 6 \]

### Step 3: Calculate the acceleration \( a \)

Acceleration is the derivative of velocity with respect to time:
\[ a = \frac{dv}{dt} = \frac{d}{dt}(4t - 6) \]
Again using the power rule:
\[ a = 4 \]

### Step 4: Determine the initial velocity

To find the initial velocity, we evaluate \( v \) at \( t = 0 \):
\[ v(0) = 4(0) - 6 = -6 \]

### Step 5: Determine the position at \( t = 0 \)

To find the initial position, we evaluate \( x \) at \( t = 0 \):
\[ x(0) = 2(0)^2 - 6(0) + 4 = 4 \]

### Conclusion

From our calculations:
- The initial velocity \( v(0) = -6 \)
- The acceleration \( a = 4 \) is constant.
- The initial position \( x(0) = 4 \) indicates that the particle is not at the origin.

Thus, the correct answer is that all options are wrong, and the correct choice is (d) none of these.

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