A particle moves along the X-axis as `x=u(t-2s)=at(t-2)^2`.

Options

A. the initial velocity of the particle is u
B. the acceleration of the particle is a
C. the acceleration of the particle is 2a
D. at t=2 s particle is at the origin.

Answer

To solve the problem, we need to analyze the motion of a particle described by the equation:

\[ x = u(t - 2) + a(t - 2)^2 \]

where \( u \) is the initial velocity, \( a \) is the acceleration, and \( t \) is the time.

### Step-by-Step Solution:

1. **Identify the Position Function**:
The position of the particle is given by:
\[ x(t) = u(t - 2) + a(t - 2)^2 \]

2. **Differentiate to Find Velocity**:
To find the velocity \( v(t) \), we differentiate the position function with respect to time \( t \):
\[ v(t) = \frac{dx}{dt} = \frac{d}{dt}[u(t - 2) + a(t - 2)^2] \]
Using the product and chain rules:
\[ v(t) = u + 2a(t - 2) \]

3. **Calculate Initial Velocity**:
To find the initial velocity at \( t = 0 \):
\[ v(0) = u + 2a(0 - 2) = u - 4a \]
Since the initial velocity is given as \( u \), we have:
\[ u - 4a = u \implies -4a = 0 \implies a = 0 \]
This indicates that the acceleration \( a \) is zero.

4. **Differentiate to Find Acceleration**:
The acceleration \( a(t) \) is the derivative of the velocity:
\[ a(t) = \frac{dv}{dt} = \frac{d}{dt}[u + 2a(t - 2)] = 2a \]

5. **Check Particle's Position at \( t = 2 \)**:
To check if the particle is at the origin when \( t = 2 \):
\[ x(2) = u(2 - 2) + a(2 - 2)^2 = 0 + 0 = 0 \]
This confirms that the particle is indeed at the origin at \( t = 2 \).

### Summary of Results:
- Initial velocity \( v(0) = u - 4a \) indicates that if \( a \neq 0 \), the initial velocity is not \( u \).
- The acceleration is \( 2a \).
- The particle is at the origin when \( t = 2 \).

### Conclusion:
- The initial velocity is not \( u \) if \( a \) is non-zero.
- The acceleration of the particle is \( 2a \).
- The particle is at the origin at \( t = 2 \).