Answer is: option1

I only

Solution:

The velocity function is the derivative of the position function: \[ v(t) = \frac{d}{dt} [(t - 2)^3 (t - 6)] \] Using the product rule: \[ v(t) = 3(t - 2)^2 (t - 6) + (t - 2)^3 (1) \] Factor out \( (t - 2)^2 \): \[ v(t) = (t - 2)^2 [3(t - 6) + (t - 2)] \] Simplify inside the brackets: \[ 3t - 18 + t - 2 = 4t - 20 \] Thus, we have: \[ v(t) = (t - 2)^2 (4t - 20) \] Factor out the 4: \[ v(t) = 4 (t - 2)^2 (t - 5) \] A particle is at rest when \( v(t) = 0 \): \[ 4 (t - 2)^2 (t - 5) = 0 \] This occurs when: \[ (t - 2)^2 = 0 \quad \text{or} \quad (t - 5) = 0 \] \[ t = 2, \quad t = 5 \]

Thus, the particle is at rest at \( t = 2 \) and \( t = 5 \), not \( t = 6 \). So statement II is false.

A particle changes direction when \( v(t) \) changes sign. Since \( (t - 2)^2 \) is always non-negative and equals zero at \( t = 2 \), the sign of \( v(t) \) around \( t = 2 \) depends on \( (t - 5) \).

1. For \( t < 5 \), \( (t - 5) \) is negative, so \( v(t) \leq 0 \).
2. For \( t > 5 \), \( (t - 5) \) is positive, so \( v(t) \geq 0 \).

Because \( (t - 2)^2 \) is always positive except at \( t = 2 \), the velocity does not change sign at \( t = 2 \), meaning the particle does not change direction at \( t = 2 \).
Thus, statement III is false. For \( t > 5 \):

1. \( (t - 2)^2 \) is positive.
2. \( (t - 5) \) is positive.

Thus, \( v(t) \) is positive, meaning the particle is moving to the right. Statement I is true.

The correct answer is: I only