A particle moves along a straight line with position \( x(t) = t^3 - 6t^2 + 9t + 2 \) (where \( x \) is in meters and \( t \) is in seconds). Determine the ratio of the total distance traveled by the particle to the magnitude of its displacement during the time interval from \( t = 0 \text{ s} \) to \( t = 2 \text{ s} \).

Velocity \( v(t) = 3t^2 - 12t + 9 = 3(t-1)(t-3) \). Velocity becomes zero at \( t = 1 \text{ s} \). Positions: \( x(0) = 2 \text{ m} \), \( x(1) = 1 - 6 + 9 + 2 = 6 \text{ m} \), \( x(2) = 8 - 24 + 18 + 2 = 4 \text{ m} \). Distance from \( t=0 \) to \( 1 \): \( |6-2| = 4 \text{ m} \); from \( t=1 \) to \( 2 \): \( |4-6| = 2 \text{ m} \). Total distance = \( 6 \text{ m} \). Displacement = \( x(2) - x(0) = 2 \text{ m} \). Ratio = \( 6 / 2 = 3 \).