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(D) The position of the particle is given by $x(t) = 2 + t - 3t^2$.$1$. Displacement:The displacement $\Delta x$ in the interval $t = 0$ to $t = 1$ is

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$-2, 2.1$

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(D) The position of the particle is given by $x(t) = 2 + t - 3t^2$.$1$. Displacement:The displacement $\Delta x$ in the interval $t = 0$ to $t = 1$ is given by $x(1) - x(0)$.$x(1) = 2 + 1 - 3(1)^2 = 0 \, m$.$x(0) = 2 + 0 - 3(0)^2 = 2 \, m$.Displacement $\Delta x = 0 - 2 = -2 \, m$.$2$. Distance:The velocity of the particle is $v(t) = \frac{dx}{dt} = 1 - 6t$.The particle changes direction when $v(t) = 0$,which occurs at $t = \frac{1}{6} \, s$.Since $t = \frac{1}{6} \, s$ lies within the interval $[0, 1]$,the total distance is the sum of the absolute displacements in the intervals $[0, 1/6]$ and $[1/6, 1]$.Distance $d = |x(1/6) - x(0)| + |x(1) - x(1/6)|$.$x(1/6) = 2 + 1/6 - 3(1/6)^2 = 2 + 1/6 - 3/36 = 2 + 1/6 - 1/12 = 2 + 1/12 = 25/12 \, m$.$|x(1/6) - x(0)| = |25/12 - 2| = |1/12| = 1/12 \, m$.$|x(1) - x(1/6)| = |0 - 25/12| = 25/12 \, m$.Total distance $d = 1/12 + 25/12 = 26/12 = 13/6 \approx 2.167 \, m$.Rounding to the nearest provided option,the distance is $2.1 \, m$.Thus,the displacement is $-2 \, m$ and the distance is $2.1 \, m$.

The position of a particle along the $x$-axis at time $t$ is given by $x = 2 + t - 3t^2$. The displacement and the distance travelled in the interval $t = 0$ to $t = 1$ are respectively:

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