This problem involves using integrals to reason about position from piecewise velocity. Compute x(2) = ∫0 to1 t dt + ∫1 to2 (2-t) dt = [t²/2]0 to1 + [2t - t²/2]1 to2 = 0.5 + 0.5 = 1. Integration accumulates change over pieces. It connects velocity segments to total position. A tempting distractor is 2, possibly from integrating t over full [0,2]. The transferable strategy is to find the signed area under the velocity curve, piecewise, for net displacement.

This problem tests the skill of using integrals to reason about motion, specifically computing net displacement from a piecewise velocity. Displacement is ∫ from 0 to 4 of (|t-2| - 1) dt, which splits into ∫0 to 2 (1-t) dt + ∫2 to 4 (t-3) dt = 0 + 0 = 0. This shows how positive and negative areas cancel in net displacement. Integration captures the overall change despite direction changes. A tempting distractor like -4 might come from ignoring the absolute value and integrating directly, but that misses the piecewise definition. A transferable strategy is to interpret the definite integral of velocity as the net area under the curve, representing displacement.

This problem involves using integrals to reason about a particle's position from its velocity function. We compute $x(\pi) = \int_0^\pi(1 + \sin t) , dt = [t - \cos t]_0^\pi = (\pi - (-1)) - (0 - 1) = \pi + 2$. Integration connects velocity to position change via the antiderivative. The initial $x(0)=0$ is inherent in the definite integral. A tempting distractor is $\pi$, perhaps from integrating only the 1 and ignoring $\sin t$. The transferable strategy is to find the signed area under the velocity curve for net displacement.

This problem involves using integrals to reason about position from piecewise velocity. We compute x(5) = x(0) + ∫0 to2 3 dt + ∫2 to5 0 dt = 4 + 6 + 0 = 10. Integration accumulates position change over each piece. Initial condition starts the calculation. A tempting distractor is 9, perhaps from multiplying 3 by 3 instead of 2. The transferable strategy is to find the signed area under the velocity curve, piecewise if necessary, for net displacement.

This problem involves using integrals to find position from velocity with a trigonometric function. We calculate position using $x(t) = x(0) + \int_0^t v(s),ds$. With $v(t) = 2\sin t$ and $x(0) = 0$, we get $x(\pi) = 0 + \int_0^\pi 2\sin t,dt = -2\cos t|_0^\pi = -2\cos(\pi) - (-2\cos(0)) = -2(-1) - (-2)(1) = 2 + 2 = 4$. Students choosing C might incorrectly evaluate $\cos(\pi)$ as 0 instead of -1, leading to an answer of 0. The key insight is that the integral of sine is negative cosine, and careful evaluation at the bounds is crucial for trigonometric integrals.

This problem requires finding total distance traveled using integrals, which differs from displacement. Total distance is $\int_a^b |v(t)|,dt$. With $v(t) = 1-t$, velocity changes sign at $t=1$ (where $v(1)=0$). For $0\le t\le 1$, $v(t)\ge 0$, and for $1\le t\le 2$, $v(t)\le 0$. Total distance = $\int_0^1 (1-t),dt + \int_1^2 |1-t|,dt = \int_0^1 (1-t),dt + \int_1^2 (t-1),dt = \frac{1}{2} + \frac{1}{2} = 1$. Students choosing A might calculate displacement (net change) instead of total distance. Remember that total distance requires splitting the integral where velocity changes sign and using absolute value.

This problem involves using integrals to reason about position from velocity. Compute x(4) = 3 + ∫0 to4 (t/2) dt = 3 + [t²/4] from 0 to 4 = 3 + 4 = 7. Integration links velocity to position change. Initial condition is added to the definite integral. A tempting distractor is 11, perhaps from integrating t instead of t/2. The transferable strategy is to find the signed area under the velocity curve for net displacement.

This problem tests the skill of using integrals to reason about motion, specifically finding position from linear velocity. x(2) = -3 + ∫0 to 2 t dt = -3 + [t²/2] from 0 to 2 = -3 + 2 = -1. Integration adds the displacement to initial position. This connects velocity directly to position change. A tempting distractor like -5 might come from subtracting the integral instead of adding, but that reverses the direction. A transferable strategy is to interpret the definite integral of velocity as the net area under the curve, representing displacement.

This problem tests the skill of using integrals to reason about motion, specifically computing displacement from cosine velocity. Displacement = ∫0 to π cos t dt = [sin t] from 0 to π = 0 - 0 = 0. This shows how symmetric positive and negative areas cancel. Integration captures the net zero change. A tempting distractor like 2 might come from total distance instead of net, but the question asks for displacement. A transferable strategy is to interpret the definite integral of velocity as the net area under the curve, representing displacement.

This problem tests the skill of using integrals to reason about motion, specifically finding position from velocity. The position function x(t) is obtained by integrating the velocity v(t) over time and adding the initial position, so x(2) = x(0) + ∫ from 0 to 2 of (3 - 2t) dt. Computing the integral gives [3t - t²] from 0 to 2, which is (6 - 4) - 0 = 2, so x(2) = 5 + 2 = 7. This connection shows how the net change in position is the area under the velocity curve. A tempting distractor like 5 might come from forgetting to add the integral to the initial position, but that ignores the displacement caused by velocity. A transferable strategy is to interpret the definite integral of velocity as the net area under the curve, representing displacement.