This problem involves using integrals to reason about a particle's position from its velocity function. The position at time t is the initial position plus the integral of velocity from the starting time to t, as integration accumulates the change in position. For v(t) = 4 - t², we compute x(2) = ∫ from 0 to 2 of (4 - t²) dt, which evaluates to [4t - t³/3] from 0 to 2 = 8 - 8/3 = 16/3. This connects velocity to position because the definite integral gives the net displacement over the interval. A tempting distractor is 8, which might come from integrating only the 4t term and ignoring the -t² part. The transferable strategy is to interpret the net displacement as the signed area under the velocity-time graph.

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 sinusoidal velocity. Position x(1) = 10 + ∫ from 0 to 1 of 2 sin(πt) dt = 10 + [- (2/π) cos(πt)] from 0 to 1 = 10 + (2/π)(2) = 10 + 4/π. Integration here accumulates the oscillating velocity into net displacement. The antiderivative directly links the periodic function to position change. A tempting distractor like 10 + 2/π might forget the factor of 2 in the amplitude, but that underestimates the integral. 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 when velocity is non-negative. Since $v(t) \geq 0$, x(3) = 2 + $\int_0^3 v(t) , dt$ = 2 + 7 = 9. The integral directly gives positive displacement. This connects the given integral to position. A tempting distractor like 5 might subtract instead of add, but positivity ensures forward motion. 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 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 tests the skill of using integrals to reason about motion, specifically finding position using arctangent integral. x(1) = 0 + ∫0 to 1 1/(1+t²) dt = [arctan t] from 0 to 1 = π/4. This connects the rational velocity to position via known antiderivative. Integration accumulates the decreasing speed. A tempting distractor like 1 might approximate without integrating, but exact is arctan. 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 displacement from acceleration through double integration. v(t) = ∫0 to t 2s ds = t², then displacement = ∫0 to 2 s² ds = [s³/3] from 0 to 2 = 8/3. This connects acceleration to velocity to position. The process accumulates changes hierarchically. A tempting distractor like 4 might come from integrating acceleration directly without finding velocity first, but that skips a step. 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 displacement from its velocity function. The displacement on [0,1] is the integral of v(t) = 2 cos(πt) dt from 0 to 1, evaluating to [ (2/π) sin(πt) ] from 0 to 1 = 0. Integration connects velocity to position change by accumulating signed areas. Here, the positive and negative parts of the cosine function cancel out over the interval. A tempting distractor is 2, possibly from integrating a constant 2 without the cosine term. The transferable strategy is to compute the signed area under the velocity curve for net displacement, considering cancellations.

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.

This problem tests the skill of using integrals to reason about motion, specifically calculating total distance from positive sine velocity. Since $\sin t \geq 0$ on $[0, \pi]$, total distance = $\int_0^\pi \sin t , dt = [-\cos t]_0^\pi = 2$. No sign change, so total equals net. Integration captures the full area. A tempting distractor like $\pi$ might confuse with the interval length, but integral is area. A transferable strategy is to interpret the definite integral of $|\text{velocity}|$ as the total area under the curve, representing distance traveled.