Position, Velocity, and Acceleration Using Integrals - AP Calculus AB

1. A particle's velocity is $v(t)=\dfrac{t}{2}$ for $0\le t\le4$, with $x(0)=3$. What is $x(4)$?
   Answer: 7 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.
2. A particle's acceleration is $a(t)=4$ (m/s$^2$) with $v(0)=-1$ (m/s). What is $v(2)$?
   Answer: $7$ m/s This problem tests using integrals to find velocity from acceleration. Since acceleration is the derivative of velocity, we integrate to find velocity: $v(t) = v(0) + \int_0^t a(s)\,ds$. With constant acceleration $a(t) = 4$ and $v(0) = -1$, we get $v(2) = -1 + \int_0^2 4\,dt = -1 + 4t|_0^2 = -1 + 8 = 7$ m/s. Students choosing B might forget to include the initial velocity and only compute the change in velocity. Remember that integrating acceleration gives the change in velocity, which must be added to the initial velocity to find the final velocity.
3. A particle has velocity $v(t)=2(t-1)$ for $0\le t\le2$ and $x(0)=0$. What is the displacement on $[0,2]$?
   Answer: 0 This problem involves using integrals to reason about displacement from velocity. The displacement on $[0,2]$ is $\int_0^2 2(t-1) \, dt = [t^2 - 2t]_0^2 = 0$. Integration links velocity to net position change, with areas canceling due to sign change. Velocity is negative then positive, returning to start. A tempting distractor is 2, possibly from ignoring the sign change and integrating |2(t-1)| incorrectly. The transferable strategy is to compute the signed area under the velocity curve for net displacement.
4. A particle has velocity $v(t)=3-2t$ (m/s) for $0\le t\le2$ and $x(0)=5$. What is $x(2)$?​​
   Answer: $7$ This problem tests using integrals to find position from velocity. Since velocity is the derivative of position, we find position by integrating velocity: $x(t) = x(0) + \int_0^t v(s)\,ds$. With $v(t) = 3-2t$ and $x(0) = 5$, we get $x(2) = 5 + \int_0^2 (3-2t)\,dt = 5 + [3t - t^2]_0^2 = 5 + (6-4) = 7$. Students choosing B might forget to add the initial position $x(0) = 5$ and only compute the integral value of 2. The key strategy is to remember that integrating velocity gives displacement (change in position), which must be added to the initial position to find the final position.
5. A particle's velocity is $v(t)=2\sin t$ for $0\le t\le\pi$. If $x(0)=1$, what is $x(\pi)$?​
   Answer: $5$ This problem requires integrating velocity to find position change. We have x(π) = x(0) + ∫[0 to π] 2sin(t) dt = 1 + [-2cos(t)]₀^π = 1 + (-2cos(π) - (-2cos(0))) = 1 + (-2(-1) - (-2(1))) = 1 + (2 + 2) = 1 + 4 = 5. The integral of sin(t) is -cos(t), and we must evaluate at the bounds carefully. A common mistake is forgetting the negative sign in the antiderivative or miscalculating cos(π) = -1. The strategy: when integrating trigonometric velocities, carefully track signs in both the antiderivative and the evaluation at bounds.
6. A particle's velocity is $v(t)=2-2\cos t$ for $0\le t\le\pi$, with $x(0)=0$. What is the displacement on $[0,\pi]$?
   Answer: $2\pi$ This problem involves using integrals to reason about displacement from velocity. The displacement on [0,π] is ∫0 to π (2 - 2 cos t) dt = [2t - 2 sin t] from 0 to π = 2π. Integration connects velocity to net change with antiderivative. The function is always positive, no cancellation. A tempting distractor is π, perhaps from integrating 1 - cos t instead. The transferable strategy is to compute the signed area under the velocity curve for net displacement.
7. A particle has acceleration $a(t)=6t$ with $v(0)=2$ and $x(0)=0$. What is $x(1)$?
   Answer: 3 This problem tests the skill of using integrals to reason about motion, specifically finding position from linear acceleration. v(t) = 2 + 3t², x(1) = ∫0 to 1 (2 + 3s²) ds = [2s + s³] from 0 to 1 = 3. Double integration links acceleration to position. It accumulates quadratic velocity. A tempting distractor like 2 might stop at velocity integral, but position needs another. A transferable strategy is to interpret the definite integral of velocity as the net area under the curve, representing displacement.
8. A particle's velocity is $v(t)=2\cos(\pi t)$ (m/s) for $0\le t\le1$, with $x(0)=0$. What is the displacement on $[0,1]$?
   Answer: 0 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.
9. A particle's velocity is $v(t)=\begin{cases}t,&0\le t\le1\\2-t,&1<t\le2\end{cases}$ with $x(0)=0$. What is $x(2)$?
   Answer: 1 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.
10. A particle's velocity is $v(t)=\sin t$ for $0\le t\le\pi$. What is the total distance traveled on $[0,\pi]$?
    Answer: 2 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.
11. A particle has velocity $v(t)=3-2t$ (m/s) for $0\le t\le2$ and $x(0)=5$. What is $x(2)$?​
    Answer: $7\text{ m}$ This problem requires using integrals to find position from velocity. Since velocity is the derivative of position, we find position by integrating velocity: x(t) = x(0) + ∫[0 to t] v(s) ds. Here, x(2) = 5 + ∫[0 to 2] (3-2t) dt = 5 + [3t - t²]₀² = 5 + (6 - 4) - 0 = 5 + 2 = 7. A common error is forgetting the initial position x(0) = 5, which would give just 2 as the answer. Remember: when integrating velocity to find position, always add the initial position to the definite integral of velocity.
12. A particle moves with velocity $v(t)=2\sin t$ for $0\le t\le\pi$ and $x(0)=0$. What is $x(\pi)$?​​
    Answer: $4$ 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.
13. Velocity is given by $v(t)=|t-2|-1$ (m/s) for $0\le t\le4$, and $x(0)=0$. What is the displacement on $[0,4]$?
    Answer: 0 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.
14. A particle's velocity is $v(t)=t-3$ (m/s) for $0 \le t \le 5$. What is the displacement on $[0,5]$?
    Answer: $-\dfrac{5}{2}$ This problem tests the skill of using integrals to reason about motion, specifically computing displacement from linear velocity. Displacement = $\int_0^5 (t-3) \, dt$ = $\left[ \frac{t^2}{2} - 3t \right]_0^5$ = $(12.5 - 15) = -2.5$. Integration captures net change with sign change. It shows backward then forward motion. A tempting distractor like -5 might double the negative area, but the integral is direct. A transferable strategy is to interpret the definite integral of velocity as the net area under the curve, representing displacement.
15. A particle's velocity is $v(t)=4-t^2$ (m/s) for $0\le t\le2$, with $x(0)=0$. What is $x(2)$?
    Answer: $\dfrac{16}{3}$ 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.
16. A particle has velocity $v(t)=\dfrac{1}{1+t^2}$ (m/s) for $0\le t\le1$, with $x(0)=0$. What is $x(1)$?
    Answer: $\dfrac{\pi}{4}$ 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.
17. A particle moves with velocity $v(t)=\cos t$ for $0\le t\le \pi$. What is the net displacement on $[0,\pi]$?​
    Answer: $0$ This problem involves finding displacement by integrating velocity. The net displacement is ∫[0 to π] cos(t) dt = [sin(t)]₀^π = sin(π) - sin(0) = 0 - 0 = 0. Since cos(t) is positive on [0, π/2] and negative on [π/2, π], the particle moves forward then backward by equal amounts. A tempting error is to think the answer should be 2 (the total distance traveled), but displacement measures net change in position, not total distance. Remember: displacement is the signed area under the velocity curve, where areas below the t-axis subtract from those above.
18. A particle has velocity $v(t)=3-2t$ (m/s) for $0\le t\le2$ and $x(0)=5$. What is $x(2)$?
    Answer: 7 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.
19. A particle has velocity $v(t)=1-t$ for $0\le t\le2$ and $x(0)=0$. What is $x(2)$?​
    Answer: $0$ This problem requires integrating velocity to find position. We have x(2) = x(0) + ∫[0 to 2] (1-t) dt = 0 + [t - t²/2]₀² = (2 - 4/2) - 0 = 2 - 2 = 0. The velocity v(t) = 1-t is positive for t < 1 and negative for t > 1, meaning the particle moves forward then backward, returning to its starting position. A common mistake is evaluating the antiderivative incorrectly, perhaps getting [t - t²]₀² = -2. The key insight: when velocity changes sign, the particle can return to its starting point if the positive and negative areas under the velocity curve are equal.
20. A particle has velocity $v(t)=2-\sqrt{t}$ (m/s) for $0\le t\le4$, with $x(0)=0$. What is $x(4)$?
    Answer: $\dfrac{8}{3}$ This problem tests the skill of using integrals to reason about motion, specifically computing position from velocity. x(4) = 0 + ∫0 to 4 (2 - √t) dt = [2t - (2/3)$t^{3/2}$] from 0 to 4 = 8 - (2/3)*8 = 8 - 16/3 = 8/3. This integral accumulates the decreasing velocity. It links the function to net displacement. A tempting distractor like 4 might ignore the square root term properly in antiderivative, but that errors the calculation. A transferable strategy is to interpret the definite integral of velocity as the net area under the curve, representing displacement.

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