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

1. A bacterium population changes at $P'(t)=500-20t$ cells/hour for $0\le t\le10$; which gives net change in population?
   Answer: $\int_0^{10} (500-20t)\,dt$ This problem requires using accumulated-change reasoning to find the net change from a given rate function. The integral of P'(t) = 500 - 20t from 0 to 10 accumulates the rate of change to find the net population increase. This positive rate over the interval means the integral gives the total growth. By the Fundamental Theorem, this equals P(10) - P(0), but the integral is the direct expression from the rate. Choice C is a tempting distractor with absolute value, but not needed for net change. In general, to convert a rate function to total net change, integrate it over the interval without absolute values.
2. A particle's velocity is $v(t)=\sqrt{t}-2$ m/s for $0\le t\le9$; what is its displacement on $[0,9]$?
   Answer: $\int_0^9 (\sqrt{t}-2)\,dt$ This problem requires using accumulated-change reasoning to find the displacement from a velocity function. The definite integral of the velocity function over the time interval [0,9] represents the net change in position, as velocity is the rate of change of position with respect to time. By the Fundamental Theorem of Calculus, ∫ from 0 to 9 of v(t) dt equals the position at t=9 minus the position at t=0. This captures the net effect with v(t) negative initially and positive later. A tempting distractor is choice A, which uses absolute value and computes total distance rather than net displacement. In general, to find the total change in a quantity from its rate of change, integrate the rate function over the interval.
3. A ship's velocity is $v(t)=\dfrac{10}{1+t^2}$ km/h for $0\le t\le3$; what distance does it travel?
   Answer: $\int_0^3 \dfrac{10}{1+t^2}\,dt$ km This problem requires accumulated-change reasoning to find distance traveled from the ship's velocity. Since velocity $v(t) = \dfrac{10}{1+t^2}$ km/h gives the rate of position change, integrating $\int_0^3 \dfrac{10}{1+t^2}\,dt$ accumulates all velocity values from t = 0 to t = 3 hours, yielding total distance in kilometers. The velocity is always positive, so distance equals displacement here. Choice D incorrectly divides by 3, which would give average velocity, not distance traveled. The key insight: when finding total change from a rate, integrate the rate function - the units automatically work out correctly (km/h × h = km).
4. A car's velocity is $v(t)=3t^2-12t+5$ m/s for $0\le t\le4$; what is its displacement over $[0,4]$?
   Answer: $\int_0^4 (3t^2-12t+5)\,dt$ This problem requires using accumulated-change reasoning to find the net change from a given rate function. The definite integral of the velocity function v(t) over [0,4] directly gives the displacement, as it accumulates the instantaneous changes in position. Since velocity can be positive or negative, the integral accounts for direction, providing the net displacement rather than total distance. Here, ∫ from 0 to 4 of (3t² - 12t + 5) dt computes this net change accurately. A tempting distractor is choice C, which uses the absolute value and would give the total distance traveled instead of net displacement. In general, to convert a rate function to total net change, integrate it over the interval without absolute values.
5. A boat's velocity is $v(t)=2\cos(t)-1$ m/s for $0\le t\le2\pi$; what is its displacement?
   Answer: $\int_0^{2\pi} (2\cos(t)-1)\,dt$ This problem requires using accumulated-change reasoning to find the displacement from a velocity function. The definite integral of the velocity function over the time interval [0,2π] represents the net change in position, as velocity is the rate of change of position with respect to time. By the Fundamental Theorem of Calculus, ∫ from 0 to 2π of v(t) dt equals the position at t=2π minus the position at t=0. This integral captures the net effect despite v(t) changing signs during the interval. A tempting distractor is choice A, which uses absolute value and calculates total distance instead of net displacement. In general, to find the total change in a quantity from its rate of change, integrate the rate function over the interval.
6. A particle moves with velocity $v(t)=\frac{2}{1+t^2}$ m/s for $0\le t\le3$; what is its displacement?
   Answer: $\int_0^3 \frac{2}{1+t^2}\,dt$ This problem requires using accumulated-change reasoning to find the displacement from a velocity function. The definite integral of the velocity function over the time interval [0,3] represents the total change in position, as velocity is the rate of change of position with respect to time. By the Fundamental Theorem of Calculus, $\int_0^3 v(t) \, dt$ equals the position at t=3 minus the position at t=0. Since v(t) is positive throughout, the integral gives both net displacement and total distance. A tempting distractor is choice C, which subtracts values of v(t) at endpoints, confusing velocity with position. In general, to find the total change in a quantity from its rate of change, integrate the rate function over the interval.
7. A train's velocity is $v(t)=4\sqrt{t}$ m/s for $0\le t\le9$; what is the total change in position?
   Answer: $\int_0^9 4\sqrt{t}\,dt$ m This problem requires accumulated-change reasoning to find the train's position change from its velocity. The velocity v(t) = 4√t represents the rate of position change, so ∫₀⁹(4√t)dt accumulates all velocity values from t = 0 to t = 9, giving total displacement in meters. Since velocity is always non-negative (√t ≥ 0), the displacement equals the distance traveled. Choice E incorrectly divides by 9, which would give average velocity over the interval, not total position change. The fundamental principle: to find accumulated change from a rate function, integrate the rate - here, integrating velocity over time yields displacement.
8. A car's velocity is $v(t)=3t^2-12t+5$ m/s for $0 \le t \le 4$; what is its net displacement?
   Answer: $\displaystyle \int_0^4 (3t^2-12t+5)\,dt$ This problem requires using accumulated-change reasoning to find net displacement from velocity. The integral of velocity over time gives the total change in position, so we need $\int_0^4 (3t^2 - 12t + 5) \, dt$. This integral represents the net displacement because velocity is the rate of change of position, and integrating a rate gives the total change. The absolute value integral (choice B) would give total distance traveled, not net displacement, which is different when velocity changes sign. When given a rate function, integrate it over the time interval to find the total accumulated change.
9. A car's acceleration is $a(t)=2t$ (m/s$^2$) and $v(0)=3$ m/s. What is the car's change in velocity on $[0,4]$?
   Answer: $\displaystyle \int_{0}^{4} 2t\,dt$ This problem uses accumulated-change reasoning to find velocity change from acceleration. Since acceleration a(t)=2t represents the rate of velocity change, the integral ∫₀⁴ 2t dt gives the total change in velocity from t=0 to t=4. The integral accumulates all instantaneous acceleration values over the time interval, yielding the net velocity change. Choice C incorrectly adds the initial velocity v(0)=3, but the question asks for change in velocity, not final velocity. To find how much a quantity changes when given its rate, integrate the rate function.
10. A train's velocity is $v(t)=10-\tfrac{t}{2}$ m/s for $2 \le t \le 8$; what is net displacement?
    Answer: $\displaystyle\int_2^8 (10-\tfrac{t}{2})\,dt$ This problem uses accumulated-change reasoning to find the train's net displacement from velocity. Since velocity $v(t) = 10 - \tfrac{t}{2}$ is the rate of change of position, integrating from 2 to 8 gives the net displacement: $\int_2^8 (10 - \tfrac{t}{2}) \, dt$. The velocity changes from positive to negative at $t = 20$, but since we're only considering $t \in [2,8]$, the velocity remains positive throughout our interval. Choice C ($v(8) - v(2)$) would incorrectly give the change in velocity rather than position. When given a velocity function, integrate it over the specified interval to find net position change.
11. A runner's velocity is $v(t)=5\sin(t)$ m/s for $0\le t\le\pi$; what is the runner's net change in position?
    Answer: $\int_0^{\pi} 5\sin(t)\,dt$ This problem requires using accumulated-change reasoning to find the net change in position from a velocity function. The definite integral of the velocity function over the time interval $[0, \pi]$ represents the net change in position, as velocity is the rate of change of position with respect to time. By the Fundamental Theorem of Calculus, $\int_0^\pi v(t) \, dt$ equals the position at $t=\pi$ minus the position at $t=0$. This integral incorporates the direction of motion through the sign of $v(t)$, providing the net displacement. A tempting distractor is choice C, which uses the absolute value and computes total distance rather than net change. In general, to find the total change in a quantity from its rate of change, integrate the rate function over the interval.
12. A cyclist's velocity is $v(t)=\sin(t^2)$ m/s for $0\le t\le2$; which expression gives net change in position?
    Answer: $\int_0^2 \sin(t^2)\,dt$ This problem requires using accumulated-change reasoning to find the net change from a given rate function. The integral of $v(t) = \sin(t^2)$ from 0 to 2 represents the net accumulation of position changes over time. This integral sums up the contributions of velocity at each instant, yielding the overall net displacement. Unlike antiderivatives that might be guessed incorrectly, this setup requires recognizing the integral form directly. Choice B is a tempting distractor because it mimics the Fundamental Theorem for a simpler function like $\sin(t)$, but here the argument is $t^2$, so it fails. In general, to convert a rate function to total net change, integrate it over the interval without absolute values.
13. A boat's velocity is given by $v(t)=t^3-6t$ m/s for $-2\le t\le2$; what is its displacement over $[-2,2]$?
    Answer: $\int_{-2}^{2} (t^3-6t)\,dt$ This problem requires using accumulated-change reasoning to find the net change from a given rate function. Integrating v(t) = t³ - 6t from -2 to 2 gives the net displacement, accounting for directional changes. As an odd function over a symmetric interval, the integral is zero, reflecting canceling movements. This highlights how the integral computes net, not gross, change. Choice B is a tempting distractor using absolute value, which would give total distance instead of net displacement. In general, to convert a rate function to total net change, integrate it over the interval without absolute values.
14. A runner's speed is $v(t)=\sqrt{t}$ (m/s) for $0\le t\le9$. What is the runner's displacement over $[0,9]$?
    Answer: $\displaystyle \int_{0}^{9} \sqrt{t}\,dt$ This problem requires accumulated-change reasoning to find displacement from velocity. The velocity v(t)=√t represents the instantaneous rate of position change, so the integral ∫₀⁹√t dt gives the total displacement from t=0 to t=9. The integral accumulates all velocity values over the time interval, and since √t is always positive for t>0, the displacement equals the distance traveled. Choice C shows the derivative of √t, which would give acceleration, not displacement. When velocity is given and displacement is needed, integrate the velocity function over the time interval.
15. A particle has velocity $v(t)=\ln(t+1)$ (m/s) for $0\le t\le3$. What is its net displacement on $[0,3]$?
    Answer: $\displaystyle \int_{0}^{3} \ln(t+1)\,dt$ This problem uses accumulated-change reasoning to find displacement from velocity. Since velocity v(t)=ln(t+1) represents the rate of position change, the integral ∫₀³ ln(t+1)dt gives the net displacement from t=0 to t=3. The integral accumulates all velocity values, and since ln(t+1)≥0 for t≥0, the particle moves in the positive direction throughout. Choice D shows the integral of 1/(t+1), which is the derivative of ln(t+1), not what we need for displacement. When given velocity and asked for displacement, integrate the velocity function over the time interval.
16. A car's velocity is $v(t)=3t^2-12t+5$ m/s for $0 \le t \le 4$; what is the displacement on $[0,4]$?
    Answer: $\int_0^4 (3t^2-12t+5)\,dt$ m This problem requires using accumulated-change reasoning to find displacement from velocity. Since velocity $v(t) = 3t^2 - 12t + 5$ represents the rate of position change, the integral $\int_0^4 (3t^2 - 12t + 5) \, dt$ gives the total change in position (displacement) from t = 0 to t = 4. The integral accumulates all the instantaneous velocity values over the time interval, yielding displacement in meters. Choice D (using absolute value) would give total distance traveled, not displacement, which can be negative if the car moves backward. The key strategy is: to find total change from a rate function, integrate the rate over the given interval.
17. A tank fills at rate $r(t)=t e^{-t}$ L/min for $0\le t\le5$; which expression gives total liters added?
    Answer: $\int_0^5 t e^{-t}\,dt$ This problem requires using accumulated-change reasoning to find the net change from a given rate function. The integral of r(t) = t $e^{-t}$ from 0 to 5 sums the instantaneous filling rates to give the total volume added. This accumulation accounts for the varying rate over time, providing the net change in tank volume. Unlike differences in rates, the integral properly aggregates the contributions. Choice A is a tempting distractor because it subtracts rates at endpoints, but rates are not the accumulated quantity. In general, to convert a rate function to total net change, integrate it over the interval without absolute values.
18. A particle's velocity is $v(t)=\dfrac{6}{t+1}$ cm/s for $0\le t\le3$; which gives its displacement on $[0,3]$?
    Answer: $\int_0^3 \dfrac{6}{t+1}\,dt$ This problem requires using accumulated-change reasoning to find the net change from a given rate function. Integrating $v(t) = \dfrac{6}{t+1}$ from 0 to 3 accumulates the velocity to find the net displacement in position. Since the velocity is positive throughout, the integral gives both net displacement and total distance, but the key is the accumulation via integration. This follows the principle that the definite integral of a rate yields the net change in the quantity. Choice C is a tempting distractor as it evaluates the velocity function at endpoints, but that gives a difference in rates, not the accumulated change. In general, to convert a rate function to total net change, integrate it over the interval without absolute values.
19. A cyclist's velocity is $v(t)=4-0.5t$ m/s for $0\le t\le6$; what is the net change in position?
    Answer: $\int_0^6 (4-0.5t)\,dt$ This problem requires using accumulated-change reasoning to find the net change in position from a velocity function. The definite integral of the velocity function over the time interval [0,6] represents the net change in position, as velocity is the rate of change of position with respect to time. By the Fundamental Theorem of Calculus, ∫ from 0 to 6 of v(t) dt equals the position at t=6 minus the position at t=0. This accounts for the continuous decrease in velocity, which remains positive over the interval. A tempting distractor is choice C, which uses absolute value unnecessarily since v(t) doesn't change sign here. In general, to find the total change in a quantity from its rate of change, integrate the rate function over the interval.
20. A tank fills at rate $r(t)=5e^{-t/2}$ L/min for $0\le t\le4$; how many liters are added?
    Answer: $\displaystyle\int_0^4 5e^{-t/2}\,dt$ This problem uses accumulated-change reasoning to find total volume added from a filling rate. Since r(t) = $5e^{-t/2}$ gives the rate of volume change in L/min, integrating it from 0 to 4 yields the total volume added: ∫₀⁴ $5e^{-t/2}$ dt. The integral accumulates all the instantaneous volume additions over the 4-minute period. Choice A (r(4)-r(0)) incorrectly finds the change in rate rather than the total volume accumulated. To find total accumulated quantity from a rate of change, integrate the rate function over the time interval.

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