AP Calculus BC Quiz: Position Velocity And Acceleration Using Integrals

What this quiz covers

This quiz focuses on Position Velocity And Acceleration Using Integrals, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus BC.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $\displaystyle\int_2^8 (10-\tfrac{t}{2})\,dt$ (correct answer)
2. $\displaystyle\int_2^8 |10-\tfrac{t}{2}|\,dt$
3. $v(8)-v(2)$
4. $\dfrac{1}{6}\displaystyle\int_2^8 (10-\tfrac{t}{2})\,dt$
5. $6\cdot v(8)$

Explanation: 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.

Question 2

1. $P(10)$
2. $P'(10)-P'(0)$
3. $\displaystyle \int_{0}^{10} 200e^{-0.1t}\,dt$ (correct answer)
4. $\displaystyle \int_{0}^{10} 200e^{-0.1t}\,dt + P(0)$
5. $\displaystyle \left|\int_{0}^{10} 200e^{-0.1t}\,dt\right|$

Explanation: This problem applies accumulated-change reasoning to find population change from growth rate. The rate P'(t)=200e^(-0.1t) represents how fast the population changes per year, so the integral ∫₀¹⁰ 200e^(-0.1t)dt gives the total population change from t=0 to t=10. The integral accumulates all instantaneous growth rates, and since the exponential is always positive, the population increases throughout. Choice D incorrectly adds the initial population P(0), but we want only the change, not the final population. To find total change from a rate of change, integrate the rate function over the given interval.

Question 3

1. $v(\pi)-v(0)$
2. $\int_0^{\pi} 5\sin(t)\,dt$ (correct answer)
3. $\int_0^{\pi} \lvert 5\sin(t)\rvert\,dt$
4. $\pi\,v\!\left(\frac{\pi}{2}\right)$
5. $\frac{1}{\pi}\int_0^{\pi} 5\sin(t)\,dt$

Explanation: 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.

Question 4

1. $\int_0^9 \lvert \sqrt{t}-2\rvert\,dt$
2. $\int_0^9 (\sqrt{t}-2)\,dt$ (correct answer)
3. $v(9)-v(0)$
4. $9\,v(9)$
5. $\frac{1}{9}\int_0^9 (\sqrt{t}-2)\,dt$

Explanation: 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.

Question 5

1. $v(6)-v(0)$
2. $\int_0^6 (4-0.5t)\,dt$ (correct answer)
3. $\int_0^6 \lvert 4-0.5t\rvert\,dt$
4. $6\,v(6)$
5. $\frac{1}{6}\int_0^6 (4-0.5t)\,dt$

Explanation: 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.

Question 6

1. $\int_0^{2\pi} \lvert 2\cos(t)-1\rvert\,dt$
2. $v(2\pi)-v(0)$
3. $\int_0^{2\pi} (2\cos(t)-1)\,dt$ (correct answer)
4. $2\pi\,v(\pi)$
5. $\frac{1}{2\pi}\int_0^{2\pi} (2\cos(t)-1)\,dt$

Explanation: 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.

Question 7

1. $\displaystyle\int_0^{2\pi} |2+\cos t|\,dt$
2. $\dfrac{1}{2\pi}\displaystyle\int_0^{2\pi} (2+\cos t)\,dt$
3. $v(2\pi)-v(0)$
4. $\displaystyle\int_0^{2\pi} (2+\cos t)\,dt$ (correct answer)
5. $(2\pi)\cdot v(2\pi)$

Explanation: This problem requires accumulated-change reasoning to find net displacement from velocity. The integral ∫₀^(2π) (2+cos t) dt gives the cyclist's net displacement because velocity is the rate of change of position. Since 2+cos t is always positive (ranging from 1 to 3), the net displacement equals the total distance traveled. Choice C (v(2π)-v(0)) would give the change in velocity rather than position, confusing the rate with the accumulated quantity. When given a velocity function, integrate it over the time interval to find the net change in position.

Question 8

1. $\displaystyle\int_1^5 (t-3)\,dt$ (correct answer)
2. $\displaystyle\int_1^5 |t-3|\,dt$
3. $v(5)-v(1)$
4. $\dfrac{1}{4}\displaystyle\int_1^5 (t-3)\,dt$
5. $4\cdot v(5)$

Explanation: This question uses accumulated-change reasoning to find net displacement from velocity. The integral ∫₁⁵(t-3)dt gives the net change in position because velocity is the derivative of position. Note that v(t) = t-3 is negative for t < 3 and positive for t > 3, so the particle changes direction at t = 3. Choice B with absolute value would give total distance traveled, not net displacement, which are different when velocity changes sign. To find net position change from velocity, integrate the velocity function (without absolute value) over the time interval.

Question 9

1. $r(4)-r(0)$
2. $\displaystyle\int_0^4 5e^{-t/2}\,dt$ (correct answer)
3. $\dfrac{1}{4}\displaystyle\int_0^4 5e^{-t/2}\,dt$
4. $4\cdot r(4)$
5. $\displaystyle\int_0^4 |5e^{-t/2}|\,dt$

Explanation: 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.

Question 10

1. $r(2)-r(0)$
2. $\displaystyle\int_0^2 (4-\ln(t+1))\,dt$ (correct answer)
3. $\dfrac{1}{2}\displaystyle\int_0^2 (4-\ln(t+1))\,dt$
4. $2\cdot r(2)$
5. $\displaystyle\int_0^2 |4-\ln(t+1)|\,dt$

Explanation: This question applies accumulated-change reasoning to find total volume drained from a drain rate. Since r(t) = 4-ln(t+1) gives the rate of volume change in L/min, integrating from 0 to 2 yields the total volume drained: ∫₀²(4-ln(t+1))dt. The integral accumulates all the instantaneous volume changes over the 2-minute period. Choice A (r(2)-r(0)) would give the change in drain rate, not the total volume drained. To find total accumulated quantity from a rate of change, integrate the rate function over the time interval.

Question 11

1. $\int_0^{\pi} (\sin t+t)\,dt$ m (correct answer)
2. $v(\pi)$ m
3. $\int_0^{\pi} (\sin t+t)\,dt$ m/s
4. $\int_0^{\pi} |\sin t+t|\,dt$ m
5. $(\sin \pi+\pi)-(\sin 0+0)$ m

Explanation: This problem uses accumulated-change reasoning to find the particle's net change in position from its velocity function. The velocity v(t) = sin t + t gives the rate of position change, so integrating ∫₀^π(sin t + t)dt accumulates all velocity values from t = 0 to t = π, yielding the net displacement. Since velocity is always positive on this interval (sin t ≥ -1 and t ≥ 0, so sin t + t > 0), the displacement equals the distance traveled. Choice E incorrectly tries to use v(π) - v(0), which would give change in velocity, not position. Remember: to find position change from velocity, integrate the velocity function over the time interval.

Question 12

1. $\int_1^5 (6-2t)\,dt$ m (correct answer)
2. $\int_1^5 (6-2t)\,dt$ m/s
3. $(6-2\cdot5)-(6-2\cdot1)$ m
4. $\int_1^5 |6-2t|\,dt$ m
5. $v(5)$ m

Explanation: This problem uses accumulated-change reasoning to find the drone's altitude change from its vertical velocity. The velocity v(t) = 6 - 2t represents the rate of altitude change, so ∫₁⁵(6 - 2t)dt gives the net change in altitude from t = 1 to t = 5. Note that v(t) changes sign at t = 3 (positive before, negative after), meaning the drone rises then falls, but the integral correctly accounts for this to give net displacement. Choice D with absolute value would give total distance traveled vertically, not net altitude change. The strategy remains: integrate the rate function to find total change, regardless of whether the rate changes sign.

Question 13

1. $\int_0^2 (e^{-t}-2)\,dt$ m (correct answer)
2. $\int_0^2 |e^{-t}-2|\,dt$ m
3. $v(2)-v(0)$ m
4. $\int_0^2 (e^{-t}-2)\,dt$ m/s
5. $v(2)$ m

Explanation: This problem uses accumulated-change reasoning to find displacement from the particle's velocity on a line. The velocity v(t) = e^(-t) - 2 gives the rate of position change, so ∫₀²(e^(-t) - 2)dt accumulates these rates from t = 0 to t = 2, yielding net displacement. Since e^(-t) < 1 for t > 0 and we subtract 2, the velocity is always negative, meaning the particle moves in the negative direction. Choice B with absolute value would give total distance traveled, not net displacement, which is negative here. The strategy remains consistent: integrate velocity to find displacement, preserving the sign to indicate direction.

Question 14

1. $\int_0^3 (2-\ln(t+1))\,dt$ m (correct answer)
2. $\int_0^3 (2-\ln(t+1))\,dt$ m/s
3. $\int_0^3 |2-\ln(t+1)|\,dt$ m
4. $v(3)-v(0)$ m
5. $v(3)$ m

Explanation: This problem uses accumulated-change reasoning to find net displacement from the particle's velocity function. The velocity $v(t) = 2 - \ln(t + 1)$ represents the rate of position change, so $\int_0^3 [2 - \ln(t + 1)]\,dt$ accumulates these rates from t = 0 to t = 3, yielding net displacement in meters. The velocity starts positive ($v(0) = 2$) but becomes negative when $\ln(t + 1) > 2$, which occurs around t ≈ 6.4, so velocity remains positive on [0,3]. Choice C with absolute value would be equivalent here but is unnecessary since $v(t) > 0$ on the interval. Remember: integrate velocity to find displacement - this fundamental relationship converts rate information into total change.

Question 15

1. $4$ m
2. $\int_0^4 (3t^2-12t+5)\,dt$ m (correct answer)
3. $v(4)-v(0)$ m
4. $\int_0^4 |3t^2-12t+5|\,dt$ m
5. $\dfrac{1}{4}\int_0^4 (3t^2-12t+5)\,dt$ m/s

Explanation: 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.

Question 16

1. $\int_0^{\pi} e^{-t}\cos t\,dt$ (correct answer)
2. $e^{-\pi}\cos\pi-e^{0}\cos0$
3. $\int_0^{\pi} e^{-t}\sin t\,dt$
4. $\pi\cdot v(\pi)$
5. $\int_0^{\pi} |e^{-t}\cos t|\,dt$

Explanation: This problem requires using accumulated-change reasoning to find the net change from a given rate function. The integral of $v(t) = e^{-t} \cos t$ from $0$ to $\pi$ gives the net displacement by accumulating velocity. The oscillating velocity leads to potential cancellations in the net change. This method captures the directional accumulation accurately. Choice B is a tempting distractor mimicking an antiderivative evaluation, but it's incorrect for this function. In general, to convert a rate function to total net change, integrate it over the interval without absolute values.

Question 17

1. $v(6)-v(0)$
2. $\displaystyle\int_0^6 (8-2t)\,dt$ (correct answer)
3. $\displaystyle\int_0^6 |8-2t|\,dt$
4. $\dfrac{1}{6}\displaystyle\int_0^6 (8-2t)\,dt$
5. $6\cdot v(6)$

Explanation: This problem uses accumulated-change reasoning to find the runner's net displacement from velocity. Since velocity v(t) = 8-2t is the rate of change of position, integrating it from 0 to 6 gives the total change in position: $\int_0^6 (8-2t)\,dt$. The integral accumulates all the instantaneous changes in position over the 6-second interval. Choice A ($v(6)-v(0)$) incorrectly finds the change in velocity rather than the change in position. To find total change from a rate function, integrate the rate over the given time interval.

Question 18

1. $\int_0^3 (2+\cos(\pi t))\,dt$ (correct answer)
2. $\int_0^3 (2-\cos(\pi t))\,dt$
3. $v(3)$
4. $3\cdot v(3)$
5. $\int_0^3 |2+\cos(\pi t)|\,dt$

Explanation: This problem requires using accumulated-change reasoning to find the net change from a given rate function. Integrating $v(t) = 2 + \cos(\pi t)$ from 0 to 3 gives the net displacement by accumulating velocity over the interval. The positive velocity ensures the integral reflects forward progress, equaling the net change in position. This method relies on the integral capturing total accumulation from the rate. Choice E is a tempting distractor with absolute value, but unnecessary here since velocity doesn't change sign. In general, to convert a rate function to total net change, integrate it over the interval without absolute values.

Question 19

1. $r(20)$
2. $\displaystyle \int_{0}^{20} (5-0.2t)\,dt$ (correct answer)
3. $r(20)-r(0)$
4. $\displaystyle \int_{0}^{20} (5-0.2t)\,dt + L(0)$
5. $\displaystyle \int_{0}^{20} (5-0.2t)^2\,dt$

Explanation: This problem applies accumulated-change reasoning to find total oil leaked from leak rate. The rate r(t)=5-0.2t L/hr represents how fast oil leaks, so the integral ∫₀²⁰(5-0.2t)dt gives the total volume leaked from t=0 to t=20. The integral accumulates all instantaneous leak rates; note that r(t) becomes negative after t=25, but we're only integrating to t=20 where it remains positive. Choice D incorrectly adds initial volume L(0), but we want total leaked, not a cumulative amount. To find total accumulation from a rate, integrate the rate function over the time period.

Question 20

1. $\int_0^3 \frac{2}{1+t^2}\,dt$ (correct answer)
2. $\int_0^3 \left(\frac{2}{1+t^2}\right)^2\,dt$
3. $\frac{2}{1+3^2}-\frac{2}{1+0^2}$
4. $3\cdot \frac{2}{1+3^2}$
5. $\int_0^3 \left\lvert \frac{2}{1+t^2}\right\rvert\,dt$

Explanation: 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.