A particle’s velocity is for and for . What is displacement on ?
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AP Calculus AB Quiz
Practice Exploring Accumulations Of Change in AP Calculus AB with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.
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A particle’s velocity is v(t)=2 for 0≤t≤1 and v(t)=−2 for 1≤t≤3. What is displacement on [0,3]?
This quiz focuses on Exploring Accumulations Of Change, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus AB.
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.
A particle’s velocity is v(t)=2 for 0≤t≤1 and v(t)=−2 for 1≤t≤3. What is displacement on [0,3]?
Explanation: This problem requires accumulation reasoning to find displacement from a piecewise constant velocity. From t = 0 to t = 1, the particle moves 2(1 - 0) = 2 units forward, and from t = 1 to t = 3, it moves -2(3 - 1) = -4 units (backward). The net displacement is 2 + (-4) = -2 units, indicating the particle ends up 2 units behind its starting position. A common error is to add the distances traveled (2 + 4 = 6) without considering direction, but displacement is a signed quantity. For piecewise constant velocities, compute the displacement for each interval and sum the signed results.
A bank balance changes at rate B′(t)=−5 dollars/day for 0≤t≤7. What is B(7)−B(0)?
Explanation: This problem uses accumulation reasoning to find the change in bank balance from a constant negative rate. The change B(7) - B(0) equals the integral of B'(t) = -5 from t = 0 to t = 7. With a constant rate, this equals -5(7 - 0) = -35 dollars. The negative result indicates the balance decreased by 35 dollars. A potential mistake is to compute 5 × 7 = 35 and select answer A, forgetting that the negative rate means the balance is decreasing. When working with constant rates of change, the net change equals rate times time interval, keeping track of the sign.
A function satisfies f′(x)=(x+1)1 for 0≤x≤1. What is f(1)−f(0)?
Explanation: This problem applies accumulation reasoning to find function change from the derivative of natural logarithm. The rate f'(x) = 1/(x+1) over [0,1] integrates to ∫₀¹ 1/(x+1) dx = [ln(x+1)]₀¹ = ln(2) - ln(1) = ln(2) - 0 = ln(2). The natural logarithm of (x+1) is the antiderivative of 1/(x+1). Students might choose 1 by evaluating ln(2) ≈ 0.693 and rounding, but the exact answer is ln(2). When the derivative involves 1/(x+a), the antiderivative is ln|x+a| plus a constant.
A function satisfies f′(x)=1−x on 0≤x≤3. What is f(3)−f(0)?
Explanation: This problem uses accumulation reasoning to find function change from a decreasing linear rate. The rate f'(x) = 1 - x starts positive at x = 0 and becomes negative for x > 1 over [0,3]. We integrate ∫₀³ (1-x) dx = [x - x²/2]₀³ = (3 - 9/2) - 0 = 3 - 4.5 = -1.5 = -3/2. The rate crosses zero at x = 1, with more negative contribution than positive. Students might choose 3 by evaluating the positive part only, but the full interval includes significant negative accumulation. When linear rates cross zero, compute the full integral to capture both positive and negative contributions.
A particle’s velocity is v(t)=cost for 0≤t≤π. What is its displacement?
Explanation: This problem requires accumulation reasoning to find displacement from cosine velocity over a half period. The velocity v(t)=cost decreases from 1 to -1 over [0,π], so we integrate ∫0πcostdt=[sint]0π=sin(π)−sin(0)=0−0=0. The cosine function creates equal positive and negative areas over this half period, resulting in zero net displacement. Students might choose 2 by confusing this with the sine integral over [0,π], but cosine over [0,π] integrates to zero. When cosine is integrated over a half period starting from zero, the result is always zero due to symmetry.
A function satisfies f′(x)=−1 for −3≤x≤1 and f(−3)=5. What is f(1)?
Explanation: This problem applies accumulation reasoning with an initial condition to find a function value. Given f'(x) = -1 is constant over [-3,1] and f(-3) = 5, the change f(1) - f(-3) = -1×(1-(-3)) = -1×4 = -4. Therefore, f(1) = f(-3) + (-4) = 5 - 4 = 1. The negative constant rate produces linear decrease from the initial value. Students might choose -1 by confusing the rate with the final value, but f(1) equals the initial value plus the accumulated change. When rates of change are negative constants, subtract the accumulated change magnitude from the initial value.
A function satisfies f′(x)=4−4x for 0≤x≤2. What is f(2)−f(0)?
Explanation: This problem uses accumulation reasoning to find function change from a linear derivative with specific symmetry. The rate f'(x) = 4 - 4x = 4(1-x) decreases linearly from 4 to -4 over [0,2], crossing zero at x = 1. We integrate ∫₀² (4-4x) dx = [4x - 2x²]₀² = (8 - 8) - 0 = 0. The equal positive and negative areas on either side of x = 1 cancel exactly. Students might choose 4 by evaluating the initial rate, or -4 by evaluating the final rate, but the symmetric accumulation yields zero net change. When linear rates cross zero at the midpoint of the interval, the positive and negative contributions cancel completely.
A tank drains at rate r(t)=−3 L/min for 2≤t≤6. What is the net change in volume?
Explanation: This problem applies accumulation reasoning to find volume change from a constant draining rate. The tank drains at r(t) = -3 L/min over the 4-minute interval [2,6], so we accumulate this negative rate over time. The net change equals -3 L/min × (6-2) min = -3 × 4 = -12 liters, indicating the tank loses 12 liters. A student might choose -3 by confusing the rate with the total change, ignoring the 4-minute duration. When dealing with constant negative rates (outflow/draining), multiply the rate by the time interval to find total volume lost.
A temperature changes at rate T′(t)=t21 for 1≤t≤2. What is T(2)−T(1)?
Explanation: This problem applies accumulation reasoning to find temperature change from a reciprocal rate function. The rate T′(t)=t21 decreases rapidly over [1,2], so we integrate ∫12t21dt=∫12t−2dt. This equals [−t−1]12=−21−(−11)=−21+1=21. The negative reciprocal antiderivative captures the decreasing contribution as t increases. Students might choose 1 by evaluating 1/12 at the left endpoint, but this gives the initial rate, not the accumulated change. When integrating power functions with negative exponents, apply the power rule with careful attention to signs.
A function satisfies f′(x)=2x−4 on 0≤x≤4. What is f(4)−f(0)?
Explanation: This problem uses accumulation reasoning to find function change from a quadratic derivative that crosses zero. The rate f′(x)=2x−4=2(x−2) is negative for x<2 and positive for x>2 over [0,4]. We integrate ∫04(2x−4)dx=[x2−4x]04=(16−16)−0=0. The symmetric quadratic creates equal negative and positive contributions that cancel exactly. Students might choose -8 by evaluating only the negative part or 8 by considering only the positive part, but the full interval yields zero net change. When quadratic rates are symmetric about their zero, the accumulated change can be zero despite significant variation.
A runner’s velocity is v(t)=4 for 0≤t≤2 and v(t)=−4 for 2≤t≤5. What is the net displacement on [0,5]?
Explanation: This problem applies accumulation reasoning to find net displacement from velocity intervals. From 0 to 2, velocity 4 gives displacement 4 × 2 = 8. From 2 to 5, velocity -4 gives displacement -4 × 3 = -12. The net displacement is 8 + (-12) = -4. A tempting error is to compute 4 × 2 - 4 × 3 = -4 without recognizing this is the same calculation. To find net change from rates, multiply each rate by its time duration and add all results.
A function satisfies f′(x)=3 for −2≤x≤5. What is f(5)−f(−2)?
Explanation: This problem requires accumulation reasoning to find function change from a constant derivative. The rate f'(x) = 3 is constant over the interval [-2,5], so we accumulate by multiplying the rate times the interval length. The change equals 3 × (5-(-2)) = 3 × 7 = 21. Constant rates of change produce linear accumulation proportional to the interval length. Students might choose 9 by computing 3×3, possibly using the wrong interval length, but the correct interval is from -2 to 5, spanning 7 units. For constant rates of change, multiply the rate by the total interval length to find accumulated change.
A particle’s velocity is v(t)=sint for 0≤t≤π. What is its displacement?
Explanation: This problem uses accumulation reasoning to find displacement from sinusoidal velocity. The velocity v(t) = sin t oscillates over [0,π], so we accumulate by integrating ∫0πsintdt. This equals [−cost]0π=−cos(π)−(−cos(0))=−(−1)−(−1)=1+1=2 meters. The sine function is entirely positive over this interval, so all motion is in the forward direction. Students might choose 0 thinking sine waves always cancel out, but this only happens over complete periods like [0,2π]. For sinusoidal velocities over partial periods, evaluate the definite integral to find net displacement.
A tank’s net flow rate is r(t)=6−2t gal/min for 0≤t≤4. Net change in volume?
Explanation: This problem requires accumulation reasoning to find volume change from a linear flow rate. The rate r(t) = 6 - 2t changes linearly from 6 gal/min at t = 0 to -2 gal/min at t = 4, so we integrate ∫₀⁴ (6-2t) dt. This equals [6t - t²]₀⁴ = (24 - 16) - 0 = 8 gallons net increase. The positive inflow initially exceeds the growing outflow, resulting in net gain. Students might choose 12 by evaluating only 6t at t = 4, ignoring the negative -t² term that reduces the accumulation. For linear rates, integrate the entire expression to capture the changing flow dynamics.
A function satisfies f′(x)=sinx for 0≤x≤2π. What is f(2π)−f(0)?
Explanation: This problem applies accumulation reasoning to find function change from the sine derivative. The rate f'(x) = sin x over [0,π/2] integrates to ∫₀^(π/2) sin x dx = [-cos x]₀^(π/2) = -cos(π/2) - (-cos(0)) = -0 - (-1) = 1. The sine function is entirely positive over this quarter period, contributing to positive accumulation. Students might choose 0 by incorrectly evaluating the cosine antiderivative, but -cos(π/2) = 0 and -cos(0) = -1. When integrating sine over a quarter period from zero, the result is always 1.
A particle’s velocity is v(t)=3t m/s on 0≤t≤2. What is its displacement over the interval?
Explanation: This problem uses accumulation reasoning to find displacement from velocity over an interval. The velocity function v(t) = 3t m/s represents the rate of change of position, so we accumulate this rate from t = 0 to t = 2 by integrating ∫₀² 3t dt. Evaluating this integral: ∫₀² 3t dt = [3t²/2]₀² = 3(4)/2 - 0 = 6 meters. A common error would be to evaluate v(2) = 6 and think that's the displacement, but that's just the velocity at t = 2, not the accumulated distance. To find displacement from velocity, integrate the velocity function over the given interval.
A population changes at rate P′(t)=−5 people/day for 0≤t≤7. What is the net change in population?
Explanation: This problem applies accumulation reasoning to find net population change from a constant rate of change. The population changes at a constant rate of P'(t) = -5 people/day over the 7-day interval [0,7], so we accumulate this rate over time. The net change equals -5 people/day × (7-0) days = -35 people, indicating a decrease. Students might incorrectly choose -5 by confusing the rate with the total change, but this ignores the 7-day duration. When the rate of change is constant and negative, multiply the rate by the time interval to find the total negative accumulation.
A particle’s velocity is v(t)=t−1 for 0≤t≤2. What is its displacement?
Explanation: This problem applies accumulation reasoning to find displacement from a linear velocity that crosses zero. The velocity v(t) = t - 1 is negative for t < 1 and positive for t > 1 over [0,2]. We integrate ∫02(t−1)dt=[2t2−t]02=(2−2)−0=0. The equal negative displacement in [0,1] and positive displacement in [1,2] cancel exactly. Students might choose 1 by evaluating at the zero crossing, but displacement requires integration over the full interval. When linear velocity functions cross zero symmetrically within the interval, the net displacement can be zero.
A particle’s velocity is v(t)={2t,2,0≤t≤11≤t≤3. Net displacement?
Explanation: This problem uses accumulation reasoning to find net displacement from piecewise velocity. Over [0,1], v(t) = 2t accumulates ∫₀¹ 2t dt = [t²]₀¹ = 1 unit. Over [1,3], v(t) = 2 accumulates 2×(3-1) = 4 units. The total displacement is 1 + 4 = 5 units. The accelerating motion in the first second followed by constant motion creates this cumulative effect. Students might choose 6 by evaluating 2t at t = 3, but this ignores that the velocity function changes at t = 1. For piecewise functions, integrate each piece over its domain then sum the results.
For 0≤t≤6, the rate of change is R(t)=2 on [0,1], −2 on [1,4], and 1 on [4,6]. What is the net change?
Explanation: This problem tests accumulation reasoning with multiple rate intervals to find net change. On [0,1], rate 2 gives change 2 × 1 = 2. On [1,4], rate -2 gives change -2 × 3 = -6. On [4,6], rate 1 gives change 1 × 2 = 2. The net change is 2 + (-6) + 2 = -2. Students might incorrectly compute 2 + 2 + 1 = 5 by adding rates without considering durations. For piecewise rates, multiply each rate by its interval length and sum algebraically.