AP Calculus AB Practice: Motion & Riemann Sum Applications (11 Step-by-Step Explanations)

AP Calculus AB · Motion & Riemann Sum Applications

Motion & Riemann Sum Applications: AP Calculus AB Practice Problems

Motion problems are one of the most natural real-world applications of integration, and they show up constantly on the AP Calculus AB exam. Since velocity is the derivative of position and acceleration is the derivative of velocity, integrating in the reverse direction recovers position from velocity and velocity from acceleration — but the exam consistently tests a subtlety that trips up even strong students: the difference between net displacement (how far the particle ends up from where it started) and total distance traveled (how far it actually moved, counting backtracking).

Net displacement is simply ∫_a^bv(t)dt — positive and negative movement cancel out. Total distance requires identifying every point where the particle changes direction (where v(t)=0 and actually changes sign), splitting the integral at those points, and adding the absolute value of each piece, since distance traveled backward still counts as distance. Beyond pure motion problems, this same "integrate a rate to get a total" idea extends to accumulation-rate word problems: a rate of arrivals, a rate of spreading, or a rate of leaking, all integrated over time, plus average value problems that divide an accumulated total by the length of the time interval to get a representative rate.

Below are 11 AP Calculus AB-style practice problems covering total distance vs. net displacement, average value of a rate function, and real-world accumulation problems involving arrival rates, drug concentration, rumor spread, and leaking rates. Work through each problem, select your answer, and check whether you got it right — every problem includes a complete, step-by-step explanation. Once you've worked through these, keep building your skills with thousands more official-style questions in our free AP Calculus AB QBank.

When a particle changes direction

A particle moves along a line in such a way that its position at time t is given by s = t³−9t²+24t+2. Its direction of motion changes when

Full Explanation

Direction changes where velocity v = s′(t) changes sign. Find v(t): v = 3t²−18t+24 = 3(t²−6t+8) = 3(t−2)(t−4).

This factors with two distinct (simple) roots at t=2 and t=4. Since each root has odd multiplicity (multiplicity 1), the sign of v genuinely flips at each one.

Checking signs confirms it: v is positive before t=2, negative between t=2 and t=4, and positive again after t=4 — two direction changes.

The correct answer is D.

Distance covered from a velocity function

A body moves along a straight line so that its velocity v at time t is given by v = 6t³+3t²+4. The distance the body covers from t=0 to t=2 equals

Full Explanation

Since v(t) is never negative on this interval (all terms are non-negative for t≥0), distance covered is simply ∫₀²v(t)dt — no need to check for direction changes.

Find the antiderivative: ∫(6t³+3t²+4)dt = 1.5t⁴+t³+4t.

Evaluate from 0 to 2: 1.5(16)+8+8 = 24+8+8 = 40.

The correct answer is B.

Total distance traveled, accounting for a direction change

A particle moves along a line with velocity v = 3t²−12t. The total distance traveled from t=0 to t=5 equals

Full Explanation

Find where v=0: 3t²−12t = 3t(t−4) = 0, giving t=0 and t=4. Since t=4 falls inside [0,5], the particle changes direction there — total distance requires splitting the integral at t=4 and adding absolute values.

The antiderivative is F(t) = t³−6t². From 0 to 4: F(4)−F(0) = (64−96)−0 = −32, so this piece contributes |−32| = 32.

From 4 to 5: F(5)−F(4) = (125−150)−(−32) = −25−(−32) = 7, contributing 7.

Total distance: 32+7 = 39. The correct answer is C.

Keep the momentum going.

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Net change in position, using the same velocity function

A particle moves along a line with velocity v(t) = 3t²−12t. The net change in position of the particle from t=0 to t=5 is

Full Explanation

Unlike total distance, net change in position is just a single definite integral — no need to worry about where the particle changes direction, since positive and negative motion are allowed to cancel.

∫₀⁵(3t²−12t)dt = [t³−6t²]₀⁵.

= (125−150)−0 = −25.

The correct answer is A. Compare this to Problem 3: the total distance (39) is larger in magnitude than the net change (−25), exactly because the particle backtracked partway through the interval.

Distance from acceleration, starting at rest

The acceleration of a particle moving on a straight line is given by a = sin t, and when t=0 the particle is at rest. The distance it covers from t=0 to t=2 is

Full Explanation

Integrate acceleration to get velocity: v(t) = ∫sin t dt = −cos t + C. Since the particle starts at rest, v(0)=0 → −1+C=0 → C=1, so v(t) = 1−cos t.

Since cos t ≤ 1 always, v(t) = 1−cos t ≥ 0 for every t — the particle never moves backward, so distance equals the plain definite integral of v.

∫₀²(1−cos t)dt = [t−sin t]₀² = (2−sin 2)−0.

= 2−sin 2. The correct answer is C.

Average value of a rate function · real-world context

During the worst 6-hour period of a storm, the wind velocity, in miles per hour, is given by v(t) = 8t−t²+90, 0 ≤ t ≤ 6. The average wind velocity during this period (in mph) is

Full Explanation

Average value = (1/(6−0))∫₀⁶v(t)dt. Find the antiderivative first: ∫(8t−t²+90)dt = 4t²−t³/3+90t.

Evaluate from 0 to 6: 4(36)−216/3+90(6) = 144−72+540 = 612.

Divide by the interval length: 612/6 = 102.

The correct answer is C.

6 down, 5 to go — keep practicing.

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Population from a radial density function

The center of a city, which we will assume is circular, is on a straight highway (along the diameter of the circle). The radius of the city is 4 miles. The density of the population, in thousands of people per square mile, is given by f(x) = 15−3x at a distance x miles from the highway. The population of the city (in thousands of people) is given by the integral

Full Explanation

Since density depends only on distance x from the highway, use thin strips parallel to the highway. At distance x from the center line, the strip spans the full chord of the circle, which has length 2√(16−x²) (from the circle equation x²+y²=16).

Each strip's population contribution is density × area: (15−3x) × 2√(16−x²) dx.

Since f(x) is defined for distance x on one side of the highway, and the city is symmetric on both sides, double the integral over x=0 to 4: total population = 2 × ∫₀⁴(15−3x) · 2√(16−x²)dx = 4∫₀⁴(15−3x)√(16−x²)dx.

The correct answer is C.

Solving for a time when an accumulated total is reached

A beach opens at 8 A.M. and people arrive at a rate of R(t) = 20+20t people per hour, where t represents the number of hours the beach has been open. Assuming no one leaves before noon, at what time will there be 150 people there?

Full Explanation

The total number of people after t hours is the accumulated integral of the arrival rate: P(t) = ∫₀^tR(s)ds = ∫₀^t(20+20s)ds = 20t+10t².

Set this equal to 150 and solve for t: 10t²+20t−150 = 0 → t²+2t−15 = 0.

Factor: (t+5)(t−3) = 0, so t=3 (rejecting the negative root, since t represents hours after opening).

3 hours after 8:00 A.M. is 11:00 A.M. The correct answer is C.

Average value of a drug concentration function

Suppose the amount of a drug in a patient's bloodstream t hours after intravenous administration is A(t) = 50/(t+1)² mg. The average amount in the bloodstream during the first 4 hours is

Full Explanation

Average value = (1/4)∫₀⁴50/(t+1)²dt = (50/4)∫₀⁴(t+1)⁻²dt.

The antiderivative of (t+1)⁻² is −1/(t+1), so: (50/4)[−1/(t+1)]₀⁴ = 12.5 × [−1/5−(−1/1)] = 12.5 × (1−1/5).

= 12.5 × 0.8 = 10.0.

The correct answer is B.

9 down, 2 to go — almost there.

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Accumulated total over a specific time window

A rumor spreads through a town at the rate of R(t) = t²+8t new people per day, t days after it was first heard. Approximately how many people hear the rumor during the second week (from the 7th to the 14th days) after it was first heard?

Full Explanation

The number of people who hear the rumor during a time window is the accumulated integral of the rate over that window: ∫₇¹⁴(t²+8t)dt = [t³/3+4t²]₇¹⁴.

At t=14: 2744/3+4(196) = 914.67+784 = 1698.67. At t=7: 343/3+4(49) = 114.33+196 = 310.33.

Subtract: 1698.67−310.33 = 1388.33.

Rounding, approximately 1388 people hear the rumor during the second week. The correct answer is B.

Total accumulated leakage from an exponential rate

Oil is leaking from a tanker at the rate of L(t) = 1200e^−0.25t gal/hr, where t is given in hours. The total number of gallons of oil that will leak out during the first 6 hours is approximately

Full Explanation

Total leaked gallons is the accumulated integral of the leak rate: ∫₀⁶1200e^−0.25tdt = 1200 × [−(1/0.25)e^−0.25t]₀⁶ = −4800[e^−1.5−1].

Compute e^−1.5 ≈ 0.2231, so e^−1.5−1 ≈ −0.7769.

−4800 × (−0.7769) ≈ 3729.

The correct answer is B.

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