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8.2 Connecting Position, Velocity, and Acceleration of Functions Using Integrals

Syllabus
2020
Topic
8.2
Level
—

Learning objectives

—CHA-4.C—Determine values for positions and rates of change using definite integrals in problems involving rectilinear motion• CHA-4.C Determine values for positions and rates of change using definite integrals in problems involving rectilinear motion.• CHA-4.C.1 For a particle in rectilinear motion over an interval of time, the definite integral of velocity represents the particle’s displacement over the interval of time, and the definite integral of speed represents the particle’s total distance traveled over the interval of time.• Enduring understanding CHA-4: Definite integrals allow us to solve problems involving the accumulation of change over an interval.

Integrals Accumulate Motion

For motion along a line, v(t)=s′(t)v(t)=s'(t) and a(t)=v′(t)a(t)=v'(t). Definite integrals reverse these derivative relationships: accumulated acceleration changes velocity, and accumulated velocity changes position.

v(b)=v(a)+\int_a^b a(t),dt,\qquad s(b)=s(a)+\int_a^b v(t),dt

Quantity on [a,b][a,b] Calculation Sign meaning
displacement ∫abv(t) dt\int_a^b v(t)\,dt forward and backward motion cancel
final position s(a)+∫abv(t) dts(a)+\int_a^b v(t)\,dt initial position plus displacement
total distance ∫ab∣v(t)∣ dt\int_a^b|v(t)|\,dt all traveled lengths are positive

Let v(t)=t−2v(t)=t-2 meters/second for 0≤t≤40\le t\le4 and s(0)=5s(0)=5 meters. Displacement is ∫04(t−2) dt=0\int_0^4(t-2)\,dt=0, so s(4)=5s(4)=5 meters. Since velocity changes sign at t=2t=2, total distance is ∫02(2−t) dt+∫24(t−2) dt=2+2=4\int_0^2(2-t)\,dt+\int_2^4(t-2)\,dt=2+2=4 meters.

Do not use ∣∫v∣|\int v| for total distance: cancellation has already occurred inside that integral. Split at every time when velocity changes sign, or integrate ∣v∣|v|. If velocity is in meters/second and time in seconds, its definite integral is in meters—not meters/second.