Mistake Master
Home› Unit 8 · Applications of Integration› 8.1·8.2·8.3·8.4·8.5·8.6·8.7·8.8·8.9·8.10·8.11·8.12·8.13

Connecting Position, Velocity, and Acceleration of Functions Using Integrals AB & BC

Integrating a rate gives a signed change, so $\int_{t_1}^{t_2} v = s(t_2) - s(t_1)$ is a displacement and $\int_{t_1}^{t_2} a = v(t_2) - v(t_1)$ is a change in velocity. Recovering a position requires the initial value as an addend: $s(t_2) = s(t_1) + \int_{t_1}^{t_2} v$, and a stated $s(0)$ appears in a problem because the answer needs it. The units settle which quantity is in hand, since the units of an integral are those of the integrand times those of the variable.

Displacement and total distance are different integrals: $\int v$ lets the trips cancel, while total distance is $\int |v|$, which instructs you to solve $v(t) = 0$, integrate each piece where $v$ keeps one sign, and add the absolute values. For $v(t) = (t - 1)(t - 3)$ on $[0, 4]$ the pieces give $\frac{4}{3}$, $-\frac{4}{3}$ and $\frac{4}{3}$, so the displacement is $\frac{4}{3}$ and the total distance is $4$. Speed is $|v|$, and it increases exactly when $v$ and $a$ share a sign, so a particle with $v = -5$ and $a = 2$ is slowing while its velocity rises.

v(t) = (t − 1)(t − 3) ON [0, 4]. IT TURNS TWICE. +4/3 −4/3 +4/3 DISPLACEMENT ADDS THE THREE SIGNED PIECES: 4/3 − 4/3 + 4/3 = 4/3 TOTAL DISTANCE ADDS THEIR SIZES: 4/3 + 4/3 + 4/3 = 4 THE DOTS ARE WHERE v = 0, AT t = 1 AND 3 THE PARTICLE GOES FORWARD, BACK THE SAME AMOUNT, THEN FORWARD. IT TRAVELS 4 UNITS AND ENDS UP ONLY 4/3 FROM WHERE IT BEGAN.
Drawn to scale at $80$ px per unit of $t$ and $45$ px per unit of $v$, with the parabola computed rather than sketched. The three shaded pieces have equal areas by construction, which is what makes the displacement exactly one third of the total distance here.
FOUR QUESTIONS THAT SOUND ALIKE AND ARE NOT. THE QUESTION WHAT TO INTEGRATE WHAT ELSE IS NEEDED HOW FAR DID IT MOVE, NET? v NOTHING HOW FAR DID IT TRAVEL? |v| THE ROOTS OF v = 0 WHERE IS IT NOW? v THE STARTING VALUE HOW FAST IS IT GOING NOW? a v AT THE START EVERY INTEGRAL HERE RETURNS A CHANGE. TWO OF THE FOUR QUESTIONS ASK FOR A VALUE INSTEAD, AND A VALUE NEEDS A STARTING POINT. THE FOURTH ROW ALSO WANTS AN ABSOLUTE VALUE AT THE VERY END.
The middle column is short on purpose: only two integrands ever appear. What separates the four questions is the third column, and three of the four rows need something the integral by itself cannot supply.

The work

3 ways in · any order
Lesson
Connecting Position, Velocity, and Acceleration of Functions Using Integrals ›

Treats every integral in the motion chain as a signed change, adds the initial value that turns a displacement into a position, splits at the roots of the velocity to separate total distance from displacement, and settles when speed increases by comparing the signs of velocity and acceleration.

Skill check · 10 scenarios
Open lesson →
Diagnostic
10-item topic check ›

Ten items on motion by integration: displacement against total distance with the interval split at every turn, positions that need the starting value added, velocity recovered from acceleration, and the sign test that decides whether speed is rising.

Not yet available · 10 items
Coming soon
Targeted Practice
Drill a single misconception ›

Pick one of the failure modes you missed and drill it on its own. The round is adaptive: two correct in a row clears the misconception and moves you to the next.

Take the diagnostic to identify your misconceptions
Diagnostic first