Connecting Position, Velocity, and Acceleration of Functions Using Integrals

AP Calculus · Applications of Integration

Position, Velocity & Acceleration Using Integrals

Integrating climbs back up the chain — but it can never recover where the motion started.

key idea$s(b)=s(a)+\int_a^b v(t)\,dt\qquad \text{distance}=\int_a^b\lvert v(t)\rvert\,dt$

Setting up the model

Differentiating moves down the chain from position to velocity to acceleration; integrating moves back up it. The one thing integration cannot recover is the starting position, which is why every problem of this kind supplies one.

Working a motion problem

Decide which direction you are travelling along the chain. Given $v$ and wanting $s$, you integrate; given $s$ and wanting $v$, you differentiate.
Write down the initial value you are given — $s(0)$, $v(0)$, or a position at some other time. Integration alone produces a family of answers without it.
Read the question for the words displacement, total distance, or position. These are three different quantities and each has its own integral.
For total distance, find where $v=0$ and split the integral there, taking absolute values of each piece. This step is the whole difficulty of the topic.
Attach units: position in the length unit, velocity in length per time, acceleration in length per time squared.

Question asked	Compute	Sign convention
displacement on $[a,b]$	$\displaystyle\int_a^b v\,dt$	signed — back-and-forth cancels
total distance travelled	$\displaystyle\int_a^b\lvert v\rvert\,dt$	unsigned — nothing cancels
position at time $b$	$s(a)+\displaystyle\int_a^b v\,dt$	needs the initial position
velocity at time $b$	$v(a)+\displaystyle\int_a^b a\,dt$	needs the initial velocity
average velocity	$\dfrac{1}{b-a}\displaystyle\int_a^b v\,dt$	displacement over elapsed time
speed at time $t$	$\lvert v(t)\rvert$	never negative

Rows 1 and 2 are the pair that decides most problems. If the particle never changes direction the two agree; the moment $v$ changes sign they part company, and the exam will choose an interval where it does.

DisplacementNet change in position: $s(b)-s(a)$. It can be negative or zero.

Total distanceHow far the particle actually travelled, counting every reversal. Never negative.

Speed$\lvert v\rvert$ — the magnitude of velocity, with the direction discarded.

Speeding up$v$ and $a$ have the same sign. Slowing down: opposite signs.

At rest$v=0$. A candidate turning point — but only if $v$ actually changes sign there.

Worked examples

EX 1

Position from velocity

the initial value matters

A particle has $v(t)=3t^{2}-4$ m/s and $s(0)=5$ m. Find $s(2)$.

Choose the right relation

$s(2)=s(0)+\displaystyle\int_{0}^{2}v(t)\,dt$.

Integrate

$\displaystyle\int_{0}^{2}\left(3t^{2}-4\right)dt=\left[t^{3}-4t\right]_{0}^{2}=8-8=0$.

Add the initial position

$s(2)=5+0=5$ m.

Interpret

The displacement is zero: the particle ended exactly where it began.

Note what this does not say

Zero displacement does not mean it never moved. It went backward, then forward the same amount.

EX 2

Total distance

split where $v=0$

For the same particle, $v(t)=3t^{2}-4$ on $[0,2]$, find the total distance travelled.

Find where $v=0$

$3t^{2}=4$, so $t=\dfrac{2}{\sqrt3}\approx1.155$, inside the interval.

Sign each piece

On $\left[0,\tfrac{2}{\sqrt3}\right)$, $v\lt 0$; after it, $v\gt 0$.

Integrate the first piece

$\displaystyle\int_{0}^{2/\sqrt3}v\,dt=\left[t^{3}-4t\right]_{0}^{2/\sqrt3}\approx-3.079$, so its contribution is $3.079$.

Integrate the second

$\displaystyle\int_{2/\sqrt3}^{2}v\,dt\approx0-(-3.079)=3.079$.

Add the magnitudes

Total distance $\approx6.16$ m — against a displacement of $0$. The split at $t=\dfrac{2}{\sqrt3}$ is what makes the difference.

EX 3

Velocity from acceleration

climbing two links

A particle has $a(t)=6t$ m/s$^{2}$, $v(0)=-9$ m/s and $s(0)=0$. Find when it is at rest, and its position then.

Integrate acceleration

$v(t)=v(0)+\displaystyle\int_{0}^{t}6u\,du=-9+3t^{2}$.

Solve for rest

$-9+3t^{2}=0$ gives $t^{2}=3$, so $t=\sqrt3$ (taking $t\ge0$).

Integrate velocity

$s(t)=0+\displaystyle\int_{0}^{t}\left(-9+3u^{2}\right)du=-9t+t^{3}$.

Evaluate

$s\!\left(\sqrt3\right)=-9\sqrt3+3\sqrt3=-6\sqrt3\approx-10.39$ m.

Read the story

It moved backward the whole time, coming to rest about $10.4$ m behind the start before reversing.

EX 4

Speeding up or slowing down

compare two signs

At $t=1$ a particle has $v(1)=-4$ m/s and $a(1)=-2$ m/s$^{2}$. Is it speeding up?

Read the velocity

Negative, so the particle is moving in the negative direction.

Read the acceleration

Also negative, pushing it further in that direction.

Apply the rule

Same signs means the speed is increasing.

Confirm with magnitude

Speed is $\lvert v\rvert=4$ and it is growing, even though $v$ itself is decreasing toward more negative values.

State the distinction

Velocity decreasing and speed increasing are perfectly compatible. Answer whichever the question actually asked about.

Units & interpretation

Expression	Units	Reads as
$\displaystyle\int_a^b v\,dt$	m	net change in position
$\displaystyle\int_a^b\lvert v\rvert\,dt$	m	path length actually travelled
$\dfrac{1}{b-a}\displaystyle\int_a^b v\,dt$	m/s	average velocity
$\dfrac{1}{b-a}\displaystyle\int_a^b\lvert v\rvert\,dt$	m/s	average speed — a different number
$\displaystyle\int_a^b a\,dt$	m/s	net change in velocity

Rows 3 and 4 differ whenever the particle reverses. Average velocity can be zero while average speed is large — and the exam asks for each by name, so read the noun carefully.

Watch out

Displacement and total distance are different questions

$\int v$ answers the first, $\int\lvert v\rvert$ the second. Using one for the other is the single most costly error in this topic.

Integration cannot supply the starting position

$\int_a^b v$ gives the change in position. Add $s(a)$ to get an actual location.

Split at every sign change of $v$

Absolute value inside an integral is handled by splitting the interval, not by taking the absolute value of the final answer.

Velocity decreasing is not the same as slowing down

Speed grows whenever $v$ and $a$ share a sign, including when both are negative. Compare the signs, not the words.

Course map. Topic 8.2 · LO CHA-4.C — determine values for positions and rates of change using definite integrals in problems involving rectilinear motion.