Position, Velocity &
Acceleration Using Integrals

Learn how integration connects the fundamental quantities of motion: recover velocity from acceleration and position from velocity.

AP Calculus AB

What Is Rectilinear Motion?

When an object moves along a straight line, we describe its motion with three related functions of time $ t $:

$ s(t) $

Position: where the object is on the line at time $ t $.

$ v(t) $

Velocity: rate of change of position: $ v(t) = s'(t) $.

$ a(t) $

Acceleration: rate of change of velocity: $ a(t) = v'(t) $.

Derivatives Go Down, Integrals Go Up

Why Do Integrals Matter Here?

In real-world problems, you often measure acceleration (e.g., from sensors) but need to find velocity or position. Integration lets you work backwards:

1

Acceleration is given as a function $ a(t) $. Integrate to find $ v(t) $.

2

Velocity is recovered. Integrate again to find $ s(t) $.

3

Initial conditions (e.g., $ v(0) = 5 $) pin down the constants of integration.

The Integral Relationships

Each constant $ C $ is determined by an initial condition: a known value of $ v $ or $ s $ at a specific time.

Definite Integral Form

The definite integral gives you the net change over an interval:

Displacement vs. Total Distance

Displacement

Net change in position: $$ \Delta s = \int_a^b v(t)\,dt $$ Can be positive, negative, or zero.

Total Distance

Total path length: $$ D = \int_a^b |v(t)|\,dt $$ Always non-negative.

Indefinite vs. Definite Integral

	Indefinite $ \int v\,dt $	Definite $ \int_a^b v\,dt $
Result	A function $ s(t) + C $	A number (net change)
Needs $+C$?	Yes	No
Use when...	You want a full formula for $ s(t) $	You want displacement over $[a,b]$
Initial condition	Needed to find $ C $	Built into the formula

Step-by-Step Method

1

Identify what you're given: $ a(t) $, $ v(t) $, or $ s(t) $? Note any initial conditions.

2

Integrate the given function. Use the antiderivative rules you know.

3

Add $ +C $ if using an indefinite integral. Substitute the initial condition to solve for $ C $.

4

Write the complete function. Check: does it match the given initial condition?

5

Repeat if you need to go up another level (e.g., $ a \to v \to s $).

Example 1: Constant Acceleration

A particle has acceleration $ a(t) = 6 $ m/s², with $ v(0) = 2 $ m/s and $ s(0) = 1 $ m. Find $ s(t) $.

1

Find $ v(t) $: $ v(t) = \int 6\,dt = 6t + C_1 $. Since $ v(0) = 2 $: $ C_1 = 2 $. So $ v(t) = 6t + 2 $.

2

Find $ s(t) $: $ s(t) = \int (6t + 2)\,dt = 3t^2 + 2t + C_2 $. Since $ s(0) = 1 $: $ C_2 = 1 $. So $ s(t) = 3t^2 + 2t + 1 $.

Example 2: Variable Acceleration

Given $ a(t) = 4t - 2 $, $ v(1) = 3 $, $ s(1) = 0 $. Find $ s(t) $.

1

Integrate for $ v(t) $: $ v(t) = \int (4t - 2)\,dt = 2t^2 - 2t + C_1 $. Use $ v(1) = 3 $: $ 2 - 2 + C_1 = 3 $, so $ C_1 = 3 $. Thus $ v(t) = 2t^2 - 2t + 3 $.

2

Integrate for $ s(t) $: $ s(t) = \int (2t^2 - 2t + 3)\,dt = \frac{2}{3}t^3 - t^2 + 3t + C_2 $. Use $ s(1) = 0 $: $ \frac{2}{3} - 1 + 3 + C_2 = 0 $, so $ C_2 = -\frac{8}{3} $.

Example 3: Displacement and Distance

A particle has $ v(t) = t^2 - 4 $ on $[0, 3]$. Find displacement and total distance.

One cart, three meanings: position, displacement, and distance

A blue dynamics cart rests on a straight laboratory rail with a photogate and end buffers; an observer stands behind the setup.

Suppose v(t) = t² − 4t + 3 on [0,4] with initial position 2. The final position is 10/3, displacement is 4/3 units, and total distance is 4 units.

Signed change and distance differ

Suppose v(t) = t² − 4t + 3 on [0,4] with initial position 2. The final position is 10/3, displacement is 4/3 units, and total distance is 4 units.

Visualizing $ v(t) = t^2 - 4 $

When Is the Particle at Rest?

A particle is at rest when $ v(t) = 0 $. This is where the velocity function crosses the $ t $-axis. These points are critical for total distance problems.

1

Set $ v(t) = 0 $ and solve for $ t $. These are the turning points.

2

Split the interval at each zero of $ v $. On each sub-interval, $ v $ has a constant sign.

3

Integrate $ |v(t)| $ on each sub-interval and add the absolute values.

Interpreting Signed Area

$\int v\,dt > 0$

Particle moves in the positive direction during this interval.

$\int v\,dt < 0$

Particle moves in the negative direction during this interval.

Common Mistakes to Avoid

✕ Forgetting $ +C $

Every indefinite integral requires a constant. Without it, your function is incomplete and initial conditions can't be applied.

✕ Displacement = Distance

Displacement is $\int v\,dt$ (signed). Total distance is $\int |v|\,dt$. They differ when velocity changes sign.

✕ Wrong Direction of Integration

Integrating position does NOT give velocity. Integration goes up the chain ($a \to v \to s$), not down.

Summary Table

Given	Operation	Result
$ a(t) $	$\int a(t)\,dt$	$ v(t) + C $
$ v(t) $	$\int v(t)\,dt$	$ s(t) + C $
$ v(t) $ on $[a,b]$	$\int_a^b v(t)\,dt$	Displacement
$ v(t) $ on $[a,b]$	$\int_a^b \|v(t)\|\,dt$	Total distance
$ s(t) $	$ s'(t) $	$ v(t) $
$ v(t) $	$ v'(t) $	$ a(t) $

Question 1 of 5

A particle has velocity $ v(t) = 3t^2 - 6t $. What is the displacement from $ t = 0 $ to $ t = 3 $?

$\int_0^3 (3t^2 - 6t)\,dt = [t^3 - 3t^2]_0^3 = (27 - 27) - 0 = 0$. The particle returns to its starting position, so displacement is zero.

Question 2 of 5

If $ a(t) = 10 $ and $ v(0) = -3 $, what is $ v(t) $?

$v(t) = \int 10\,dt = 10t + C$. Using $v(0) = -3$: $C = -3$. So $v(t) = 10t - 3$. Choice (d) is the position if you integrated again.

Question 3 of 5

A particle has $ v(t) = t - 2 $ on $[0, 4]$. What is the total distance traveled?

$v(t) = 0$ at $t = 2$. Total distance $= \int_0^2 |t-2|\,dt + \int_2^4 |t-2|\,dt = \int_0^2 (2-t)\,dt + \int_2^4 (t-2)\,dt = 2 + 2 = 4$. Note: displacement is $0$, but total distance is $4$.

Question 4 of 5

Given $ a(t) = 6t $, $ v(0) = 4 $, and $ s(0) = 1 $, find $ s(2) $.

$v(t) = \int 6t\,dt = 3t^2 + C_1$. $v(0)=4 \Rightarrow C_1=4$. So $v(t) = 3t^2 + 4$. Then $s(t) = \int(3t^2+4)\,dt = t^3 + 4t + C_2$. $s(0)=1 \Rightarrow C_2=1$. So $s(t) = t^3 + 4t + 1$. $s(2) = 8 + 8 + 1 = 17$.

Question 5 of 5

If $ v(t) = \cos(t) $ and $ s(0) = 2 $, what is $ s(\pi) $?

$s(t) = \int \cos(t)\,dt = \sin(t) + C$. $s(0) = \sin(0) + C = C = 2$. So $s(t) = \sin(t) + 2$. $s(\pi) = \sin(\pi) + 2 = 0 + 2 = 2$.

Given	Operation	Result
\( a(t) \)	\(\int a(t)\,dt\)	\( v(t) + C \)
\( v(t) \)	\(\int v(t)\,dt\)	\( s(t) + C \)
\( v(t) \) on \([a,b]\)	\(\int_a^b v(t)\,dt\)	Displacement
\( v(t) \) on \([a,b]\)	\(\int_a^b \|v(t)\|\,dt\)	Total distance
\( s(t) \)	\( s'(t) \)	\( v(t) \)
\( v(t) \)	\( v'(t) \)	\( a(t) \)

	Indefinite \( \int v\,dt \)	Definite \( \int_a^b v\,dt \)
Result	A function \( s(t) + C \)	A number (net change)
Needs \(+C\)?	Yes	No
Use when...	You want a full formula for \( s(t) \)	You want displacement over \([a,b]\)
Initial condition	Needed to find \( C \)	Built into the formula

Position, Velocity &Acceleration Using Integrals

What Is Rectilinear Motion?

Derivatives Go Down, Integrals Go Up

Why Do Integrals Matter Here?

The Integral Relationships

Definite Integral Form

Displacement vs. Total Distance

Indefinite vs. Definite Integral

Step-by-Step Method

Example 1: Constant Acceleration

Example 2: Variable Acceleration

Example 3: Displacement and Distance

One cart, three meanings: position, displacement, and distance

Signed change and distance differ

Visualizing \( v(t) = t^2 - 4 \)

When Is the Particle at Rest?

Interpreting Signed Area

Common Mistakes to Avoid

Summary Table

Question 1 of 5

Question 2 of 5

Question 3 of 5

Question 4 of 5

Question 5 of 5

Position, Velocity &
Acceleration Using Integrals