Position, Velocity &
Acceleration Using Integrals

Learn how to recover position from velocity and velocity from acceleration using the power of integration and initial conditions.

AP Calculus BC

What is Rectilinear Motion?

An object moving along a straight line has three fundamental kinematic quantities: position, velocity, and acceleration. Each is a function of time $ t $.

$ s(t) $

Position: where the object is at time $ t $

$ v(t) $

Velocity: rate of change of position

$ a(t) $

Acceleration: rate of change of velocity

The Derivative Chain

In AP Calculus, you already know derivatives take you forward (position to velocity to acceleration). This unit focuses on using integrals to reverse that process.

Why Integrals Matter Here

In many real-world problems, you are given acceleration (from a force) or velocity (from a speedometer) and must find the object's position. Integration is the tool that recovers the original function from its rate of change.

The Fundamental Recovery Formulas

1

Velocity from acceleration: $$ v(t) = v(t_0) + \int_{t_0}^{t} a(\tau)\,d\tau $$

2

Position from velocity: $$ s(t) = s(t_0) + \int_{t_0}^{t} v(\tau)\,d\tau $$

Displacement vs. Total Distance

Displacement

$ \int_{a}^{b} v(t)\,dt $ : signed area (positive minus negative). Tells you net change in position.

Total Distance

$ \int_{a}^{b} |v(t)|\,dt $ : sum of all areas. Tells you how far the object actually traveled.

Indefinite vs. Definite Integrals

	Indefinite Integral	Definite Integral
Form	$ \int v(t)\,dt = s(t) + C $	$ \int_{a}^{b} v(t)\,dt $
Result	A family of position functions	A number: net change in position
Needs	Initial condition to find $ C $	Bounds of integration
Use case	Find $ s(t) $ for all $ t $	Find displacement on $[a, b]$

How to Apply the Formulas

1

Identify what you are given: $ a(t) $, $ v(t) $, or $ s(t) $, and what initial conditions are provided.

2

Integrate the given function. If given $ a(t) $, integrate to get $ v(t) + C_1 $. If given $ v(t) $, integrate to get $ s(t) + C_2 $.

3

Apply initial conditions to solve for the constant(s) of integration.

4

Write the final function. For displacement or total distance, evaluate with a definite integral instead.

5

Check units and signs. Negative velocity means moving backward: this affects displacement vs. distance.

Example 1: Velocity to Position

1

Integrate $ v(t) $: $ s(t) = \int (3t^2 - 6t)\,dt = t^3 - 3t^2 + C $

2

Apply $ s(0) = 4 $: $ 0 - 0 + C = 4 $, so $ C = 4 $.

3

Result: $ s(t) = t^3 - 3t^2 + 4 $.

Example 2: Acceleration to Position

1

Find $ v(t) $: $ v(t) = \int (6t - 2)\,dt = 3t^2 - 2t + C_1 $. Since $ v(0) = 5 $: $ C_1 = 5 $. So $ v(t) = 3t^2 - 2t + 5 $.

2

Find $ s(t) $: $ s(t) = \int (3t^2 - 2t + 5)\,dt = t^3 - t^2 + 5t + C_2 $. Since $ s(0) = -1 $: $ C_2 = -1 $.

3

Result: $ s(t) = t^3 - t^2 + 5t - 1 $.

Example 3: Displacement & Distance

1

Displacement: $ \int_0^3 (t^2 - 4)\,dt = \left[\frac{t^3}{3} - 4t\right]_0^3 = (9 - 12) - 0 = -3 $.

2

Find where $ v = 0 $: $ t^2 - 4 = 0 \Rightarrow t = 2 $ (in $[0,3]$).

3

Total distance: $ \int_0^2 |t^2 - 4|\,dt + \int_2^3 |t^2 - 4|\,dt = \int_0^2 (4 - t^2)\,dt + \int_2^3 (t^2 - 4)\,dt $.

4

Evaluate: $ \left[4t - \frac{t^3}{3}\right]_0^2 + \left[\frac{t^3}{3} - 4t\right]_2^3 = \frac{16}{3} + \frac{7}{3} = \frac{23}{3} \approx 7.67 $.

Graphical Interpretation

The signed area under $ v(t) $ gives displacement. Taking the absolute value $ |v(t)| $ before integrating counts all area as positive, giving total distance.

The Accumulation Form

A powerful alternative to finding $ +C $ is the accumulation form, which builds the initial condition directly into a definite integral:

Follow the salt through a working mixing rig

A stirred transparent tank has one upper inlet and one lower outlet with a valve. The exact diagram links its 100 L salt balance to a computed approach to equilibrium.

The distinction between net change and total movement also appears in flowing water. Suppose a tank receives and releases 4 gal/min for 5 minutes. Its 200 gal inventory stays constant while 20 gal enters and 20 gal leaves: net change is zero and total throughput is 40 gal.

Net inventory versus throughput

The distinction between net change and total movement also appears in flowing water. Suppose a tank receives and releases 4 gal/min for 5 minutes. Its 200 gal inventory stays constant while 20 gal enters and 20 gal leaves: net change is zero and total throughput is 40 gal.

When Velocity Changes Sign

When $ v(t) $ changes sign, the particle reverses direction. This is critical for computing total distance: you must split the integral at each zero of $ v(t) $.

1

Find all zeros of $ v(t) $ on the interval $[a, b]$.

2

Split the interval at those zeros: $[a, c_1], [c_1, c_2], \ldots, [c_n, b]$.

3

Integrate $ |v(t)| $ on each subinterval and add the results.

Common Mistakes

✕ Forgetting $ +C $

When using indefinite integrals, always include the constant. Without an initial condition, you have infinitely many solutions.

✕ Confusing Displacement & Distance

$ \int v(t)\,dt $ gives displacement (signed). Total distance requires $ \int |v(t)|\,dt $: split at zeros of $ v $.

✕ Skipping the Double Integration

Going from $ a(t) $ to $ s(t) $ requires two integrations with two constants: don't jump from acceleration directly to position in one step.

Summary Table

Quantity	Formula	What It Gives You
$ v(t) $ from $ a(t) $	$ v(t_0) + \int_{t_0}^t a\,d\tau $	Velocity at time $ t $
$ s(t) $ from $ v(t) $	$ s(t_0) + \int_{t_0}^t v\,d\tau $	Position at time $ t $
Displacement	$ \int_a^b v(t)\,dt $	Net change in position
Total Distance	$ \int_a^b \|v(t)\|\,dt $	Actual path length
$ s(t) $ from $ a(t) $	Integrate twice with 2 ICs	Position function

Question 1 of 5

A particle has velocity $ v(t) = 4t - 3 $ and $ s(0) = 2 $. What is $ s(t) $?

$ \int (4t - 3)\,dt = 2t^2 - 3t + C $. Using $ s(0) = 2 $: $ C = 2 $. So $ s(t) = 2t^2 - 3t + 2 $.

Question 2 of 5

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

$ v(t) = \int 10\,dt = 10t + C $. Since $ v(0) = -3 $: $ C = -3 $. So $ v(t) = 10t - 3 $. Option (d) gives position, not velocity.

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 (2 - t)\,dt + \int_2^4 (t - 2)\,dt = 2 + 2 = 4 $. Note: the displacement is $ 0 $, but the total distance is $ 4 $.

Question 4 of 5

Given $ v(t) = \cos t $ and $ s(0) = 5 $, what is $ s\!\left(\frac{\pi}{2}\right) $?

$ s(t) = 5 + \int_0^t \cos\tau\,d\tau = 5 + \sin t $. At $ t = \frac{\pi}{2} $: $ s = 5 + \sin\frac{\pi}{2} = 5 + 1 = 6 $.

Question 5 of 5

Which expression represents the displacement of a particle on $[1, 5]$ given velocity $ v(t) $?

Displacement is the net change in position: $ s(5) - s(1) = \int_1^5 v(t)\,dt $. Option (a) gives total distance. Option (d) gives change in velocity, not position.

	Indefinite Integral	Definite Integral
Form	\( \int v(t)\,dt = s(t) + C \)	\( \int_{a}^{b} v(t)\,dt \)
Result	A family of position functions	A number: net change in position
Needs	Initial condition to find \( C \)	Bounds of integration
Use case	Find \( s(t) \) for all \( t \)	Find displacement on \([a, b]\)

Quantity	Formula	What It Gives You
\( v(t) \) from \( a(t) \)	\( v(t_0) + \int_{t_0}^t a\,d\tau \)	Velocity at time \( t \)
\( s(t) \) from \( v(t) \)	\( s(t_0) + \int_{t_0}^t v\,d\tau \)	Position at time \( t \)
Displacement	\( \int_a^b v(t)\,dt \)	Net change in position
Total Distance	\( \int_a^b \|v(t)\|\,dt \)	Actual path length
\( s(t) \) from \( a(t) \)	Integrate twice with 2 ICs	Position function

Position, Velocity &Acceleration Using Integrals

What is Rectilinear Motion?

The Derivative Chain

Why Integrals Matter Here

The Fundamental Recovery Formulas

Displacement vs. Total Distance

Indefinite vs. Definite Integrals

How to Apply the Formulas

Example 1: Velocity to Position

Example 2: Acceleration to Position

Example 3: Displacement & Distance

Graphical Interpretation

The Accumulation Form

Follow the salt through a working mixing rig

Net inventory versus throughput

When Velocity Changes Sign

Common Mistakes

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