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

Review: Derivatives and Motion

s(t) \xrightarrow{\text{derivative}} v(t) \xrightarrow{\text{derivative}} a(t)

Or, in second-derivative terms,

s''(t) = a(t)

Working Backwards with Integration

a(t) \xrightarrow{\text{integrate}} v(t) \xrightarrow{\text{integrate}} s(t)

Key principle: to get from acceleration to position, integrate twice.

Example 1: Position from Acceleration

Problem: A particle on the x-axis has acceleration

a(t) = 12t - 4

. Its velocity is 18 cm/s at

t = 2

seconds, and its initial position is

x(0) = 8

cm. Find

x(t)

Step 1: Integrate acceleration to get velocity.

v(t) = \int (12t - 4)\,dt = 6t^2 - 4t + C_v

Step 2: Use the given condition to find $C_v$ . At

t = 2

v = 18

18 = 24 - 8 + C_v \quad \Rightarrow \quad C_v = 2

v(t) = 6t^2 - 4t + 2

Step 3: Integrate velocity to get position.

x(t) = \int (6t^2 - 4t + 2)\,dt = 2t^3 - 2t^2 + 2t + C_x

Step 4: Use the initial condition. At

t = 0

x = 8

, so

C_x = 8

x(t) = 2t^3 - 2t^2 + 2t + 8

Displacement vs. Total Distance

Displacement is the net change in position, and it can be positive or negative:

\text{Displacement} = \int_{t_1}^{t_2} v(t)\,dt

Total distance is how far the object actually traveled, and it’s always positive:

\text{Total Distance} = \int_{t_1}^{t_2} |v(t)|\,dt

Important Warning

\left|\int_{t_1}^{t_2} v(t)\,dt\right| \neq \text{Total Distance}

Taking the absolute value of the result just gives you the absolute value of displacement. The absolute value has to go inside the integral.

Speed vs. Velocity

Velocity: $v(t)$ , can be negative

Speed: $|v(t)|$ , never negative

So displacement is the integral of velocity, and total distance is the integral of speed, $\int_{t_1}^{t_2} |v(t)|\,dt$

Example 2: Drive with a Change of Direction

Problem: Someone drives for 30 minutes with velocity

v(t) = 0.3t + \ln(t+1) - 2

miles per minute.

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

Notice the velocity is negative at first (wrong direction) and crosses zero at about

t \approx 2.495

minutes, after which it stays positive.

Part (a): Displacement after 10 minutes

\int_0^{10} [0.3t + \ln(t+1) - 2]\,dt \approx 11.38 \text{ miles}

The small negative area from the early backtracking partially cancels the positive area.

Part (b): Displacement after 15 minutes

\int_0^{15} v(t)\,dt \approx 33.11 \text{ miles}

Part (c): Displacement after 30 minutes

\int_0^{30} v(t)\,dt \approx 151.45 \text{ miles}

That’s how far the destination is from the starting point.

Part (d): Total distance actually driven

This needs the absolute value, since we want every mile traveled including the backtracking:

\int_0^{30} |v(t)|\,dt \approx 155.81 \text{ miles}

Calculator steps (TI-84):

Press MATH, then 9 (for $\int$ )

Enter the bounds, 0 to 30

For the integrand, press MATH, scroll to NUM, select 1:abs(. (Shortcut: press 2nd + 0 for CATALOG, then select abs(.)

Enter the velocity function

Specify the variable

Interpretation: the driver covered about 155.8 miles but ended up only 151.5 miles from home, because of the wrong-direction stretch at the beginning.

Position Using Accumulation

When you know the initial position and the velocity function:

x(t) = x(t_0) + \int_{t_0}^{t} v(s)\,ds

This is an accumulation function: initial position plus the accumulated change.

The same idea works one level up. Acceleration is the rate of change of velocity, so if you know the velocity at one time:

v(t) = v(t_0) + \int_{t_0}^{t} a(s)\,ds

The definite integral of acceleration over an interval is the change in velocity over that interval.

Example: a particle has acceleration

a(t) = 3\sqrt{t}

, and

v(1) = -5

. Find

v(4)

v(4) = v(1) + \int_{1}^{4} 3\sqrt{t}\,dt = -5 + \left[2t^{3/2}\right]_{1}^{4} = -5 + (16 - 2) = 9

The velocity went up by 14, from

-5

9

Example 3: Using the Accumulation Formula

A particle has velocity

v(t) = 4t^3 - 6t^2 + 1

Part (a): Position at $t = 3$ if $x(0) = 5$

\int_0^3 (4t^3 - 6t^2 + 1)\,dt = \left[t^4 - 2t^3 + t\right]_0^3 = 81 - 54 + 3 = 30

x(3) = 5 + 30 = 35

Part (b): Position at $t = 2$ if $x(1) = -2$

\int_1^2 (4t^3 - 6t^2 + 1)\,dt = \left[t^4 - 2t^3 + t\right]_1^2 = (16 - 16 + 2) - (1 - 2 + 1) = 2

x(2) = -2 + 2 = 0

Example 4: Total Distance with a Sign Change

Problem: A particle has velocity

v(t) = -\frac{3}{2}t + 3

for

0 \leq t \leq 4

. Find the total distance traveled.

Step 1: Find when velocity is zero.

0 = -\frac{3}{2}t + 3 \quad \Rightarrow \quad t = 2

Step 2: Split the integral at $t = 2$ , since velocity changes sign there.

\text{Total Distance} = \left|\int_0^2 v(t)\,dt\right| + \left|\int_2^4 v(t)\,dt\right|

Step 3: Evaluate each piece.

\int_0^2 \left(-\frac{3}{2}t + 3\right)dt = \left[-\frac{3}{4}t^2 + 3t\right]_0^2 = (-3 + 6) - 0 = 3

\int_2^4 \left(-\frac{3}{2}t + 3\right)dt = \left[-\frac{3}{4}t^2 + 3t\right]_2^4 = (-12 + 12) - (-3 + 6) = -3

Step 4: Absolute values, then add.

\text{Total Distance} = |3| + |-3| = 6

Alternative: just compute

\int_0^4 |v(t)|\,dt = 6

directly. It handles the sign change automatically.

Calculator Tips for Absolute Value Integrals

Getting to absolute value on a TI-84:

MATH → scroll right to NUM → 1:abs(

Or 2nd + 0 (CATALOG) → A → abs(

Setting up the integral:


∫(abs(function), x, lower, upper)

Using stored functions: store the function in Y1, then reference it with VARS → Y-VARS → 1:Function → Y1.

Recalling a previous entry: press 2nd + ENTER to pull up your last calculation and edit it. Saves a ton of time.

Summary of Key Formulas

Given	Find	Formula
Acceleration $a(t)$	Velocity	$v(t) = \int a(t)\,dt + C$
Velocity $v(t)$	Position	$x(t) = \int v(t)\,dt + C$

Quantity	Formula	Notes
Displacement	$\int_{t_1}^{t_2} v(t)\,dt$	Net change, can be negative
Total Distance	$\int_{t_1}^{t_2} \lvert v(t)\rvert\,dt$	Actual distance, always positive
Speed	$\lvert v(t)\rvert$	Absolute value of velocity

Position from an initial condition:

x(t) = x(t_0) + \int_{t_0}^t v(s)\,ds

Problem-Solving Strategy

Position from acceleration:

Integrate acceleration to get velocity, with $+C_v$

Use the given velocity condition to find $C_v$

Integrate velocity to get position, with $+C_x$

Use the initial position to find $C_x$

Total distance traveled:

Find where $v(t) = 0$ , which is where velocity changes sign

Split the time interval at those points

Integrate over each subinterval separately

Take the absolute value of each result and add

Or just use

\int |v(t)|\,dt

directly on the calculator.

Common Mistakes

Taking the absolute value of the final result for total distance. The absolute value goes inside the integral.

Confusing speed with velocity. Speed is $|v(t)|$ and is never negative.

Forgetting the constant when integrating. Always include $+C$ and use initial conditions to pin it down.

Not checking for sign changes in velocity when finding total distance.

Key Takeaways

Integration reverses differentiation: acceleration → velocity → position

Two integrations to get from acceleration to position

Displacement: $\int v(t)\,dt$ , the net change in position

Total distance: $\int |v(t)|\,dt$ , the actual distance traveled

Absolute value placement matters: inside the integral, not outside

Solve $v(t) = 0$ to find where to split the integrals

Accumulation formula: $x(t) = x(t_0) + \int_{t_0}^t v(s)\,ds$

Same idea one level up: $v(t) = v(t_0) + \int_{t_0}^t a(s)\,ds$

Use $C_v$ for velocity and $C_x$ for position, and find each from its own condition

Store values on the calculator to avoid rounding errors, and use 2nd + ENTER to recall entries

Displacement can cancel out, but total distance adds up all the movement

Back Next