8.2 Motion with integrals

Achievable AP Calculus AB

8. Applications of integrals

Our AP Calculus AB course is currently in development and is a work-in-progress.

Motion with integrals

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What you’ll learn:

How to recover velocity and position from acceleration
Compute displacement and total distance using integration

Motion along a line is one of the most common applications of differential and integral calculus. Recall from section 4.2 that derivatives connect position, velocity, and acceleration. Here we go in the reverse direction: if you know an acceleration function (and an initial condition), you can integrate to find velocity, and integrate again to find position.

Displacement

Integrating an object’s velocity over a time interval gives its displacement, meaning the net change in position.

$Displacement = \int_{a}^{b} v (t) d t$

For example, let the velocity (in meters per second) be

$v (t) = 3 t - 6$

Find the change in position from $t = 0$ to $t = 4$ seconds:

$\int_{0}^{4} v (t) d t$

$= \int_{0}^{4} (3 t - 6) d t$

$= (\frac{3}{2} t^{2} - 6 t)_{0}^{4}$

$= 0$

A displacement of $0$ means the object ends where it started after the first 4 seconds.

Total distance

Total distance traveled counts how far the object moves regardless of direction. To do that, we integrate the absolute value of velocity.

$Total distance = \int_{a}^{b} ∣ v (t) ∣ d t$

To evaluate:

Find when the velocity changes sign (first find where it equals $0$ ).
Write the absolute value as a piecewise function.
Split the integral at those points.

Using the same velocity function and the interval $0 \leq t \leq 4$ :

$v (t) = 3 t - 6$

The velocity is $0$ when $3 t - 6 = 0$ , so $t = 2$ . That’s where the direction changes, so it’s the breakpoint for the absolute value.

$∣3 t - 6∣ = ⎩ ⎨ ⎧ 3 t - 6 - (3 t - 6) if t \geq 2 if t < 2$

Then the total distance is

$\int_{0}^{2} - (3 t - 6) d t + \int_{2}^{4} (3 t - 6) d t$

$= \int_{2}^{4} (3 t - 6) d t - \int_{0}^{2} (3 t - 6) d t$

$= 12$

Because an integral gives signed area, any region where $v (t) < 0$ contributes negative area. Taking $∣ v (t) ∣$ (or splitting the integral and flipping the sign where needed) turns that negative area into positive distance.

For problems that allow a calculator/Desmos, you can enter the absolute value directly, like

$\int_{0}^{4} ∣3 t - 6∣ d t$

Accumulating position from velocity

If you know the velocity function and the position at one time, you can find the position at another time using the Fundamental Theorem of Calculus.

The velocity, in units per second, of an object moving along a number line is $v (t) = t^{2} - 6 t + 5$ . Its position at time $t = 0$ is $s (0) = 10$ .

a) Find the particle’s position at $t = 6$ seconds.
b) Find the total distance the particle traveled on $0 \leq t \leq 6$ .

Solutions

a) $s (6)$

(spoiler)

By the FTC,

$\int_{0}^{6} v (t) d t = s (6) - s (0)$

$s (6) = s (0) + \int_{0}^{6} (t^{2} - 6 t + 5) d t$

$= 10 - 6$

$= 4$

So the particle starts at $10$ units and is at $4$ units when $t = 6$ seconds.

b) Total distance over $[0, 6]$

(spoiler)

The total distance traveled is

$\int_{0}^{6} ∣ v (t) ∣ d t$

$= \int_{0}^{6} ∣ t^{2} - 6 t + 5∣ d t$

On $[0, 6]$ , the velocity is positive for $0 \leq t < 1$ and $5 < t \leq 6$ , and negative for $1 < t < 5$ . So the absolute value in piecewise form is

$∣ t^{2} - 6 t + 5∣ = ⎩ ⎨ ⎧ t^{2} - 6 t + 5 - (t^{2} - 6 t + 5) t^{2} - 6 t + 5 if 0 \leq t < 1 if 1 \leq t \leq 5 if 5 < t \leq 6$

So the integral set-up for the total distance is

$\int_{0}^{1} (t^{2} - 6 t + 5) d t - \int_{1}^{5} (t^{2} - 6 t + 5) d t + \int_{5}^{6} (t^{2} - 6 t + 5) d t$

$= 15.333 units$

From acceleration to position

Sometimes you’re given acceleration and initial conditions for both velocity and position. In that case, you integrate twice, solving for the constant of integration $+ C$ each time.

Given:

$a (t) = 2 t + 1$

$v (1) = 3$

$s (0) = 5$

Find $s (t)$ .

Solution

(spoiler)

$v (t) = \int a (t) d t$

$= \int (2 t + 1) d t$

$= t^{2} + t + C$

Use $v (1) = 3$ to find $C$ :

$3 = 1^{2} + 1 + C$

$C = 1$

So $v (t) = t^{2} + t + 1$ . Then

$s (t) = \int v (t) d t$

$= \int (t^{2} + t + 1) d t$

$= \frac{t ^{3}}{3} + \frac{t ^{2}}{2} + t + C$

Use $s (0) = 5$ to find $C$ :

$5 = \frac{0 ^{3}}{3} + \frac{0 ^{2}}{2} + 0 + C$

$C = 5$

Therefore

$s (t) = \frac{t ^{3}}{3} + \frac{t ^{2}}{2} + t + 5$

Velocity graphs

Sometimes velocity is given as a graph rather than a formula. Since integrating velocity gives change in position, you can find:

displacement by computing net signed area under $v (t)$
total distance by computing total area (treating areas below the axis as positive)

The velocity of a particle moving along a line is shown in the graph below for $0 \leq t \leq 6$ .

Graph of v(t)

a) Find the displacement of the particle from $t = 0$ to $t = 6$ .
b) Find the total distance traveled over this time.
c) If the particle’s position at $t = 2$ is $s (2) = 1$ , find its position at $t = 6$ .

Solutions

a) Displacement

(spoiler)

To find displacement, add the signed areas (areas above the axis are positive; below are negative).

On $0 < t < 1.5$ :

The triangle above the $x$ -axis has a base of length $1.5$ and a height of $1$ , so the signed area is

$A_{1} = \frac{1}{2} (1.5) (1)$

$= 0.75$

On $1.5 < t < 5$ :

The trapezoid below the $x$ -axis has a height of $1$ and two bases of length $3.5$ and $2$ , so the signed area is

$A_{2} = \frac{1}{2} (- 1) (3.5 + 2)$

$= - 2.75$

On $5 < t < 6$ :

The triangle above the $x$ -axis has a base of length $1$ and a height of $1$ , so the signed area is

$A_{3} = \frac{1}{2} (1) (1)$

$= 0.5$

Adding these values gives a displacement of

$0.75 - 2.75 + 0.5 = - 1.5$

b) Total distance

(spoiler)

Total distance is the sum of the absolute values of those areas:

$∣0.75∣ + ∣ - 2.75∣ + ∣0.5∣ = 4$

c) $s (6)$ given $s (2) = 1$

(spoiler)

The displacement from $t = 2$ to $t = 6$ is

$\int_{2}^{6} v (t) d t = s (6) - s (2)$

$s (6) = s (2) + \int_{2}^{6} v (t) d t$

Now compute the net signed area from $t = 2$ to $t = 6$ :

On $2 < t < 4$ :

The rectangle below the $x$ -axis has a width of $1$ and a length of $2$ , so the signed area is

$A = - 2$

On $4 < t < 6$ :

The triangle from $t = 4$ to $t = 5$ matches the triangle from $t = 5$ to $t = 6$ with opposite sign, so these signed areas cancel.

So $\int_{2}^{6} v (t) d t = - 2$ . Substitute this and $s (2) = 1$ :

$s (6) = s (2) + \int_{2}^{6} v (t) d t$

$= 1 - 2$

$= - 1$

Displacement

Displacement = definite integral of velocity: $\int_{a}^{b} v (t) d t$
Represents net change in position over $[a, b]$
Signed value; can be zero if object returns to start

Total distance

Total distance = $\int_{a}^{b} ∣ v (t) ∣ d t$
Integrate absolute value of velocity
Requires splitting integral at points where $v (t) = 0$

Accumulating position from velocity

Position at time $t$ : $s (t) = s (a) + \int_{a}^{t} v (x) d x$
Use Fundamental Theorem of Calculus
Displacement = change in position, total distance = integral of $∣ v (t) ∣$

From acceleration to position

Integrate acceleration to get velocity: $v (t) = \int a (t) d t + C$
Integrate velocity to get position: $s (t) = \int v (t) d t + C$
Use initial conditions to solve for constants

Velocity graphs

Displacement = net signed area under $v (t)$ curve
Total distance = sum of absolute values of areas under $v (t)$
For position at a later time: add displacement over interval to known position

Key formulas

Displacement: $\int_{a}^{b} v (t) d t$
Total distance: $\int_{a}^{b} ∣ v (t) ∣ d t$
Position from velocity: $s (t) = s (a) + \int_{a}^{t} v (x) d x$
Velocity from acceleration: $v (t) = \int a (t) d t + C$

Motion with integrals

What you’ll learn:

How to recover velocity and position from acceleration
Compute displacement and total distance using integration

Displacement

Integrating an object’s velocity over a time interval gives its displacement, meaning the net change in position.

$Displacement = \int_{a}^{b} v (t) d t$

For example, let the velocity (in meters per second) be

$v (t) = 3 t - 6$

Find the change in position from $t = 0$ to $t = 4$ seconds:

$\int_{0}^{4} v (t) d t$

$= \int_{0}^{4} (3 t - 6) d t$

$= (\frac{3}{2} t^{2} - 6 t)_{0}^{4}$

$= 0$

A displacement of $0$ means the object ends where it started after the first 4 seconds.

Total distance

Total distance traveled counts how far the object moves regardless of direction. To do that, we integrate the absolute value of velocity.

$Total distance = \int_{a}^{b} ∣ v (t) ∣ d t$

To evaluate:

Find when the velocity changes sign (first find where it equals $0$ ).
Write the absolute value as a piecewise function.
Split the integral at those points.

Using the same velocity function and the interval $0 \leq t \leq 4$ :

$v (t) = 3 t - 6$

The velocity is $0$ when $3 t - 6 = 0$ , so $t = 2$ . That’s where the direction changes, so it’s the breakpoint for the absolute value.

$∣3 t - 6∣ = ⎩ ⎨ ⎧ 3 t - 6 - (3 t - 6) if t \geq 2 if t < 2$

Then the total distance is

$\int_{0}^{2} - (3 t - 6) d t + \int_{2}^{4} (3 t - 6) d t$

$= \int_{2}^{4} (3 t - 6) d t - \int_{0}^{2} (3 t - 6) d t$

$= 12$

For problems that allow a calculator/Desmos, you can enter the absolute value directly, like

$\int_{0}^{4} ∣3 t - 6∣ d t$

Accumulating position from velocity

If you know the velocity function and the position at one time, you can find the position at another time using the Fundamental Theorem of Calculus.

The velocity, in units per second, of an object moving along a number line is $v (t) = t^{2} - 6 t + 5$ . Its position at time $t = 0$ is $s (0) = 10$ .

a) Find the particle’s position at $t = 6$ seconds.
b) Find the total distance the particle traveled on $0 \leq t \leq 6$ .

Solutions

a) $s (6)$

(spoiler)

By the FTC,

$\int_{0}^{6} v (t) d t = s (6) - s (0)$

$s (6) = s (0) + \int_{0}^{6} (t^{2} - 6 t + 5) d t$

$= 10 - 6$

$= 4$

So the particle starts at $10$ units and is at $4$ units when $t = 6$ seconds.

b) Total distance over $[0, 6]$

(spoiler)

The total distance traveled is

$\int_{0}^{6} ∣ v (t) ∣ d t$

$= \int_{0}^{6} ∣ t^{2} - 6 t + 5∣ d t$

On $[0, 6]$ , the velocity is positive for $0 \leq t < 1$ and $5 < t \leq 6$ , and negative for $1 < t < 5$ . So the absolute value in piecewise form is

$∣ t^{2} - 6 t + 5∣ = ⎩ ⎨ ⎧ t^{2} - 6 t + 5 - (t^{2} - 6 t + 5) t^{2} - 6 t + 5 if 0 \leq t < 1 if 1 \leq t \leq 5 if 5 < t \leq 6$

So the integral set-up for the total distance is

$\int_{0}^{1} (t^{2} - 6 t + 5) d t - \int_{1}^{5} (t^{2} - 6 t + 5) d t + \int_{5}^{6} (t^{2} - 6 t + 5) d t$

$= 15.333 units$

From acceleration to position

Sometimes you’re given acceleration and initial conditions for both velocity and position. In that case, you integrate twice, solving for the constant of integration $+ C$ each time.

Given:

$a (t) = 2 t + 1$

$v (1) = 3$

$s (0) = 5$

Find $s (t)$ .

Solution

(spoiler)

$v (t) = \int a (t) d t$

$= \int (2 t + 1) d t$

$= t^{2} + t + C$

Use $v (1) = 3$ to find $C$ :

$3 = 1^{2} + 1 + C$

$C = 1$

So $v (t) = t^{2} + t + 1$ . Then

$s (t) = \int v (t) d t$

$= \int (t^{2} + t + 1) d t$

$= \frac{t ^{3}}{3} + \frac{t ^{2}}{2} + t + C$

Use $s (0) = 5$ to find $C$ :

$5 = \frac{0 ^{3}}{3} + \frac{0 ^{2}}{2} + 0 + C$

$C = 5$

Therefore

$s (t) = \frac{t ^{3}}{3} + \frac{t ^{2}}{2} + t + 5$

Velocity graphs

Sometimes velocity is given as a graph rather than a formula. Since integrating velocity gives change in position, you can find:

displacement by computing net signed area under $v (t)$
total distance by computing total area (treating areas below the axis as positive)

The velocity of a particle moving along a line is shown in the graph below for $0 \leq t \leq 6$ .

a) Find the displacement of the particle from $t = 0$ to $t = 6$ .
b) Find the total distance traveled over this time.
c) If the particle’s position at $t = 2$ is $s (2) = 1$ , find its position at $t = 6$ .

Solutions

a) Displacement

(spoiler)

To find displacement, add the signed areas (areas above the axis are positive; below are negative).

On $0 < t < 1.5$ :

The triangle above the $x$ -axis has a base of length $1.5$ and a height of $1$ , so the signed area is

$A_{1} = \frac{1}{2} (1.5) (1)$

$= 0.75$

On $1.5 < t < 5$ :

The trapezoid below the $x$ -axis has a height of $1$ and two bases of length $3.5$ and $2$ , so the signed area is

$A_{2} = \frac{1}{2} (- 1) (3.5 + 2)$

$= - 2.75$

On $5 < t < 6$ :

The triangle above the $x$ -axis has a base of length $1$ and a height of $1$ , so the signed area is

$A_{3} = \frac{1}{2} (1) (1)$

$= 0.5$

Adding these values gives a displacement of

$0.75 - 2.75 + 0.5 = - 1.5$

b) Total distance

(spoiler)

Total distance is the sum of the absolute values of those areas:

$∣0.75∣ + ∣ - 2.75∣ + ∣0.5∣ = 4$

c) $s (6)$ given $s (2) = 1$

(spoiler)

The displacement from $t = 2$ to $t = 6$ is

$\int_{2}^{6} v (t) d t = s (6) - s (2)$

$s (6) = s (2) + \int_{2}^{6} v (t) d t$

Now compute the net signed area from $t = 2$ to $t = 6$ :

On $2 < t < 4$ :

The rectangle below the $x$ -axis has a width of $1$ and a length of $2$ , so the signed area is

$A = - 2$

On $4 < t < 6$ :

The triangle from $t = 4$ to $t = 5$ matches the triangle from $t = 5$ to $t = 6$ with opposite sign, so these signed areas cancel.

So $\int_{2}^{6} v (t) d t = - 2$ . Substitute this and $s (2) = 1$ :

$s (6) = s (2) + \int_{2}^{6} v (t) d t$

$= 1 - 2$

$= - 1$

Achievable AP Calculus AB

8. Applications of integrals

Our AP Calculus AB course is currently in development and is a work-in-progress.

Motion with integrals

7 min read

Font

Discuss

Feedback

What you’ll learn:

How to recover velocity and position from acceleration
Compute displacement and total distance using integration

Displacement

Integrating an object’s velocity over a time interval gives its displacement, meaning the net change in position.

$Displacement = \int_{a}^{b} v (t) d t$

For example, let the velocity (in meters per second) be

$v (t) = 3 t - 6$

Find the change in position from $t = 0$ to $t = 4$ seconds:

$\int_{0}^{4} v (t) d t$

$= \int_{0}^{4} (3 t - 6) d t$

$= (\frac{3}{2} t^{2} - 6 t)_{0}^{4}$

$= 0$

A displacement of $0$ means the object ends where it started after the first 4 seconds.

Total distance

Total distance traveled counts how far the object moves regardless of direction. To do that, we integrate the absolute value of velocity.

$Total distance = \int_{a}^{b} ∣ v (t) ∣ d t$

To evaluate:

Find when the velocity changes sign (first find where it equals $0$ ).
Write the absolute value as a piecewise function.
Split the integral at those points.

Using the same velocity function and the interval $0 \leq t \leq 4$ :

$v (t) = 3 t - 6$

The velocity is $0$ when $3 t - 6 = 0$ , so $t = 2$ . That’s where the direction changes, so it’s the breakpoint for the absolute value.

$∣3 t - 6∣ = ⎩ ⎨ ⎧ 3 t - 6 - (3 t - 6) if t \geq 2 if t < 2$

Then the total distance is

$\int_{0}^{2} - (3 t - 6) d t + \int_{2}^{4} (3 t - 6) d t$

$= \int_{2}^{4} (3 t - 6) d t - \int_{0}^{2} (3 t - 6) d t$

$= 12$

For problems that allow a calculator/Desmos, you can enter the absolute value directly, like

$\int_{0}^{4} ∣3 t - 6∣ d t$

Accumulating position from velocity

If you know the velocity function and the position at one time, you can find the position at another time using the Fundamental Theorem of Calculus.

The velocity, in units per second, of an object moving along a number line is $v (t) = t^{2} - 6 t + 5$ . Its position at time $t = 0$ is $s (0) = 10$ .

a) Find the particle’s position at $t = 6$ seconds.
b) Find the total distance the particle traveled on $0 \leq t \leq 6$ .

Solutions

a) $s (6)$

(spoiler)

By the FTC,

$\int_{0}^{6} v (t) d t = s (6) - s (0)$

$s (6) = s (0) + \int_{0}^{6} (t^{2} - 6 t + 5) d t$

$= 10 - 6$

$= 4$

So the particle starts at $10$ units and is at $4$ units when $t = 6$ seconds.

b) Total distance over $[0, 6]$

(spoiler)

The total distance traveled is

$\int_{0}^{6} ∣ v (t) ∣ d t$

$= \int_{0}^{6} ∣ t^{2} - 6 t + 5∣ d t$

On $[0, 6]$ , the velocity is positive for $0 \leq t < 1$ and $5 < t \leq 6$ , and negative for $1 < t < 5$ . So the absolute value in piecewise form is

$∣ t^{2} - 6 t + 5∣ = ⎩ ⎨ ⎧ t^{2} - 6 t + 5 - (t^{2} - 6 t + 5) t^{2} - 6 t + 5 if 0 \leq t < 1 if 1 \leq t \leq 5 if 5 < t \leq 6$

So the integral set-up for the total distance is

$\int_{0}^{1} (t^{2} - 6 t + 5) d t - \int_{1}^{5} (t^{2} - 6 t + 5) d t + \int_{5}^{6} (t^{2} - 6 t + 5) d t$

$= 15.333 units$

From acceleration to position

Sometimes you’re given acceleration and initial conditions for both velocity and position. In that case, you integrate twice, solving for the constant of integration $+ C$ each time.

Given:

$a (t) = 2 t + 1$

$v (1) = 3$

$s (0) = 5$

Find $s (t)$ .

Solution

(spoiler)

$v (t) = \int a (t) d t$

$= \int (2 t + 1) d t$

$= t^{2} + t + C$

Use $v (1) = 3$ to find $C$ :

$3 = 1^{2} + 1 + C$

$C = 1$

So $v (t) = t^{2} + t + 1$ . Then

$s (t) = \int v (t) d t$

$= \int (t^{2} + t + 1) d t$

$= \frac{t ^{3}}{3} + \frac{t ^{2}}{2} + t + C$

Use $s (0) = 5$ to find $C$ :

$5 = \frac{0 ^{3}}{3} + \frac{0 ^{2}}{2} + 0 + C$

$C = 5$

Therefore

$s (t) = \frac{t ^{3}}{3} + \frac{t ^{2}}{2} + t + 5$

Velocity graphs

Sometimes velocity is given as a graph rather than a formula. Since integrating velocity gives change in position, you can find:

displacement by computing net signed area under $v (t)$
total distance by computing total area (treating areas below the axis as positive)

The velocity of a particle moving along a line is shown in the graph below for $0 \leq t \leq 6$ .

Graph of v(t)

a) Find the displacement of the particle from $t = 0$ to $t = 6$ .
b) Find the total distance traveled over this time.
c) If the particle’s position at $t = 2$ is $s (2) = 1$ , find its position at $t = 6$ .

Solutions

a) Displacement

(spoiler)

To find displacement, add the signed areas (areas above the axis are positive; below are negative).

On $0 < t < 1.5$ :

The triangle above the $x$ -axis has a base of length $1.5$ and a height of $1$ , so the signed area is

$A_{1} = \frac{1}{2} (1.5) (1)$

$= 0.75$

On $1.5 < t < 5$ :

The trapezoid below the $x$ -axis has a height of $1$ and two bases of length $3.5$ and $2$ , so the signed area is

$A_{2} = \frac{1}{2} (- 1) (3.5 + 2)$

$= - 2.75$

On $5 < t < 6$ :

The triangle above the $x$ -axis has a base of length $1$ and a height of $1$ , so the signed area is

$A_{3} = \frac{1}{2} (1) (1)$

$= 0.5$

Adding these values gives a displacement of

$0.75 - 2.75 + 0.5 = - 1.5$

b) Total distance

(spoiler)

Total distance is the sum of the absolute values of those areas:

$∣0.75∣ + ∣ - 2.75∣ + ∣0.5∣ = 4$

c) $s (6)$ given $s (2) = 1$

(spoiler)

The displacement from $t = 2$ to $t = 6$ is

$\int_{2}^{6} v (t) d t = s (6) - s (2)$

$s (6) = s (2) + \int_{2}^{6} v (t) d t$

Now compute the net signed area from $t = 2$ to $t = 6$ :

On $2 < t < 4$ :

The rectangle below the $x$ -axis has a width of $1$ and a length of $2$ , so the signed area is

$A = - 2$

On $4 < t < 6$ :

The triangle from $t = 4$ to $t = 5$ matches the triangle from $t = 5$ to $t = 6$ with opposite sign, so these signed areas cancel.

So $\int_{2}^{6} v (t) d t = - 2$ . Substitute this and $s (2) = 1$ :

$s (6) = s (2) + \int_{2}^{6} v (t) d t$

$= 1 - 2$

$= - 1$

Displacement

Displacement = definite integral of velocity: $\int_{a}^{b} v (t) d t$
Represents net change in position over $[a, b]$
Signed value; can be zero if object returns to start

Total distance

Total distance = $\int_{a}^{b} ∣ v (t) ∣ d t$
Integrate absolute value of velocity
Requires splitting integral at points where $v (t) = 0$

Accumulating position from velocity

Position at time $t$ : $s (t) = s (a) + \int_{a}^{t} v (x) d x$
Use Fundamental Theorem of Calculus
Displacement = change in position, total distance = integral of $∣ v (t) ∣$

From acceleration to position

Integrate acceleration to get velocity: $v (t) = \int a (t) d t + C$
Integrate velocity to get position: $s (t) = \int v (t) d t + C$
Use initial conditions to solve for constants

Velocity graphs

Displacement = net signed area under $v (t)$ curve
Total distance = sum of absolute values of areas under $v (t)$
For position at a later time: add displacement over interval to known position

Key formulas

Displacement: $\int_{a}^{b} v (t) d t$
Total distance: $\int_{a}^{b} ∣ v (t) ∣ d t$
Position from velocity: $s (t) = s (a) + \int_{a}^{t} v (x) d x$
Velocity from acceleration: $v (t) = \int a (t) d t + C$

Motion with integrals

What you’ll learn:

How to recover velocity and position from acceleration
Compute displacement and total distance using integration

Displacement

Integrating an object’s velocity over a time interval gives its displacement, meaning the net change in position.

$Displacement = \int_{a}^{b} v (t) d t$

For example, let the velocity (in meters per second) be

$v (t) = 3 t - 6$

Find the change in position from $t = 0$ to $t = 4$ seconds:

$\int_{0}^{4} v (t) d t$

$= \int_{0}^{4} (3 t - 6) d t$

$= (\frac{3}{2} t^{2} - 6 t)_{0}^{4}$

$= 0$

A displacement of $0$ means the object ends where it started after the first 4 seconds.

Total distance

Total distance traveled counts how far the object moves regardless of direction. To do that, we integrate the absolute value of velocity.

$Total distance = \int_{a}^{b} ∣ v (t) ∣ d t$

To evaluate:

Find when the velocity changes sign (first find where it equals $0$ ).
Write the absolute value as a piecewise function.
Split the integral at those points.

Using the same velocity function and the interval $0 \leq t \leq 4$ :

$v (t) = 3 t - 6$

The velocity is $0$ when $3 t - 6 = 0$ , so $t = 2$ . That’s where the direction changes, so it’s the breakpoint for the absolute value.

$∣3 t - 6∣ = ⎩ ⎨ ⎧ 3 t - 6 - (3 t - 6) if t \geq 2 if t < 2$

Then the total distance is

$\int_{0}^{2} - (3 t - 6) d t + \int_{2}^{4} (3 t - 6) d t$

$= \int_{2}^{4} (3 t - 6) d t - \int_{0}^{2} (3 t - 6) d t$

$= 12$

For problems that allow a calculator/Desmos, you can enter the absolute value directly, like

$\int_{0}^{4} ∣3 t - 6∣ d t$

Accumulating position from velocity

If you know the velocity function and the position at one time, you can find the position at another time using the Fundamental Theorem of Calculus.

The velocity, in units per second, of an object moving along a number line is $v (t) = t^{2} - 6 t + 5$ . Its position at time $t = 0$ is $s (0) = 10$ .

a) Find the particle’s position at $t = 6$ seconds.
b) Find the total distance the particle traveled on $0 \leq t \leq 6$ .

Solutions

a) $s (6)$

(spoiler)

By the FTC,

$\int_{0}^{6} v (t) d t = s (6) - s (0)$

$s (6) = s (0) + \int_{0}^{6} (t^{2} - 6 t + 5) d t$

$= 10 - 6$

$= 4$

So the particle starts at $10$ units and is at $4$ units when $t = 6$ seconds.

b) Total distance over $[0, 6]$

(spoiler)

The total distance traveled is

$\int_{0}^{6} ∣ v (t) ∣ d t$

$= \int_{0}^{6} ∣ t^{2} - 6 t + 5∣ d t$

On $[0, 6]$ , the velocity is positive for $0 \leq t < 1$ and $5 < t \leq 6$ , and negative for $1 < t < 5$ . So the absolute value in piecewise form is

$∣ t^{2} - 6 t + 5∣ = ⎩ ⎨ ⎧ t^{2} - 6 t + 5 - (t^{2} - 6 t + 5) t^{2} - 6 t + 5 if 0 \leq t < 1 if 1 \leq t \leq 5 if 5 < t \leq 6$

So the integral set-up for the total distance is

$\int_{0}^{1} (t^{2} - 6 t + 5) d t - \int_{1}^{5} (t^{2} - 6 t + 5) d t + \int_{5}^{6} (t^{2} - 6 t + 5) d t$

$= 15.333 units$

From acceleration to position

Sometimes you’re given acceleration and initial conditions for both velocity and position. In that case, you integrate twice, solving for the constant of integration $+ C$ each time.

Given:

$a (t) = 2 t + 1$

$v (1) = 3$

$s (0) = 5$

Find $s (t)$ .

Solution

(spoiler)

$v (t) = \int a (t) d t$

$= \int (2 t + 1) d t$

$= t^{2} + t + C$

Use $v (1) = 3$ to find $C$ :

$3 = 1^{2} + 1 + C$

$C = 1$

So $v (t) = t^{2} + t + 1$ . Then

$s (t) = \int v (t) d t$

$= \int (t^{2} + t + 1) d t$

$= \frac{t ^{3}}{3} + \frac{t ^{2}}{2} + t + C$

Use $s (0) = 5$ to find $C$ :

$5 = \frac{0 ^{3}}{3} + \frac{0 ^{2}}{2} + 0 + C$

$C = 5$

Therefore

$s (t) = \frac{t ^{3}}{3} + \frac{t ^{2}}{2} + t + 5$

Velocity graphs

Sometimes velocity is given as a graph rather than a formula. Since integrating velocity gives change in position, you can find:

displacement by computing net signed area under $v (t)$
total distance by computing total area (treating areas below the axis as positive)

The velocity of a particle moving along a line is shown in the graph below for $0 \leq t \leq 6$ .

a) Find the displacement of the particle from $t = 0$ to $t = 6$ .
b) Find the total distance traveled over this time.
c) If the particle’s position at $t = 2$ is $s (2) = 1$ , find its position at $t = 6$ .

Solutions

a) Displacement

(spoiler)

To find displacement, add the signed areas (areas above the axis are positive; below are negative).

On $0 < t < 1.5$ :

The triangle above the $x$ -axis has a base of length $1.5$ and a height of $1$ , so the signed area is

$A_{1} = \frac{1}{2} (1.5) (1)$

$= 0.75$

On $1.5 < t < 5$ :

The trapezoid below the $x$ -axis has a height of $1$ and two bases of length $3.5$ and $2$ , so the signed area is

$A_{2} = \frac{1}{2} (- 1) (3.5 + 2)$

$= - 2.75$

On $5 < t < 6$ :

The triangle above the $x$ -axis has a base of length $1$ and a height of $1$ , so the signed area is

$A_{3} = \frac{1}{2} (1) (1)$

$= 0.5$

Adding these values gives a displacement of

$0.75 - 2.75 + 0.5 = - 1.5$

b) Total distance

(spoiler)

Total distance is the sum of the absolute values of those areas:

$∣0.75∣ + ∣ - 2.75∣ + ∣0.5∣ = 4$

c) $s (6)$ given $s (2) = 1$

(spoiler)

The displacement from $t = 2$ to $t = 6$ is

$\int_{2}^{6} v (t) d t = s (6) - s (2)$

$s (6) = s (2) + \int_{2}^{6} v (t) d t$

Now compute the net signed area from $t = 2$ to $t = 6$ :

On $2 < t < 4$ :

The rectangle below the $x$ -axis has a width of $1$ and a length of $2$ , so the signed area is

$A = - 2$

On $4 < t < 6$ :

The triangle from $t = 4$ to $t = 5$ matches the triangle from $t = 5$ to $t = 6$ with opposite sign, so these signed areas cancel.

So $\int_{2}^{6} v (t) d t = - 2$ . Substitute this and $s (2) = 1$ :

$s (6) = s (2) + \int_{2}^{6} v (t) d t$

$= 1 - 2$

$= - 1$