Motion with integrals | Applications of integrals

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 were used to relate position, velocity, and acceleration. Now we explore the reverse - given an acceleration function and an initial condition, we can find the velocity of an object, and integrating again gives the position.

Displacement

Integrating the velocity function of an object over a time interval gives its displacement (net change in position).

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

For example, let the velocity of an object in meters per second be

$v (t) = 3 t - 6$

and suppose we want to find its change in position from $t = 0$ to $t = 4$ seconds. Integrating,

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

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

$= \frac{3}{2} t^{2} - 6 t_{0}^{4}$

$= 0$

This means that the object returned to its starting position after the 1st four seconds.

Total distance

On the other hand, finding the total distance traveled requires taking the absolute value of the velocity function to account for changes in direction.

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

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.

Let’s use the same velocity function from before and find the total distance traveled from $0 \leq t \leq 4$ . The function

$v (t) = 3 t - 6$

changes sign when it crosses the $t$ -axis at $t = 2$ , which is the breakpoint for the absolute value in piecewise form.

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

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 integrals give signed area, we essentially have to subtract the area of any regions below the $x$ -axis to add it to the total distance.

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

$\int_{0}^{4} ∣3 t - 6∣ d t$

Accumulating position from velocity

Given the velocity function and the position at one time, we can find the position at another 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,

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

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

$= 10 - 6$

$= 4$

The particle starts at $10$ units and ends up 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 function 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

In some cases, you’re given acceleration and initial conditions for both velocity and position. These require integrating twice and 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 the given $v (1) = 3$ 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 the given $s (0) = 5$ to solve for $C$ :

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

$C = 5$

Therefore

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

Velocity graphs

Sometimes the velocity of an object is given graphically rather than as a function. Since integrating the velocity gives position, displacement and total distance can be found by computing the net signed area and total area, respectively.

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 the displacement, calculate the net signed area.

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 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)

The total distance traveled can be found by adding up the absolute values of the areas found in the previous part:

$∣0.75∣ + ∣ - 2.75∣ + ∣0.5∣ = 4$

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

(spoiler)

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

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

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

We calculate 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 $t = 5$ is the same as the one from $t = 5$ to $t = 6$ , so the signed areas cancel out.

Then the net signed area is $- 2$ , as is $\int_{2}^{6} v (t) d t$ . Substituting this and $s (2) = 1$ ,

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

$= 1 - 2$

$= - 1$

What you’ll learn:

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

Displacement

Integrating the velocity function of an object over a time interval gives its displacement (net change in position).

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

For example, let the velocity of an object in meters per second be

$v (t) = 3 t - 6$

and suppose we want to find its change in position from $t = 0$ to $t = 4$ seconds. Integrating,

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

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

$= \frac{3}{2} t^{2} - 6 t_{0}^{4}$

$= 0$

This means that the object returned to its starting position after the 1st four seconds.

Total distance

On the other hand, finding the total distance traveled requires taking the absolute value of the velocity function to account for changes in direction.

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

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.

Let’s use the same velocity function from before and find the total distance traveled from $0 \leq t \leq 4$ . The function

$v (t) = 3 t - 6$

changes sign when it crosses the $t$ -axis at $t = 2$ , which is the breakpoint for the absolute value in piecewise form.

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

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 integrals give signed area, we essentially have to subtract the area of any regions below the $x$ -axis to add it to the total distance.

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

$\int_{0}^{4} ∣3 t - 6∣ d t$

Accumulating position from velocity

Given the velocity function and the position at one time, we can find the position at another 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,

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

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

$= 10 - 6$

$= 4$

The particle starts at $10$ units and ends up 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 function 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

In some cases, you’re given acceleration and initial conditions for both velocity and position. These require integrating twice and 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 the given $v (1) = 3$ 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 the given $s (0) = 5$ to solve for $C$ :

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

$C = 5$

Therefore

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

Velocity graphs

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 the displacement, calculate the net signed area.

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 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)

The total distance traveled can be found by adding up the absolute values of the areas found in the previous part:

$∣0.75∣ + ∣ - 2.75∣ + ∣0.5∣ = 4$

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

(spoiler)

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

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

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

We calculate 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 $t = 5$ is the same as the one from $t = 5$ to $t = 6$ , so the signed areas cancel out.

Then the net signed area is $- 2$ , as is $\int_{2}^{6} v (t) d t$ . Substituting this and $s (2) = 1$ ,

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

$= 1 - 2$

$= - 1$