Motion with integrals | AP Calc AB Applications of integrals

What you’ll learn

Motion analysis: Recover velocity and position from acceleration using integration.
Displacement vs. distance: Integrate velocity to find net displacement and speed to find total distance.

Motion along a line is one of the most fundamental applications of calculus. Recall from section 4.2 that derivatives connect position, velocity, and acceleration.

Now, we work in reverse: if you have an acceleration function and an initial condition, you can integrate to find velocity, and integrate again to find position.

Displacement vs. total distance

While displacement measures the net change in an object’s position, total distance measures all ground covered, regardless of direction.

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

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

Example 1

A particle moves along the $x$ -axis so that its velocity at time $t$ is given by $v (t) = 3 t - 6$ .

a) Find the displacement of the particle from time $t = 0$ to $t = 4$ .

b) Find the total distance traveled by the particle from time $t = 0$ to $t = 4$ .

Solutions

a) Displacement

(spoiler)

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

A displacement of $0$ means the particle ended exactly where it started.

b) Total distance:

(spoiler)

To integrate $∣3 t - 6∣$ without a calculator, find where the velocity changes sign by setting $v (t) = 0$ :

$3 t - 6 = 0 ⟹ t = 2$

Test intervals: Since $v (t)$ is negative on $[0, 2]$ and positive on $[2, 4]$ , multiply the first interval by $- 1$ :

$\int_{0}^{4} ∣ v (t) ∣ d t = - \int_{0}^{2} (3 t - 6) d t + \int_{2}^{4} (3 t - 6) d t = - [\frac{3}{2} t^{2} - 6 t]_{0}^{2} + [\frac{3}{2} t^{2} - 6 t]_{2}^{4} = - (- 6) + (0 - (- 6)) = 12$

The particle traveled 6 units backward and 6 units forward, covering a total distance of 12 units.

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

$\int_{0}^{4} ∣3 t - 6∣ d t or \int_{0}^{4} abs (3 t - 6) d t$

Recovering position from velocity or acceleration

On the AP exam, questions frequently test your ability to navigate back and forth along the motion chain:

$a (t) integrate v (t) integrate x (t)$

Finding position at a specific time

When given the velocity function and the object’s position at a given starting time, use the Fundamental theorem of calculus to find the position at any other time.

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

Conceptually: Current position = previous position + displacement.

This approach is common on the AP exam, particularly on the calculator-active section where you can evaluate the definite integral numerically.

Example 2

(Calculator-active question)

A particle moves along the $x$ -axis so that its velocity at time $t$ is given by $v (t) = 10 sin (e^{0.5 t}) - 4$ . The position of the particle at time $t = 0$ is $x (0) = 5$ . Find the position of the particle at time $t = 3$ .

Solution

(spoiler)

Answer: $x (3) = 7.241$

To find the final position, add the given position to the net change in position (displacement).

$Setup: x (3) = x (0) + \int_{0}^{3} v (t) d t$

In Desmos, define the velocity function:

$v (t) = 10 sin (e^{0.5 t}) - 4$

Then substitute $x (0) = 5$ and evaluate $x (3)$ by entering:

$5 + \int_{0}^{3} v (t) d t$

This gives approximately $7.241$ .

Recovering the position function

Sometimes the goal is to find a formula for the particle’s position. If you’re given acceleration and enough initial conditions, integrate twice.

Integrate acceleration to find velocity.
Use the velocity condition to determine the first constant of integration.
Integrate velocity to find position.
Use the position condition to determine the second constant of integration.

Example 3

A particle moves along the $x$ -axis so that its acceleration at time $t$ is given by $a (t) = 2 t + 1$ . At time $t = 1$ , the velocity of the particle is $v (1) = 3$ , and at time $t = 0$ , the position of the particle is $x (0) = 5$ . Write an expression for the position function $x (t)$ .

Solution

(spoiler)

1. Integrate acceleration for velocity:

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

Use the initial condition $v (1) = 3$ to find $C_{1}$ :

$3 = 1^{2} + 1 + C_{1} ⟹ C_{1} = 1$

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

2. Integrate velocity for position:

$x (t) = \int (t^{2} + t + 1) d t = \frac{t ^{3}}{3} + \frac{t ^{2}}{2} + t + C_{2}$

Use $x (0) = 5$ to find $C_{2} ⟹ C_{2} = 5$ .

Therefore the position function is:

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

Velocity graphs

Sometimes velocity is given as a graph rather than an algebraic formula. Because the definite integral represents the area under the curve, you can navigate the motion chain geometrically:

Displacement: Calculate the net signed area (areas above the time axis are positive, areas below are negative).
Total Distance: Calculate the total area (treat all areas as positive).
Acceleration: Read the slope of the velocity graph at that point ( $a (t) = v^{'} (t)$ ).

Example 4

The graph of the velocity $v (t)$ , consisting of line segments, is shown below for $0 \leq t \leq 6$ . At time $t = 2$ , the position of the particle is $x (2) = 1$ .

Graph of v(t)

a) Find the displacement of the particle from $t = 0$ to $t = 6$ .

b) Find the total distance traveled by the particle over $0 \leq t \leq 6$ .

c) Find the position of the particle at $t = 6$ .

d) Find the acceleration of the particle at $t = 5$ .

e) Is the speed of the particle increasing, decreasing, or not changing at $t = 3$ ?

Answers

(spoiler)

a) Displacement: $- 1.5$
b) Total distance: $4$
c) $x (6) = - 1$
d) Acceleration: $1$
e) Not changing

Solutions

a) Displacement

(spoiler)

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

On the interval $[0, 1.5]$ : Triangle (above axis)
- $A_{1} = \frac{1}{2} (1.5) (1) = 0.75$
On $[1.5, 5]$ : Trapezoid (below axis)
- $A_{2} = \frac{- 1}{2} (3.5 + 2) = - 2.75$
On $[5, 6]$ : Triangle (above axis)
- $A_{3} = \frac{1}{2} (1) (1) = 0.5$

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) Position at $t = 6$

(spoiler)

The final position is the initial position plus the displacement from $t = 2$ to $t = 6$ :

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

$\int_{2}^{6} v (t) d t$ is the net signed area from $t = 2$ to $t = 6$ :

On $[2, 4]$ : $A = - 2$ (rectangle below axis)
On $[4, 6]$ : $A = 0$ (triangles above and below cancel out)

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

$x (6) = 1 - 2 = - 1$

d) Acceleration at $t = 5$

(spoiler)

Acceleration is the derivative of velocity. To find $a (5) = v^{'} (5)$ , look at the slope of the line segment that connects points $(4, - 1)$ and $(6, 1)$ .

$v^{'} (5) = \frac{1 - ( - 1 )}{6 - 4} = 1$

e) Speed behavior at $t = 3$

(spoiler)

Speed increases when $v (t)$ and $a (t)$ have the same sign.
It decreases when $v (t)$ and $a (t)$ have opposing signs.
It’s constant when $a (t) = 0$ .

At $t = 3$ , $v (3) = - 1 < 0$ and $a (3) = v^{'} (3) = 0$ (horizontal tangent line).

Therefore, the speed of the particle is not changing at $t = 3$ .

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: $v (t) = \int a (t) d t + C$

What you’ll learn

Motion analysis: Recover velocity and position from acceleration using integration.
Displacement vs. distance: Integrate velocity to find net displacement and speed to find total distance.

Motion along a line is one of the most fundamental applications of calculus. Recall from section 4.2 that derivatives connect position, velocity, and acceleration.

Now, we work in reverse: if you have an acceleration function and an initial condition, you can integrate to find velocity, and integrate again to find position.

Displacement vs. total distance

While displacement measures the net change in an object’s position, total distance measures all ground covered, regardless of direction.

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

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

Example 1

A particle moves along the $x$ -axis so that its velocity at time $t$ is given by $v (t) = 3 t - 6$ .

a) Find the displacement of the particle from time $t = 0$ to $t = 4$ .

b) Find the total distance traveled by the particle from time $t = 0$ to $t = 4$ .

Solutions

a) Displacement

(spoiler)

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

A displacement of $0$ means the particle ended exactly where it started.

b) Total distance:

(spoiler)

To integrate $∣3 t - 6∣$ without a calculator, find where the velocity changes sign by setting $v (t) = 0$ :

$3 t - 6 = 0 ⟹ t = 2$

Test intervals: Since $v (t)$ is negative on $[0, 2]$ and positive on $[2, 4]$ , multiply the first interval by $- 1$ :

$\int_{0}^{4} ∣ v (t) ∣ d t = - \int_{0}^{2} (3 t - 6) d t + \int_{2}^{4} (3 t - 6) d t = - [\frac{3}{2} t^{2} - 6 t]_{0}^{2} + [\frac{3}{2} t^{2} - 6 t]_{2}^{4} = - (- 6) + (0 - (- 6)) = 12$

The particle traveled 6 units backward and 6 units forward, covering a total distance of 12 units.

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

$\int_{0}^{4} ∣3 t - 6∣ d t or \int_{0}^{4} abs (3 t - 6) d t$

Recovering position from velocity or acceleration

On the AP exam, questions frequently test your ability to navigate back and forth along the motion chain:

$a (t) integrate v (t) integrate x (t)$

Finding position at a specific time

When given the velocity function and the object’s position at a given starting time, use the Fundamental theorem of calculus to find the position at any other time.

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

Conceptually: Current position = previous position + displacement.

This approach is common on the AP exam, particularly on the calculator-active section where you can evaluate the definite integral numerically.

Example 2

(Calculator-active question)

A particle moves along the $x$ -axis so that its velocity at time $t$ is given by $v (t) = 10 sin (e^{0.5 t}) - 4$ . The position of the particle at time $t = 0$ is $x (0) = 5$ . Find the position of the particle at time $t = 3$ .

Solution

(spoiler)

Answer: $x (3) = 7.241$

To find the final position, add the given position to the net change in position (displacement).

$Setup: x (3) = x (0) + \int_{0}^{3} v (t) d t$

In Desmos, define the velocity function:

$v (t) = 10 sin (e^{0.5 t}) - 4$

Then substitute $x (0) = 5$ and evaluate $x (3)$ by entering:

$5 + \int_{0}^{3} v (t) d t$

This gives approximately $7.241$ .

Recovering the position function

Sometimes the goal is to find a formula for the particle’s position. If you’re given acceleration and enough initial conditions, integrate twice.

Integrate acceleration to find velocity.
Use the velocity condition to determine the first constant of integration.
Integrate velocity to find position.
Use the position condition to determine the second constant of integration.

Example 3

A particle moves along the $x$ -axis so that its acceleration at time $t$ is given by $a (t) = 2 t + 1$ . At time $t = 1$ , the velocity of the particle is $v (1) = 3$ , and at time $t = 0$ , the position of the particle is $x (0) = 5$ . Write an expression for the position function $x (t)$ .

Solution

(spoiler)

1. Integrate acceleration for velocity:

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

Use the initial condition $v (1) = 3$ to find $C_{1}$ :

$3 = 1^{2} + 1 + C_{1} ⟹ C_{1} = 1$

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

2. Integrate velocity for position:

$x (t) = \int (t^{2} + t + 1) d t = \frac{t ^{3}}{3} + \frac{t ^{2}}{2} + t + C_{2}$

Use $x (0) = 5$ to find $C_{2} ⟹ C_{2} = 5$ .

Therefore the position function is:

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

Velocity graphs

Sometimes velocity is given as a graph rather than an algebraic formula. Because the definite integral represents the area under the curve, you can navigate the motion chain geometrically:

Displacement: Calculate the net signed area (areas above the time axis are positive, areas below are negative).
Total Distance: Calculate the total area (treat all areas as positive).
Acceleration: Read the slope of the velocity graph at that point ( $a (t) = v^{'} (t)$ ).

Example 4

The graph of the velocity $v (t)$ , consisting of line segments, is shown below for $0 \leq t \leq 6$ . At time $t = 2$ , the position of the particle is $x (2) = 1$ .

Graph of v(t)

a) Find the displacement of the particle from $t = 0$ to $t = 6$ .

b) Find the total distance traveled by the particle over $0 \leq t \leq 6$ .

c) Find the position of the particle at $t = 6$ .

d) Find the acceleration of the particle at $t = 5$ .

e) Is the speed of the particle increasing, decreasing, or not changing at $t = 3$ ?

Answers

(spoiler)

a) Displacement: $- 1.5$
b) Total distance: $4$
c) $x (6) = - 1$
d) Acceleration: $1$
e) Not changing

Solutions

a) Displacement

(spoiler)

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

On the interval $[0, 1.5]$ : Triangle (above axis)
- $A_{1} = \frac{1}{2} (1.5) (1) = 0.75$
On $[1.5, 5]$ : Trapezoid (below axis)
- $A_{2} = \frac{- 1}{2} (3.5 + 2) = - 2.75$
On $[5, 6]$ : Triangle (above axis)
- $A_{3} = \frac{1}{2} (1) (1) = 0.5$

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) Position at $t = 6$

(spoiler)

The final position is the initial position plus the displacement from $t = 2$ to $t = 6$ :

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

$\int_{2}^{6} v (t) d t$ is the net signed area from $t = 2$ to $t = 6$ :

On $[2, 4]$ : $A = - 2$ (rectangle below axis)
On $[4, 6]$ : $A = 0$ (triangles above and below cancel out)

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

$x (6) = 1 - 2 = - 1$

d) Acceleration at $t = 5$

(spoiler)

Acceleration is the derivative of velocity. To find $a (5) = v^{'} (5)$ , look at the slope of the line segment that connects points $(4, - 1)$ and $(6, 1)$ .

$v^{'} (5) = \frac{1 - ( - 1 )}{6 - 4} = 1$

e) Speed behavior at $t = 3$

(spoiler)

Speed increases when $v (t)$ and $a (t)$ have the same sign.
It decreases when $v (t)$ and $a (t)$ have opposing signs.
It’s constant when $a (t) = 0$ .

At $t = 3$ , $v (3) = - 1 < 0$ and $a (3) = v^{'} (3) = 0$ (horizontal tangent line).

Therefore, the speed of the particle is not changing at $t = 3$ .

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: $v (t) = \int a (t) d t + C$

What you’ll learn

Motion analysis: Recover velocity and position from acceleration using integration.
Displacement vs. distance: Integrate velocity to find net displacement and speed to find total distance.

Motion along a line is one of the most fundamental applications of calculus. Recall from section 4.2 that derivatives connect position, velocity, and acceleration.

Now, we work in reverse: if you have an acceleration function and an initial condition, you can integrate to find velocity, and integrate again to find position.

Displacement vs. total distance

While displacement measures the net change in an object’s position, total distance measures all ground covered, regardless of direction.

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

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

Example 1

A particle moves along the $x$ -axis so that its velocity at time $t$ is given by $v (t) = 3 t - 6$ .

a) Find the displacement of the particle from time $t = 0$ to $t = 4$ .

b) Find the total distance traveled by the particle from time $t = 0$ to $t = 4$ .

Solutions

a) Displacement

(spoiler)

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

A displacement of $0$ means the particle ended exactly where it started.

b) Total distance:

(spoiler)

To integrate $∣3 t - 6∣$ without a calculator, find where the velocity changes sign by setting $v (t) = 0$ :

$3 t - 6 = 0 ⟹ t = 2$

Test intervals: Since $v (t)$ is negative on $[0, 2]$ and positive on $[2, 4]$ , multiply the first interval by $- 1$ :

$\int_{0}^{4} ∣ v (t) ∣ d t = - \int_{0}^{2} (3 t - 6) d t + \int_{2}^{4} (3 t - 6) d t = - [\frac{3}{2} t^{2} - 6 t]_{0}^{2} + [\frac{3}{2} t^{2} - 6 t]_{2}^{4} = - (- 6) + (0 - (- 6)) = 12$

The particle traveled 6 units backward and 6 units forward, covering a total distance of 12 units.

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

$\int_{0}^{4} ∣3 t - 6∣ d t or \int_{0}^{4} abs (3 t - 6) d t$

Recovering position from velocity or acceleration

On the AP exam, questions frequently test your ability to navigate back and forth along the motion chain:

$a (t) integrate v (t) integrate x (t)$

Finding position at a specific time

When given the velocity function and the object’s position at a given starting time, use the Fundamental theorem of calculus to find the position at any other time.

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

Conceptually: Current position = previous position + displacement.

This approach is common on the AP exam, particularly on the calculator-active section where you can evaluate the definite integral numerically.

Example 2

(Calculator-active question)

A particle moves along the $x$ -axis so that its velocity at time $t$ is given by $v (t) = 10 sin (e^{0.5 t}) - 4$ . The position of the particle at time $t = 0$ is $x (0) = 5$ . Find the position of the particle at time $t = 3$ .

Solution

(spoiler)

Answer: $x (3) = 7.241$

To find the final position, add the given position to the net change in position (displacement).

$Setup: x (3) = x (0) + \int_{0}^{3} v (t) d t$

In Desmos, define the velocity function:

$v (t) = 10 sin (e^{0.5 t}) - 4$

Then substitute $x (0) = 5$ and evaluate $x (3)$ by entering:

$5 + \int_{0}^{3} v (t) d t$

This gives approximately $7.241$ .

Recovering the position function

Sometimes the goal is to find a formula for the particle’s position. If you’re given acceleration and enough initial conditions, integrate twice.

Integrate acceleration to find velocity.
Use the velocity condition to determine the first constant of integration.
Integrate velocity to find position.
Use the position condition to determine the second constant of integration.

Example 3

A particle moves along the $x$ -axis so that its acceleration at time $t$ is given by $a (t) = 2 t + 1$ . At time $t = 1$ , the velocity of the particle is $v (1) = 3$ , and at time $t = 0$ , the position of the particle is $x (0) = 5$ . Write an expression for the position function $x (t)$ .

Solution

(spoiler)

1. Integrate acceleration for velocity:

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

Use the initial condition $v (1) = 3$ to find $C_{1}$ :

$3 = 1^{2} + 1 + C_{1} ⟹ C_{1} = 1$

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

2. Integrate velocity for position:

$x (t) = \int (t^{2} + t + 1) d t = \frac{t ^{3}}{3} + \frac{t ^{2}}{2} + t + C_{2}$

Use $x (0) = 5$ to find $C_{2} ⟹ C_{2} = 5$ .

Therefore the position function is:

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

Velocity graphs

Sometimes velocity is given as a graph rather than an algebraic formula. Because the definite integral represents the area under the curve, you can navigate the motion chain geometrically:

Displacement: Calculate the net signed area (areas above the time axis are positive, areas below are negative).
Total Distance: Calculate the total area (treat all areas as positive).
Acceleration: Read the slope of the velocity graph at that point ( $a (t) = v^{'} (t)$ ).

Example 4

The graph of the velocity $v (t)$ , consisting of line segments, is shown below for $0 \leq t \leq 6$ . At time $t = 2$ , the position of the particle is $x (2) = 1$ .

a) Find the displacement of the particle from $t = 0$ to $t = 6$ .

b) Find the total distance traveled by the particle over $0 \leq t \leq 6$ .

c) Find the position of the particle at $t = 6$ .

d) Find the acceleration of the particle at $t = 5$ .

e) Is the speed of the particle increasing, decreasing, or not changing at $t = 3$ ?

Answers

(spoiler)

a) Displacement: $- 1.5$
b) Total distance: $4$
c) $x (6) = - 1$
d) Acceleration: $1$
e) Not changing

Solutions

a) Displacement

(spoiler)

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

On the interval $[0, 1.5]$ : Triangle (above axis)
- $A_{1} = \frac{1}{2} (1.5) (1) = 0.75$
On $[1.5, 5]$ : Trapezoid (below axis)
- $A_{2} = \frac{- 1}{2} (3.5 + 2) = - 2.75$
On $[5, 6]$ : Triangle (above axis)
- $A_{3} = \frac{1}{2} (1) (1) = 0.5$

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) Position at $t = 6$

(spoiler)

The final position is the initial position plus the displacement from $t = 2$ to $t = 6$ :

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

$\int_{2}^{6} v (t) d t$ is the net signed area from $t = 2$ to $t = 6$ :

On $[2, 4]$ : $A = - 2$ (rectangle below axis)
On $[4, 6]$ : $A = 0$ (triangles above and below cancel out)

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

$x (6) = 1 - 2 = - 1$

d) Acceleration at $t = 5$

(spoiler)

Acceleration is the derivative of velocity. To find $a (5) = v^{'} (5)$ , look at the slope of the line segment that connects points $(4, - 1)$ and $(6, 1)$ .

$v^{'} (5) = \frac{1 - ( - 1 )}{6 - 4} = 1$

e) Speed behavior at $t = 3$

(spoiler)

Speed increases when $v (t)$ and $a (t)$ have the same sign.
It decreases when $v (t)$ and $a (t)$ have opposing signs.
It’s constant when $a (t) = 0$ .

At $t = 3$ , $v (3) = - 1 < 0$ and $a (3) = v^{'} (3) = 0$ (horizontal tangent line).

Therefore, the speed of the particle is not changing at $t = 3$ .

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: $v (t) = \int a (t) d t + C$