4.2.1 Position, velocity, & acceleration

Achievable AP Calculus AB

4. Contextual uses

4.2. Straight-line motion

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Position, velocity, & acceleration

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What you’ll learn

Motion: Use derivatives to relate position, velocity, and acceleration.
Average rates: Calculate average velocity and acceleration on an interval.
Distance vs. displacement: Distinguish total distance traveled from net change in position.

In motion problems, an object’s position is modeled by a function $p (t)$ or $x (t)$ , where $t$ is time. Its derivatives describe how the position changes:

1st derivative: Velocity
2nd derivative: Acceleration

Velocity

$v (t) = x^{'} (t)$

Velocity is the 1st derivative of position. Its magnitude describes how fast the object moves, and its sign indicates direction:

$v (t) > 0 \Rightarrow$ Moving right/forward/up
$v (t) < 0 \Rightarrow$ Moving left/backward/down
$v (t) = 0 \Rightarrow$ The object is momentarily at rest (not moving). Acceleration determines the direction it will start moving.

Sidenote

Speed vs. velocity

Speed is $∣ v (t) ∣$ , the magnitude of velocity. It measures only how fast the object is moving.

Velocity includes both speed and direction (through its sign).

Acceleration

$a (t) = v^{'} (t) = p^{''} (t)$

Acceleration is the derivative of velocity (or the second derivative of position). To interpret acceleration, compare the signs of $v (t)$ and $a (t)$ :

Same sign $\Rightarrow$ Object is speeding up (its speed is increasing).
Opposite signs $\Rightarrow$ Slowing down (speed is decreasing).
$a (t) = 0 \Rightarrow$ neither speeding up nor slowing down at that instant.

Think of acceleration as pointing in the direction velocity is changing: with the motion when speeding up, against the motion when slowing down.

Example 1: Finding velocity and acceleration

The position of a particle moving along the $x$ -axis is given by

$x (t) = - t^{3} + 6 t^{2} - 9 t$

where $x$ is measured in meters and $t$ is measured in seconds.

Find the particle’s velocity and acceleration at $t = 2$ and interpret the particle’s motion at that instant.

Solution

(spoiler)

1. Differentiate twice:

$v (t) a (t) = x^{'} (t) = - 3 t^{2} + 12 t - 9 = v^{'} (t) = x^{''} (t) = - 6 t + 12$

2. Evaluate each at $t = 2$ :

Velocity: $v (2) = - 3 (2)^{2} + 12 (2) - 9 = 3$
Acceleration: $a (2) = - 6 (2) + 12 = 0$

3. Interpretation:

At $t = 2$ , the particle is moving right (positive velocity) at a speed of $3$ m/s. Since $a (2) = 0$ , it is neither speeding up nor slowing down at that instant.

Example 2: Particle at rest

Using the same position function $x (t) = - t^{3} + 6 t^{2} - 9 t$ from example 1, determine at what time(s) $t$ the particle is at rest.

Solution

(spoiler)

Find when the velocity equals $0$ :

$v (t) = x^{'} (t) = - 3 t^{2} + 12 t - 9$

Set $v (t) = 0$ and solve for $t$ :

$- 3 t^{2} + 12 t - 9 = 0 - 3 (t^{2} - 4 t + 3) = 0 - 3 (t - 1) (t - 3) = 0 t = 1, 3$

The particle is at rest at $t = 1$ and $t = 3$ seconds.

Example 3: Changes in direction

Using the same position function $x (t) = - t^{3} + 6 t^{2} - 9 t$ , determine at what time(s) $t$ the particle changes direction.

Solution

(spoiler)

Since the particle moves right when $v (t) > 0$ and left when $v (t) < 0$ , its velocity must equal $0$ at some time $t$ to change sign.

From example 2, these times $t = 1$ and $t = 3$ . Check the sign of $v (t)$ in the surrounding intervals. Since $v (t)$ is a downward-opening parabola with zeros at $t = 1$ and $t = 3$ , it’s negative outside those roots and positive between them.

Interval $t$	Test point	$v (t)$	Motion
$(0, 1)$	$t = 0$	$- 9$	Left
$(1, 3)$	$t = 2$	$3$	Right
$(3, \infty)$	$t = 4$	$- 9$	Left

The particle moves left, stops at $t = 1$ second and changes direction to move right, then stops again at $t = 3$ seconds and changes direction to move left.

Example 4: Speed increasing or decreasing

Using the same position function $x (t) = - t^{3} + 6 t^{2} - 9 t$ , determine at what intervals of time the particle’s speed increases or decreases.

Solution

(spoiler)

Speed increases when $v (t)$ and $a (t)$ have the same sign, and decreases when they have opposite signs. Use the functions:

$v (t) a (t) = - 3 t^{2} + 12 t - 9 = - 6 t + 12$

To build intervals, use the times when either function is $0$ :

$v (t) = 0$ at $t = 1$ and $t = 3$
$a (t) = 0$ at $t = 2$

Now test signs on each interval.

Interval $t$	$v (t)$	$a (t)$	Motion
$(0, 1)$	$-$	$+$	Slowing down
$(1, 2)$	$+$	$+$	Speeding up
$(2, 3)$	$+$	$-$	Slowing down
$(3, \infty)$	$-$	$-$	Speeding up

The particle’s speed increases for $t$ in $(1, 2)$ and $(3, \infty)$ .

Its speed decreases for $t$ in $(0, 1)$ and $(2, 3)$ .

Average velocity and acceleration

A common AP exam task is to find average velocity over a time interval given a position function. Average velocity is the change in position divided by the change in time, or the slope (average rate of change) of the position function over the interval.

$v_{avg} = \frac{x ( t _{2} ) - x ( t _{1} )}{t _{2} - t _{1}}$

Similarly, average acceleration is the change in velocity divided by the change in time.

$a_{avg} = \frac{v ( t _{2} ) - v ( t _{1} )}{t _{2} - t _{1}}$

Example

If a particle’s position in feet is given by $x (t) = t^{3} - 4 t^{2} + 2 t + 5$ , find the average velocity and average acceleration from $t = 2$ to $t = 6$ seconds.

(spoiler)

The average velocity over $[2, 6]$ is

$v_{avg} = \frac{x ( 6 ) - x ( 2 )}{6 - 2} = \frac{88}{4} = 22 ft/s$

To find average acceleration, first write the velocity function by differentiating $x (t)$ :

$v (t) = 3 t^{2} - 8 t + 2$

Then compute the average acceleration over $[2, 6]$ :

$a_{avg} = \frac{v ( 6 ) - v ( 2 )}{6 - 2} = \frac{62 - ( - 2 )}{4} = 16 ft/s^{2}$

Distance traveled

A position function can tell you how far an object has traveled, in two ways:

Displacement: The net change in position.
Total distance traveled: How much ground was covered, regardless of direction.

Displacement

If an object moves from point A to point B and then returns to point A, its net change in position is $0$ . Displacement depends only on the initial and final positions.

$Displacement = x (t_{final}) - x (t_{initial})$

Total distance

To find the total distance traveled over a time interval when given a position function:

Find where the velocity is $0$ (times when the object might change direction).
Find the displacement on each sub-interval and take the absolute values (so each piece is a positive distance).
Add those positive distances.

Example

A particle moves along the $x$ -axis with its position given by $x (t) = sin (2 t)$ . Find the displacement and the total distance traveled over the interval $[\frac{π}{2}, π]$ .

Solution

(spoiler)

The displacement is the difference between the final and initial positions:

$x (π) - x (\frac{π}{2}) = sin (2 π) - sin (π) = 0$

So the particle moves but returns to its original position.

For the total distance, find when the particle stops momentarily (when $v (t) = 0$ ).

First compute velocity:

$v (t) = 2 cos (2 t)$

Now solve $v (t) = 0$ :

$2 cos (2 t) = 0$

$2 t = \frac{π}{2}, \frac{3 π}{2}, ...$

$t = \frac{π}{4}, \frac{3 π}{4}, ...$

On the interval $[\frac{π}{2}, π]$ , the only solution is $t = \frac{3 π}{4}$ .

Split the interval into two parts: $[\frac{π}{2}, \frac{3 π}{4}]$ and $[\frac{3 π}{4}, π]$ .

Next, find the distance traveled over each interval.

On $[\frac{π}{2}, \frac{3 π}{4}]$ :

$x (\frac{3 π}{4}) - x (\frac{π}{2}) = ∣ - 1 - 0∣ = 1$

On $[\frac{3 π}{4}, π]$ :

$x (π) - x (\frac{3 π}{4}) = ∣0 - (- 1) ∣ = 1$

Adding these gives a total distance of $2$ units traveled.

Position, velocity, & acceleration

What you’ll learn

Motion: Use derivatives to relate position, velocity, and acceleration.
Average rates: Calculate average velocity and acceleration on an interval.
Distance vs. displacement: Distinguish total distance traveled from net change in position.

In motion problems, an object’s position is modeled by a function $p (t)$ or $x (t)$ , where $t$ is time. Its derivatives describe how the position changes:

1st derivative: Velocity
2nd derivative: Acceleration

Velocity

$v (t) = x^{'} (t)$

Velocity is the 1st derivative of position. Its magnitude describes how fast the object moves, and its sign indicates direction:

$v (t) > 0 \Rightarrow$ Moving right/forward/up
$v (t) < 0 \Rightarrow$ Moving left/backward/down
$v (t) = 0 \Rightarrow$ The object is momentarily at rest (not moving). Acceleration determines the direction it will start moving.

Sidenote

Speed vs. velocity

Speed is $∣ v (t) ∣$ , the magnitude of velocity. It measures only how fast the object is moving.

Velocity includes both speed and direction (through its sign).

Acceleration

$a (t) = v^{'} (t) = p^{''} (t)$

Acceleration is the derivative of velocity (or the second derivative of position). To interpret acceleration, compare the signs of $v (t)$ and $a (t)$ :

Same sign $\Rightarrow$ Object is speeding up (its speed is increasing).
Opposite signs $\Rightarrow$ Slowing down (speed is decreasing).
$a (t) = 0 \Rightarrow$ neither speeding up nor slowing down at that instant.

Think of acceleration as pointing in the direction velocity is changing: with the motion when speeding up, against the motion when slowing down.

Example 1: Finding velocity and acceleration

The position of a particle moving along the $x$ -axis is given by

$x (t) = - t^{3} + 6 t^{2} - 9 t$

where $x$ is measured in meters and $t$ is measured in seconds.

Find the particle’s velocity and acceleration at $t = 2$ and interpret the particle’s motion at that instant.

Solution

(spoiler)

1. Differentiate twice:

$v (t) a (t) = x^{'} (t) = - 3 t^{2} + 12 t - 9 = v^{'} (t) = x^{''} (t) = - 6 t + 12$

2. Evaluate each at $t = 2$ :

Velocity: $v (2) = - 3 (2)^{2} + 12 (2) - 9 = 3$
Acceleration: $a (2) = - 6 (2) + 12 = 0$

3. Interpretation:

At $t = 2$ , the particle is moving right (positive velocity) at a speed of $3$ m/s. Since $a (2) = 0$ , it is neither speeding up nor slowing down at that instant.

Example 2: Particle at rest

Using the same position function $x (t) = - t^{3} + 6 t^{2} - 9 t$ from example 1, determine at what time(s) $t$ the particle is at rest.

Solution

(spoiler)

Find when the velocity equals $0$ :

$v (t) = x^{'} (t) = - 3 t^{2} + 12 t - 9$

Set $v (t) = 0$ and solve for $t$ :

$- 3 t^{2} + 12 t - 9 = 0 - 3 (t^{2} - 4 t + 3) = 0 - 3 (t - 1) (t - 3) = 0 t = 1, 3$

The particle is at rest at $t = 1$ and $t = 3$ seconds.

Example 3: Changes in direction

Using the same position function $x (t) = - t^{3} + 6 t^{2} - 9 t$ , determine at what time(s) $t$ the particle changes direction.

Solution

(spoiler)

Since the particle moves right when $v (t) > 0$ and left when $v (t) < 0$ , its velocity must equal $0$ at some time $t$ to change sign.

Interval $t$	Test point	$v (t)$	Motion
$(0, 1)$	$t = 0$	$- 9$	Left
$(1, 3)$	$t = 2$	$3$	Right
$(3, \infty)$	$t = 4$	$- 9$	Left

The particle moves left, stops at $t = 1$ second and changes direction to move right, then stops again at $t = 3$ seconds and changes direction to move left.

Example 4: Speed increasing or decreasing

Using the same position function $x (t) = - t^{3} + 6 t^{2} - 9 t$ , determine at what intervals of time the particle’s speed increases or decreases.

Solution

(spoiler)

Speed increases when $v (t)$ and $a (t)$ have the same sign, and decreases when they have opposite signs. Use the functions:

$v (t) a (t) = - 3 t^{2} + 12 t - 9 = - 6 t + 12$

To build intervals, use the times when either function is $0$ :

$v (t) = 0$ at $t = 1$ and $t = 3$
$a (t) = 0$ at $t = 2$

Now test signs on each interval.

Interval $t$	$v (t)$	$a (t)$	Motion
$(0, 1)$	$-$	$+$	Slowing down
$(1, 2)$	$+$	$+$	Speeding up
$(2, 3)$	$+$	$-$	Slowing down
$(3, \infty)$	$-$	$-$	Speeding up

The particle’s speed increases for $t$ in $(1, 2)$ and $(3, \infty)$ .

Its speed decreases for $t$ in $(0, 1)$ and $(2, 3)$ .

Average velocity and acceleration

$v_{avg} = \frac{x ( t _{2} ) - x ( t _{1} )}{t _{2} - t _{1}}$

Similarly, average acceleration is the change in velocity divided by the change in time.

$a_{avg} = \frac{v ( t _{2} ) - v ( t _{1} )}{t _{2} - t _{1}}$

Example

If a particle’s position in feet is given by $x (t) = t^{3} - 4 t^{2} + 2 t + 5$ , find the average velocity and average acceleration from $t = 2$ to $t = 6$ seconds.

(spoiler)

The average velocity over $[2, 6]$ is

$v_{avg} = \frac{x ( 6 ) - x ( 2 )}{6 - 2} = \frac{88}{4} = 22 ft/s$

To find average acceleration, first write the velocity function by differentiating $x (t)$ :

$v (t) = 3 t^{2} - 8 t + 2$

Then compute the average acceleration over $[2, 6]$ :

$a_{avg} = \frac{v ( 6 ) - v ( 2 )}{6 - 2} = \frac{62 - ( - 2 )}{4} = 16 ft/s^{2}$

Distance traveled

A position function can tell you how far an object has traveled, in two ways:

Displacement: The net change in position.
Total distance traveled: How much ground was covered, regardless of direction.

Displacement

If an object moves from point A to point B and then returns to point A, its net change in position is $0$ . Displacement depends only on the initial and final positions.

$Displacement = x (t_{final}) - x (t_{initial})$

Total distance

To find the total distance traveled over a time interval when given a position function:

Find where the velocity is $0$ (times when the object might change direction).
Find the displacement on each sub-interval and take the absolute values (so each piece is a positive distance).
Add those positive distances.

Example

A particle moves along the $x$ -axis with its position given by $x (t) = sin (2 t)$ . Find the displacement and the total distance traveled over the interval $[\frac{π}{2}, π]$ .

Solution

(spoiler)

The displacement is the difference between the final and initial positions:

$x (π) - x (\frac{π}{2}) = sin (2 π) - sin (π) = 0$

So the particle moves but returns to its original position.

For the total distance, find when the particle stops momentarily (when $v (t) = 0$ ).

First compute velocity:

$v (t) = 2 cos (2 t)$

Now solve $v (t) = 0$ :

$2 cos (2 t) = 0$

$2 t = \frac{π}{2}, \frac{3 π}{2}, ...$

$t = \frac{π}{4}, \frac{3 π}{4}, ...$

On the interval $[\frac{π}{2}, π]$ , the only solution is $t = \frac{3 π}{4}$ .

Split the interval into two parts: $[\frac{π}{2}, \frac{3 π}{4}]$ and $[\frac{3 π}{4}, π]$ .

Next, find the distance traveled over each interval.

On $[\frac{π}{2}, \frac{3 π}{4}]$ :

$x (\frac{3 π}{4}) - x (\frac{π}{2}) = ∣ - 1 - 0∣ = 1$

On $[\frac{3 π}{4}, π]$ :

$x (π) - x (\frac{3 π}{4}) = ∣0 - (- 1) ∣ = 1$

Adding these gives a total distance of $2$ units traveled.

Achievable AP Calculus AB

4. Contextual uses

4.2. Straight-line motion

Our AP Calculus AB course is now in "early access" - get 50% off for a limited time.

Position, velocity, & acceleration

7 min read

Font

Discuss

Feedback

What you’ll learn

Motion: Use derivatives to relate position, velocity, and acceleration.
Average rates: Calculate average velocity and acceleration on an interval.
Distance vs. displacement: Distinguish total distance traveled from net change in position.

In motion problems, an object’s position is modeled by a function $p (t)$ or $x (t)$ , where $t$ is time. Its derivatives describe how the position changes:

1st derivative: Velocity
2nd derivative: Acceleration

Velocity

$v (t) = x^{'} (t)$

Velocity is the 1st derivative of position. Its magnitude describes how fast the object moves, and its sign indicates direction:

$v (t) > 0 \Rightarrow$ Moving right/forward/up
$v (t) < 0 \Rightarrow$ Moving left/backward/down
$v (t) = 0 \Rightarrow$ The object is momentarily at rest (not moving). Acceleration determines the direction it will start moving.

Sidenote

Speed vs. velocity

Speed is $∣ v (t) ∣$ , the magnitude of velocity. It measures only how fast the object is moving.

Velocity includes both speed and direction (through its sign).

Acceleration

$a (t) = v^{'} (t) = p^{''} (t)$

Acceleration is the derivative of velocity (or the second derivative of position). To interpret acceleration, compare the signs of $v (t)$ and $a (t)$ :

Same sign $\Rightarrow$ Object is speeding up (its speed is increasing).
Opposite signs $\Rightarrow$ Slowing down (speed is decreasing).
$a (t) = 0 \Rightarrow$ neither speeding up nor slowing down at that instant.

Think of acceleration as pointing in the direction velocity is changing: with the motion when speeding up, against the motion when slowing down.

Example 1: Finding velocity and acceleration

The position of a particle moving along the $x$ -axis is given by

$x (t) = - t^{3} + 6 t^{2} - 9 t$

where $x$ is measured in meters and $t$ is measured in seconds.

Find the particle’s velocity and acceleration at $t = 2$ and interpret the particle’s motion at that instant.

Solution

(spoiler)

1. Differentiate twice:

$v (t) a (t) = x^{'} (t) = - 3 t^{2} + 12 t - 9 = v^{'} (t) = x^{''} (t) = - 6 t + 12$

2. Evaluate each at $t = 2$ :

Velocity: $v (2) = - 3 (2)^{2} + 12 (2) - 9 = 3$
Acceleration: $a (2) = - 6 (2) + 12 = 0$

3. Interpretation:

At $t = 2$ , the particle is moving right (positive velocity) at a speed of $3$ m/s. Since $a (2) = 0$ , it is neither speeding up nor slowing down at that instant.

Example 2: Particle at rest

Using the same position function $x (t) = - t^{3} + 6 t^{2} - 9 t$ from example 1, determine at what time(s) $t$ the particle is at rest.

Solution

(spoiler)

Find when the velocity equals $0$ :

$v (t) = x^{'} (t) = - 3 t^{2} + 12 t - 9$

Set $v (t) = 0$ and solve for $t$ :

$- 3 t^{2} + 12 t - 9 = 0 - 3 (t^{2} - 4 t + 3) = 0 - 3 (t - 1) (t - 3) = 0 t = 1, 3$

The particle is at rest at $t = 1$ and $t = 3$ seconds.

Example 3: Changes in direction

Using the same position function $x (t) = - t^{3} + 6 t^{2} - 9 t$ , determine at what time(s) $t$ the particle changes direction.

Solution

(spoiler)

Since the particle moves right when $v (t) > 0$ and left when $v (t) < 0$ , its velocity must equal $0$ at some time $t$ to change sign.

Interval $t$	Test point	$v (t)$	Motion
$(0, 1)$	$t = 0$	$- 9$	Left
$(1, 3)$	$t = 2$	$3$	Right
$(3, \infty)$	$t = 4$	$- 9$	Left

The particle moves left, stops at $t = 1$ second and changes direction to move right, then stops again at $t = 3$ seconds and changes direction to move left.

Example 4: Speed increasing or decreasing

Using the same position function $x (t) = - t^{3} + 6 t^{2} - 9 t$ , determine at what intervals of time the particle’s speed increases or decreases.

Solution

(spoiler)

Speed increases when $v (t)$ and $a (t)$ have the same sign, and decreases when they have opposite signs. Use the functions:

$v (t) a (t) = - 3 t^{2} + 12 t - 9 = - 6 t + 12$

To build intervals, use the times when either function is $0$ :

$v (t) = 0$ at $t = 1$ and $t = 3$
$a (t) = 0$ at $t = 2$

Now test signs on each interval.

Interval $t$	$v (t)$	$a (t)$	Motion
$(0, 1)$	$-$	$+$	Slowing down
$(1, 2)$	$+$	$+$	Speeding up
$(2, 3)$	$+$	$-$	Slowing down
$(3, \infty)$	$-$	$-$	Speeding up

The particle’s speed increases for $t$ in $(1, 2)$ and $(3, \infty)$ .

Its speed decreases for $t$ in $(0, 1)$ and $(2, 3)$ .

Average velocity and acceleration

$v_{avg} = \frac{x ( t _{2} ) - x ( t _{1} )}{t _{2} - t _{1}}$

Similarly, average acceleration is the change in velocity divided by the change in time.

$a_{avg} = \frac{v ( t _{2} ) - v ( t _{1} )}{t _{2} - t _{1}}$

Example

If a particle’s position in feet is given by $x (t) = t^{3} - 4 t^{2} + 2 t + 5$ , find the average velocity and average acceleration from $t = 2$ to $t = 6$ seconds.

(spoiler)

The average velocity over $[2, 6]$ is

$v_{avg} = \frac{x ( 6 ) - x ( 2 )}{6 - 2} = \frac{88}{4} = 22 ft/s$

To find average acceleration, first write the velocity function by differentiating $x (t)$ :

$v (t) = 3 t^{2} - 8 t + 2$

Then compute the average acceleration over $[2, 6]$ :

$a_{avg} = \frac{v ( 6 ) - v ( 2 )}{6 - 2} = \frac{62 - ( - 2 )}{4} = 16 ft/s^{2}$

Distance traveled

A position function can tell you how far an object has traveled, in two ways:

Displacement: The net change in position.
Total distance traveled: How much ground was covered, regardless of direction.

Displacement

If an object moves from point A to point B and then returns to point A, its net change in position is $0$ . Displacement depends only on the initial and final positions.

$Displacement = x (t_{final}) - x (t_{initial})$

Total distance

To find the total distance traveled over a time interval when given a position function:

Find where the velocity is $0$ (times when the object might change direction).
Find the displacement on each sub-interval and take the absolute values (so each piece is a positive distance).
Add those positive distances.

Example

A particle moves along the $x$ -axis with its position given by $x (t) = sin (2 t)$ . Find the displacement and the total distance traveled over the interval $[\frac{π}{2}, π]$ .

Solution

(spoiler)

The displacement is the difference between the final and initial positions:

$x (π) - x (\frac{π}{2}) = sin (2 π) - sin (π) = 0$

So the particle moves but returns to its original position.

For the total distance, find when the particle stops momentarily (when $v (t) = 0$ ).

First compute velocity:

$v (t) = 2 cos (2 t)$

Now solve $v (t) = 0$ :

$2 cos (2 t) = 0$

$2 t = \frac{π}{2}, \frac{3 π}{2}, ...$

$t = \frac{π}{4}, \frac{3 π}{4}, ...$

On the interval $[\frac{π}{2}, π]$ , the only solution is $t = \frac{3 π}{4}$ .

Split the interval into two parts: $[\frac{π}{2}, \frac{3 π}{4}]$ and $[\frac{3 π}{4}, π]$ .

Next, find the distance traveled over each interval.

On $[\frac{π}{2}, \frac{3 π}{4}]$ :

$x (\frac{3 π}{4}) - x (\frac{π}{2}) = ∣ - 1 - 0∣ = 1$

On $[\frac{3 π}{4}, π]$ :

$x (π) - x (\frac{3 π}{4}) = ∣0 - (- 1) ∣ = 1$

Adding these gives a total distance of $2$ units traveled.

Position, velocity, & acceleration

What you’ll learn

Motion: Use derivatives to relate position, velocity, and acceleration.
Average rates: Calculate average velocity and acceleration on an interval.
Distance vs. displacement: Distinguish total distance traveled from net change in position.

In motion problems, an object’s position is modeled by a function $p (t)$ or $x (t)$ , where $t$ is time. Its derivatives describe how the position changes:

1st derivative: Velocity
2nd derivative: Acceleration

Velocity

$v (t) = x^{'} (t)$

Velocity is the 1st derivative of position. Its magnitude describes how fast the object moves, and its sign indicates direction:

$v (t) > 0 \Rightarrow$ Moving right/forward/up
$v (t) < 0 \Rightarrow$ Moving left/backward/down
$v (t) = 0 \Rightarrow$ The object is momentarily at rest (not moving). Acceleration determines the direction it will start moving.

Sidenote

Speed vs. velocity

Speed is $∣ v (t) ∣$ , the magnitude of velocity. It measures only how fast the object is moving.

Velocity includes both speed and direction (through its sign).

Acceleration

$a (t) = v^{'} (t) = p^{''} (t)$

Acceleration is the derivative of velocity (or the second derivative of position). To interpret acceleration, compare the signs of $v (t)$ and $a (t)$ :

Same sign $\Rightarrow$ Object is speeding up (its speed is increasing).
Opposite signs $\Rightarrow$ Slowing down (speed is decreasing).
$a (t) = 0 \Rightarrow$ neither speeding up nor slowing down at that instant.

Think of acceleration as pointing in the direction velocity is changing: with the motion when speeding up, against the motion when slowing down.

Example 1: Finding velocity and acceleration

The position of a particle moving along the $x$ -axis is given by

$x (t) = - t^{3} + 6 t^{2} - 9 t$

where $x$ is measured in meters and $t$ is measured in seconds.

Find the particle’s velocity and acceleration at $t = 2$ and interpret the particle’s motion at that instant.

Solution

(spoiler)

1. Differentiate twice:

$v (t) a (t) = x^{'} (t) = - 3 t^{2} + 12 t - 9 = v^{'} (t) = x^{''} (t) = - 6 t + 12$

2. Evaluate each at $t = 2$ :

Velocity: $v (2) = - 3 (2)^{2} + 12 (2) - 9 = 3$
Acceleration: $a (2) = - 6 (2) + 12 = 0$

3. Interpretation:

At $t = 2$ , the particle is moving right (positive velocity) at a speed of $3$ m/s. Since $a (2) = 0$ , it is neither speeding up nor slowing down at that instant.

Example 2: Particle at rest

Using the same position function $x (t) = - t^{3} + 6 t^{2} - 9 t$ from example 1, determine at what time(s) $t$ the particle is at rest.

Solution

(spoiler)

Find when the velocity equals $0$ :

$v (t) = x^{'} (t) = - 3 t^{2} + 12 t - 9$

Set $v (t) = 0$ and solve for $t$ :

$- 3 t^{2} + 12 t - 9 = 0 - 3 (t^{2} - 4 t + 3) = 0 - 3 (t - 1) (t - 3) = 0 t = 1, 3$

The particle is at rest at $t = 1$ and $t = 3$ seconds.

Example 3: Changes in direction

Using the same position function $x (t) = - t^{3} + 6 t^{2} - 9 t$ , determine at what time(s) $t$ the particle changes direction.

Solution

(spoiler)

Since the particle moves right when $v (t) > 0$ and left when $v (t) < 0$ , its velocity must equal $0$ at some time $t$ to change sign.

Interval $t$	Test point	$v (t)$	Motion
$(0, 1)$	$t = 0$	$- 9$	Left
$(1, 3)$	$t = 2$	$3$	Right
$(3, \infty)$	$t = 4$	$- 9$	Left

The particle moves left, stops at $t = 1$ second and changes direction to move right, then stops again at $t = 3$ seconds and changes direction to move left.

Example 4: Speed increasing or decreasing

Using the same position function $x (t) = - t^{3} + 6 t^{2} - 9 t$ , determine at what intervals of time the particle’s speed increases or decreases.

Solution

(spoiler)

Speed increases when $v (t)$ and $a (t)$ have the same sign, and decreases when they have opposite signs. Use the functions:

$v (t) a (t) = - 3 t^{2} + 12 t - 9 = - 6 t + 12$

To build intervals, use the times when either function is $0$ :

$v (t) = 0$ at $t = 1$ and $t = 3$
$a (t) = 0$ at $t = 2$

Now test signs on each interval.

Interval $t$	$v (t)$	$a (t)$	Motion
$(0, 1)$	$-$	$+$	Slowing down
$(1, 2)$	$+$	$+$	Speeding up
$(2, 3)$	$+$	$-$	Slowing down
$(3, \infty)$	$-$	$-$	Speeding up

The particle’s speed increases for $t$ in $(1, 2)$ and $(3, \infty)$ .

Its speed decreases for $t$ in $(0, 1)$ and $(2, 3)$ .

Average velocity and acceleration

$v_{avg} = \frac{x ( t _{2} ) - x ( t _{1} )}{t _{2} - t _{1}}$

Similarly, average acceleration is the change in velocity divided by the change in time.

$a_{avg} = \frac{v ( t _{2} ) - v ( t _{1} )}{t _{2} - t _{1}}$

Example

If a particle’s position in feet is given by $x (t) = t^{3} - 4 t^{2} + 2 t + 5$ , find the average velocity and average acceleration from $t = 2$ to $t = 6$ seconds.

(spoiler)

The average velocity over $[2, 6]$ is

$v_{avg} = \frac{x ( 6 ) - x ( 2 )}{6 - 2} = \frac{88}{4} = 22 ft/s$

To find average acceleration, first write the velocity function by differentiating $x (t)$ :

$v (t) = 3 t^{2} - 8 t + 2$

Then compute the average acceleration over $[2, 6]$ :

$a_{avg} = \frac{v ( 6 ) - v ( 2 )}{6 - 2} = \frac{62 - ( - 2 )}{4} = 16 ft/s^{2}$

Distance traveled

A position function can tell you how far an object has traveled, in two ways:

Displacement: The net change in position.
Total distance traveled: How much ground was covered, regardless of direction.

Displacement

If an object moves from point A to point B and then returns to point A, its net change in position is $0$ . Displacement depends only on the initial and final positions.

$Displacement = x (t_{final}) - x (t_{initial})$

Total distance

To find the total distance traveled over a time interval when given a position function:

Find where the velocity is $0$ (times when the object might change direction).
Find the displacement on each sub-interval and take the absolute values (so each piece is a positive distance).
Add those positive distances.

Example

A particle moves along the $x$ -axis with its position given by $x (t) = sin (2 t)$ . Find the displacement and the total distance traveled over the interval $[\frac{π}{2}, π]$ .

Solution

(spoiler)

The displacement is the difference between the final and initial positions:

$x (π) - x (\frac{π}{2}) = sin (2 π) - sin (π) = 0$

So the particle moves but returns to its original position.

For the total distance, find when the particle stops momentarily (when $v (t) = 0$ ).

First compute velocity:

$v (t) = 2 cos (2 t)$

Now solve $v (t) = 0$ :

$2 cos (2 t) = 0$

$2 t = \frac{π}{2}, \frac{3 π}{2}, ...$

$t = \frac{π}{4}, \frac{3 π}{4}, ...$

On the interval $[\frac{π}{2}, π]$ , the only solution is $t = \frac{3 π}{4}$ .

Split the interval into two parts: $[\frac{π}{2}, \frac{3 π}{4}]$ and $[\frac{3 π}{4}, π]$ .

Next, find the distance traveled over each interval.

On $[\frac{π}{2}, \frac{3 π}{4}]$ :

$x (\frac{3 π}{4}) - x (\frac{π}{2}) = ∣ - 1 - 0∣ = 1$

On $[\frac{3 π}{4}, π]$ :

$x (π) - x (\frac{3 π}{4}) = ∣0 - (- 1) ∣ = 1$

Adding these gives a total distance of $2$ units traveled.