Position, velocity, & acceleration | AP Calc AB Straight-line motion

What you’ll learn

How position, velocity, and acceleration are related
Finding average velocity and acceleration
Displacement vs. total distance

In motion problems, an object’s position is usually given by a function $p (t)$ or $x (t)$ , where $t$ represents time. The derivatives of the position function describe how the position changes over time:

The 1st derivative gives velocity.
The 2nd derivative gives acceleration.

Velocity

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

Velocity is the 1st derivative of position. It tells you:

How fast the object is moving
The direction of movement (based on the sign)
- When $v (t) > 0 \to$ The object is moving right/forward/up
- When $v (t) < 0 \to$ The object is moving left/backward/down
- When $v (t) = 0 \to$ The object is momentarily at rest (not moving)

Sidenote

Speed vs. velocity

Speed is the absolute value (magnitude) of velocity. Speed is always non-negative and tells you 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 (equivalently, the 2nd derivative of position). It tells you how velocity is changing. A practical way to interpret acceleration is to compare the signs of $v (t)$ and $a (t)$ at the same time $t$ :

When $a (t)$ and $v (t)$ have the same sign, the object is speeding up at that moment.
When $a (t)$ and $v (t)$ have opposite signs, the object is slowing down.
When $a (t) = 0$ , the object could be moving at constant velocity $(v (t) \neq = 0)$ or it could be at rest $(v (t) = 0)$ .
- In either case, it’s neither accelerating nor decelerating at that instant.

Although acceleration is not a force, it can help to picture it as an arrow that influences the velocity:

If $a (t)$ has the same sign as $v (t)$ , think of acceleration pointing in the same direction as the motion, increasing speed.
If the signs are opposite, think of acceleration pointing against the motion, reducing speed.

Example

Suppose the position of an object along the $x$ -axis is given by $x (t) = - t^{3} + 6 t^{2} - 9 t$ .

a) Find the velocity and acceleration at time $t = 2$ seconds.

b) At what time $t$ is the object at rest?

c) At what time does the object change direction?

d) At what time is the speed increasing or decreasing?

Solutions

a) Find the velocity and acceleration at time $t = 2$ seconds.

(spoiler)

Start by differentiating the position function to get velocity, then differentiate again to get acceleration. Finally, evaluate both at $t = 2$ .

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

At $t = 2$ :

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

So at $t = 2$ , the object is moving right with a speed of $3$ units per second, and its velocity is not changing at that instant because $a (2) = 0$ .

b) At what time $t$ is the object at rest?

(spoiler)

The object is at rest when its velocity is $0$ . 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 object is at rest at $t = 1$ and $t = 3$ seconds.

c) At what time does the object change direction?

(spoiler)

The object moves right when $v (t) > 0$ and left when $v (t) < 0$ . To change sign, $v (t)$ must equal $0$ at some time $t$ .

From part (b), the critical times are $t = 1$ and $t = 3$ . Check the sign of $v (t)$ on the intervals around these times. 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	Test point	$v (t)$	Motion
$t < 1$	$t = 0$	$- 9$	Left
$1 < t < 3$	$t = 2$	$3$	Right
$t > 3$	$t = 4$	$- 9$	Left

The object 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.

d) At what time is the speed increasing or decreasing?

(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	$v (t)$	$a (t)$	Motion
$t < 1$	$-$	$+$	Decelerating
$1 < t < 2$	$+$	$+$	Accelerating
$2 < t < 3$	$+$	$-$	Decelerating
$t > 3$	$-$	$-$	Accelerating

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.

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

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 $p (t) = t^{3} - 4 t^{2} + 2 t + 5$ , find the average velocity and average acceleration from $t = 2$ to $t = 6$ seconds.

(spoiler)

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

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

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

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

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, but there are two related ideas to keep separate:

Displacement is the net change in position.
Total distance traveled shows 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 $= p (t_{final}) - p (t_{initial})$

Total distance

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

Find where the velocity is $0$ (these are the times when the object may 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 $[\frac{π}{2}, π]$ .

(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 $t = \frac{3 π}{4}$

So 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.

Between $[\frac{π}{2}, \frac{3 π}{4}]$ :

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

Between $[\frac{3 π}{4}, π]$ :

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

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

What you’ll learn

How position, velocity, and acceleration are related
Finding average velocity and acceleration
Displacement vs. total distance

The 1st derivative gives velocity.
The 2nd derivative gives acceleration.

Velocity

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

Velocity is the 1st derivative of position. It tells you:

How fast the object is moving
The direction of movement (based on the sign)
- When $v (t) > 0 \to$ The object is moving right/forward/up
- When $v (t) < 0 \to$ The object is moving left/backward/down
- When $v (t) = 0 \to$ The object is momentarily at rest (not moving)

Sidenote

Speed vs. velocity

Speed is the absolute value (magnitude) of velocity. Speed is always non-negative and tells you only how fast the object is moving.

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

Acceleration

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

When $a (t)$ and $v (t)$ have the same sign, the object is speeding up at that moment.
When $a (t)$ and $v (t)$ have opposite signs, the object is slowing down.
When $a (t) = 0$ , the object could be moving at constant velocity $(v (t) \neq = 0)$ or it could be at rest $(v (t) = 0)$ .
- In either case, it’s neither accelerating nor decelerating at that instant.

Although acceleration is not a force, it can help to picture it as an arrow that influences the velocity:

If $a (t)$ has the same sign as $v (t)$ , think of acceleration pointing in the same direction as the motion, increasing speed.
If the signs are opposite, think of acceleration pointing against the motion, reducing speed.

Example

Suppose the position of an object along the $x$ -axis is given by $x (t) = - t^{3} + 6 t^{2} - 9 t$ .

a) Find the velocity and acceleration at time $t = 2$ seconds.

b) At what time $t$ is the object at rest?

c) At what time does the object change direction?

d) At what time is the speed increasing or decreasing?

Solutions

a) Find the velocity and acceleration at time $t = 2$ seconds.

(spoiler)

Start by differentiating the position function to get velocity, then differentiate again to get acceleration. Finally, evaluate both at $t = 2$ .

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

At $t = 2$ :

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

So at $t = 2$ , the object is moving right with a speed of $3$ units per second, and its velocity is not changing at that instant because $a (2) = 0$ .

b) At what time $t$ is the object at rest?

(spoiler)

The object is at rest when its velocity is $0$ . 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 object is at rest at $t = 1$ and $t = 3$ seconds.

c) At what time does the object change direction?

(spoiler)

The object moves right when $v (t) > 0$ and left when $v (t) < 0$ . To change sign, $v (t)$ must equal $0$ at some time $t$ .

Interval	Test point	$v (t)$	Motion
$t < 1$	$t = 0$	$- 9$	Left
$1 < t < 3$	$t = 2$	$3$	Right
$t > 3$	$t = 4$	$- 9$	Left

The object 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.

d) At what time is the speed increasing or decreasing?

(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	$v (t)$	$a (t)$	Motion
$t < 1$	$-$	$+$	Decelerating
$1 < t < 2$	$+$	$+$	Accelerating
$2 < t < 3$	$+$	$-$	Decelerating
$t > 3$	$-$	$-$	Accelerating

Average velocity and acceleration

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

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 $p (t) = t^{3} - 4 t^{2} + 2 t + 5$ , find the average velocity and average acceleration from $t = 2$ to $t = 6$ seconds.

(spoiler)

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

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

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

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

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, but there are two related ideas to keep separate:

Displacement is the net change in position.
Total distance traveled shows 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 $= p (t_{final}) - p (t_{initial})$

Total distance

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

Find where the velocity is $0$ (these are the times when the object may 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 $[\frac{π}{2}, π]$ .

(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 $t = \frac{3 π}{4}$

So 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.

Between $[\frac{π}{2}, \frac{3 π}{4}]$ :

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

Between $[\frac{3 π}{4}, π]$ :

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

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