Position, velocity, & acceleration | Straight-line motion | Contextual uses

What you’ll learn:

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

In motion problems, the position of an object is usually given by a function $s (t)$ or $x (t)$ , where the input $t$ represents time. The 1st and 2nd derivatives of the position function provide insight into how an object moves over time.

Velocity

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

The 1st derivative of the position function gives the velocity function, which tells us:

How fast an object is moving
In which direction - by convention,
- When $v (t) > 0$ (positive velocity): The object is moving right/forward/up
- When $v (t) < 0$ (negative velocity): The object is moving left/backward/down
- When $v (t) = 0$ , the particle is momentarily at rest (not moving)

Sidenote

Speed vs. velocity

Speed is the absolute value (magnitude) of the velocity - a positive value that denotes how fast an object is going.

The velocity gives both speed and direction (depending on the sign).

Acceleration

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

The derivative of the velocity function, or the 2nd derivative of position function, gives the acceleration function, which tells us if the object is speeding up or slowing down:

When $a (t)$ and $v (t)$ have the same sign at some time $t$ , 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 $(v (t) = 0)$ .
- It’s neither accelerating or decelerating.

Although acceleration is not a force itself, you can think of it as one that attempts to influence the velocity’s direction. If $a (t)$ has the same sign as $v (t)$ , imagine an arrow pointing in the same direction as velocity, adding to the speed. When the signs are opposite, imagine the acceleration arrow “pulling” the object back in the other direction, slowing it down.

Examples

1. 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) When is the object at rest?
c) When does the object change direction?
d) When is the speed increasing or decreasing?

Solutions

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

Differentiate the position function for the velocity function, and differentiate again for the acceleration function. Then evaluate at $t = 2$ :

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

At $t = 2$ :

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

$a (2) = - 6 (2) + 12 = 0$

The object is moving right at a speed of $3$ m/s and at constant velocity because the acceleration is $0$ m/s $^{2}$ .

b) When is the object at rest?

The object is at rest when its velocity is $0$ . Setting the velocity function equal to $0$ and solving 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) When does the object change direction?

The object is moving right/forward when $v (t) > 0$ and left/backward when $v (t) < 0$ . In order to change direction, it has to stop momentarily.

The “at rest” times from part b) when $v (t) = 0$ are the critical points. Test a point in each of the regions around those two times for the sign of $v (t)$ . Usually you can just use a sign diagram, but here we’ll display the data in a table.

As a shortcut, if $v (t)$ is a parabola or a line, you could also just picture its graph and consider where it’s above or below the $x$ -axis (which would tell you the sign). In this case, it’s a parabola that opens downward with its roots at $1$ and $3$ .

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 first moves left, then stops at $t = 1$ second to change direction and moves right. It stops again to change direction at $t = 3$ seconds and then moves left from that point onward.

d) When is the speed increasing or decreasing?

The speed is increasing (object is accelerating) when $v (t)$ and $a (t)$ have the same sign and decreasing (object is decelerating) when the signs are opposite. Here are the two functions:

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

When $t = 2$ , the acceleration is $0$ and the speed is not changing. Use this point in addition to the “at reset” points from the previous parts (when $v (t) = 0$ ) to create intervals and test points with a sign chart or diagram:

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 problem involves finding the average velocity over an interval of time, given a position function or a table of values for the position. For these problems, just divide the change in distance by the change in time. This is the basic slope formula, or the average rate of change.

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

Similarly, the average acceleration is the change in velocity over the change in time.

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

Example

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

Solution

(spoiler)

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

$v_{avg} = \frac{s ( 6 ) - s ( 2 )}{6 - 2}$

$= \frac{88}{4}$

$= 22 ft/s$

The velocity is the derivative of the position function:

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

Then the average acceleration is

$a_{avg} = \frac{v ( 6 ) - v ( 2 )}{6 - 2}$

$= \frac{62 - ( - 2 )}{4}$

$= 16 ft/s^{2}$

Distance traveled

With a position function, you can determine how far an object has traveled. There are two ways to define distance - the displacement and total distance traveled.

Displacement

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

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

Total distance

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

Find where the velocity is $0$ (when the object may change direction and affect the displacement).
Find the displacement in each region separately and take the absolute values - these are the net distances (positive values only) over those time intervals.
Sum the net distances.

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

Solution

(spoiler)

For the displacement:

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

The particle moves but returns to its original position shortly.

For the total distance, find when the particle stops momentarily:

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

$v (t) = 0$ when

$2 cos (2 t) = 0$

$2 t = \frac{π}{2}, \frac{3 π}{2} ... + πk, k \in Z$

$t = \frac{π}{4}, \frac{3 π}{4} ... + \frac{π}{2} k, k \in Z$

In the interval $[\frac{π}{2}, π]$ , it stops at $t = \frac{3 π}{4}$ only.

So there are two regions to consider: $[\frac{π}{2}, \frac{3 π}{4}]$ and $[\frac{3 π}{4}, π]$ .

The net distances in each of these two regions are:

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

$∣ s (\frac{3 π}{4}) - s (\frac{π}{2}) ∣$

$= ∣ sin (\frac{3 π}{2}) - sin (π) ∣$

$= ∣ - 1 - 0∣ = 1$

Between $[\frac{3 π}{4}, π]$ :

$∣ s (π) - s (\frac{3 π}{4}) ∣$

$= ∣ sin (2 π) - sin (\frac{3 π}{2}) ∣$

$= ∣0 - (- 1) ∣ = 1$

Adding up the net distances gives a total distance of $2$ units traveled.

Key points

By convention, positive means to the right and negative means to the left
To figure out if an object is speeding up or slowing down, find when $a (t)$ and $v (t)$ each equal $0$ and create a sign chart or diagram in the regions around those times to compare their signs.

What you’ll learn:

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

Velocity

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

The 1st derivative of the position function gives the velocity function, which tells us:

How fast an object is moving
In which direction - by convention,
- When $v (t) > 0$ (positive velocity): The object is moving right/forward/up
- When $v (t) < 0$ (negative velocity): The object is moving left/backward/down
- When $v (t) = 0$ , the particle is momentarily at rest (not moving)

Sidenote

Speed vs. velocity

Speed is the absolute value (magnitude) of the velocity - a positive value that denotes how fast an object is going.

The velocity gives both speed and direction (depending on the sign).

Acceleration

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

The derivative of the velocity function, or the 2nd derivative of position function, gives the acceleration function, which tells us if the object is speeding up or slowing down:

When $a (t)$ and $v (t)$ have the same sign at some time $t$ , 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 $(v (t) = 0)$ .
- It’s neither accelerating or decelerating.

Examples

1. 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) When is the object at rest?
c) When does the object change direction?
d) When is the speed increasing or decreasing?

Solutions

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

Differentiate the position function for the velocity function, and differentiate again for the acceleration function. Then evaluate at $t = 2$ :

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

At $t = 2$ :

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

$a (2) = - 6 (2) + 12 = 0$

The object is moving right at a speed of $3$ m/s and at constant velocity because the acceleration is $0$ m/s $^{2}$ .

b) When is the object at rest?

The object is at rest when its velocity is $0$ . Setting the velocity function equal to $0$ and solving 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) When does the object change direction?

The object is moving right/forward when $v (t) > 0$ and left/backward when $v (t) < 0$ . In order to change direction, it has to stop momentarily.

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 first moves left, then stops at $t = 1$ second to change direction and moves right. It stops again to change direction at $t = 3$ seconds and then moves left from that point onward.

d) When is the speed increasing or decreasing?

The speed is increasing (object is accelerating) when $v (t)$ and $a (t)$ have the same sign and decreasing (object is decelerating) when the signs are opposite. Here are the two functions:

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

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

Average velocity and acceleration

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

Similarly, the average acceleration is the change in velocity over the change in time.

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

Example

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

Solution

(spoiler)

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

$v_{avg} = \frac{s ( 6 ) - s ( 2 )}{6 - 2}$

$= \frac{88}{4}$

$= 22 ft/s$

The velocity is the derivative of the position function:

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

Then the average acceleration is

$a_{avg} = \frac{v ( 6 ) - v ( 2 )}{6 - 2}$

$= \frac{62 - ( - 2 )}{4}$

$= 16 ft/s^{2}$

Distance traveled

With a position function, you can determine how far an object has traveled. There are two ways to define distance - the displacement and total distance traveled.

Displacement

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

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

Total distance

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

Find where the velocity is $0$ (when the object may change direction and affect the displacement).
Find the displacement in each region separately and take the absolute values - these are the net distances (positive values only) over those time intervals.
Sum the net distances.

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

Solution

(spoiler)

For the displacement:

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

The particle moves but returns to its original position shortly.

For the total distance, find when the particle stops momentarily:

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

$v (t) = 0$ when

$2 cos (2 t) = 0$

$2 t = \frac{π}{2}, \frac{3 π}{2} ... + πk, k \in Z$

$t = \frac{π}{4}, \frac{3 π}{4} ... + \frac{π}{2} k, k \in Z$

In the interval $[\frac{π}{2}, π]$ , it stops at $t = \frac{3 π}{4}$ only.

So there are two regions to consider: $[\frac{π}{2}, \frac{3 π}{4}]$ and $[\frac{3 π}{4}, π]$ .

The net distances in each of these two regions are:

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

$∣ s (\frac{3 π}{4}) - s (\frac{π}{2}) ∣$

$= ∣ sin (\frac{3 π}{2}) - sin (π) ∣$

$= ∣ - 1 - 0∣ = 1$

Between $[\frac{3 π}{4}, π]$ :

$∣ s (π) - s (\frac{3 π}{4}) ∣$

$= ∣ sin (2 π) - sin (\frac{3 π}{2}) ∣$

$= ∣0 - (- 1) ∣ = 1$

Adding up the net distances gives a total distance of $2$ units traveled.

Key points

By convention, positive means to the right and negative means to the left
To figure out if an object is speeding up or slowing down, find when $a (t)$ and $v (t)$ each equal $0$ and create a sign chart or diagram in the regions around those times to compare their signs.