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 $s(t)$ 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

$$v(t)=s^{\prime }(t)$$

The 1st derivative of the position function is the velocity function. It tells you:

1. How fast the object is moving
2. Which direction it’s moving (based on the sign)
   • 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 velocity. It’s always nonnegative and tells you how fast the object is moving.

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

Acceleration

$$a(t)=v^{\prime }(t)=s^{\prime \prime }(t)$$

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 $t$:

1. When $a(t)$ and $v(t)$ have the same sign, the object is speeding up at that moment.
2. When $a(t)$ and $v(t)$ have opposite signs, the object is slowing down.
3. When $a(t)=0$, the object could be moving at constant velocity $(v(t)\ne 0)$ 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.

Examples

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

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.

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

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

At $t=2$:

$$\begin{gathered} v(2)=-3(2{)}^2+12(2)-9 \\ =3 \end{gathered}$$

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

So at $t=2$, the object is moving right with speed $3$ m/s, and its velocity is not changing at that instant because $a(2)=0$ m/s${}^2$.

b) When is the object at rest?

The object is at rest when its velocity is $0$. Set $v(t)=0$ and solve for $t$:

$$-3t^2+12t-9=0$$

$$-3(t^2-4t+3)=0$$

$$-3(t-1)(t-3)=0$$

$$t=1,3$$

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

c) When does the object change direction?

The object moves right when $v(t)>0$ and left when $v(t)<0$. To change direction, it must pass through a time when $v(t)=0$.

From part (b), the critical times are $t=1$ and $t=3$. Check the sign of $v(t)$ on the intervals around these times. A sign diagram works well; here’s the same information in a table.

As a shortcut, when $v(t)$ is a line or parabola, you can also use its graph to see where it’s above or below the $t$-axis (which tells you the sign). Here, $v(t)$ is a downward-opening parabola with zeros at $t=1$ and $t=3$.

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 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) When is the speed increasing or decreasing?

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

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

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 when you’re given a position function (or a table of position values). Average velocity is the change in position divided by the change in time - this is the slope (average rate of change) of the position function over the interval.

$$v_{\text{avg}}=\frac{s(t_2)-s(t_1)}{t_2-t_1}$$

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

$$a_{\text{avg}}=\frac{v(t_2)-v(t_1)}{t_2-t_1}$$

Example

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

Solution

Distance traveled

A position function can tell you how far an object has traveled, but there are two related ideas to keep separate:

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

Total distance

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

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

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

Solution