What is the difference between velocity and speed?

Velocity v(t) is signed and tells you both how fast and in which direction the particle moves. Speed is its absolute value |v(t)| and is never negative — it strips away direction. A velocity of -5 m/s means moving left at a speed of 5 m/s.

How do I know when a particle changes direction?

A particle changes direction only where velocity v(t) changes sign. Find where v(t)=0 or is undefined, then check the sign of v on each side. If it switches from positive to negative or vice versa, the direction reverses. If v touches zero without switching signs, no reversal occurs.

Does negative acceleration always mean slowing down?

No. Slowing down depends on velocity and acceleration having opposite signs. If the particle already moves left (v<0) and acceleration is also negative, both signs match and it speeds up. Always compare the sign of v with the sign of a; equivalently, check whether v· a is positive (speeding up) or negative (slowing down).

How is total distance different from displacement?

Displacement is the net change in position, s(b)-s(a), and can be negative. Total distance adds the absolute change in position over each leg between turning points, so it is always nonnegative and counts backtracking. When a particle reverses direction, total distance exceeds the magnitude of displacement.

AP-CALCBC-4.2

U4.2 Straight-Line Motion

Q: How is total distance different from displacement?

Displacement is the net change in position, s(b)-s(a), and can be negative. Total distance adds the absolute change in position over each leg between turning points, so it is always nonnegative and counts backtracking. When a particle reverses direction, total distance exceeds the magnitude of displacement.

Master AP Calculus BC straight-line motion: connect position, velocity, and acceleration, tell speed from velocity, and find when a particle changes direction.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on U4.2 Straight-Line Motion, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

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What this lesson covers

When a particle slides along a number line, calculus lets you decode its entire story from a single function. Position

s(t)

tells you where it is, its derivative

v(t)=s'(t)

tells you how fast and which way it's going, and the second derivative

a(t)=s''(t)

tells you how the velocity itself is changing.

This lesson ties those three functions together and clears up the trickiest exam traps: the difference between speed and velocity, how to know when the particle moves left versus right, and how to decide whether it is speeding up or slowing down. These ideas show up constantly on both the multiple-choice and free-response sections, so getting the vocabulary and sign analysis exact pays off directly.

The Three Functions and How They Connect

Straight-line motion (also called rectilinear motion) tracks a particle along a single axis. Everything flows from one position function

s(t)

.

Velocity is the rate of change of position:

v(t)=s'(t)

. It carries a sign. Positive velocity means the particle moves in the positive direction (usually right or up); negative velocity means it moves in the negative direction (left or down).

Acceleration is the rate of change of velocity:

a(t)=v'(t)=s''(t)

. It tells you whether velocity is increasing or decreasing, not directly whether the particle is getting faster.

Quantity	Symbol	Derivative relationship	Sign meaning
Position	$s(t)$	—	where the particle is
Velocity	$v(t)=s'(t)$	first derivative of $s$	direction of motion
Acceleration	$a(t)=s''(t)$	derivative of $v$	how velocity changes

Going the other direction, if you are given velocity you recover position with an antiderivative (a preview of the integral techniques later in the course). For now, focus on differentiating: given

s(t)

, compute

v(t)

and

a(t)

cleanly and keep track of units — position in meters, velocity in meters per second, acceleration in meters per second squared.

Velocity vs. Speed, and Left vs. Right

This is the single most tested distinction in the topic. Velocity is signed; speed is its absolute value,

\text{speed}=|v(t)|

. Speed is never negative.

To find when the particle moves right, solve

v(t)>0

. To find when it moves left, solve

v(t)<0

. The particle is momentarily at rest wherever

v(t)=0

.

A particle changes direction only when

v(t)

changes sign — that is,

v(t)

passes through zero and switches from positive to negative (or vice versa). A common mistake is claiming the particle changes direction every time

v(t)=0

. If velocity touches zero but does not change sign (like

v(t)=(t-2)^2

t=2

), the particle does not reverse.

To analyze direction, build a sign chart for

v(t)

: find all

t

where

v(t)=0

or is undefined, then test intervals between them. This is exactly the same technique you used for increasing/decreasing functions, because

s(t)

increases when

v(t)>0

.

Remember the phrasing on the exam: "How far right does the particle move?" asks about position, while "What is the particle's speed at

t=3

?" asks for

|v(3)|

. Read carefully — plenty of points are lost to reading errors, not calculus errors.

Speeding Up or Slowing Down

Whether a particle is speeding up or slowing down depends on the relationship between velocity and acceleration, not on acceleration alone.

The rule: the particle is speeding up when

v(t)

and

a(t)

have the same sign, and slowing down when they have opposite signs. Intuitively, if acceleration pushes in the same direction the particle already moves, it gains speed; if acceleration opposes the motion, it loses speed.

$v(t)$	$a(t)$	Speed
$+$	$+$	speeding up
$+$	$-$	slowing down
$-$	$-$	speeding up
$-$	$+$	slowing down

Notice that a negative acceleration does NOT automatically mean slowing down. If the particle already moves left (

v<0

) and acceleration is also negative, it speeds up in the negative direction.

A clean equivalent test: the speed

|v(t)|

is increasing exactly when the product

v(t)\cdot a(t)>0

. On the exam, once you compute both

v

and

a

at a given time, just multiply and check the sign. If asked over an interval, build sign charts for both

v

and

a

and compare.

How the Exam Tests This Topic

Straight-line motion appears in multiple-choice questions using a given formula for

s(t)

, and in free-response questions using a graph or a table of velocity values.

Typical tasks include: find velocity or acceleration at a specific time; determine intervals where the particle moves left; find total distance traveled; and decide whether the particle is speeding up. Total distance is different from displacement. Displacement over

[a,b]

equals

s(b)-s(a)

, the net change. Total distance adds up the magnitude of every leg of the trip, so you must locate each time the particle turns around, then sum the absolute changes in position between turning points.

Calculator-active questions often give a velocity graph and ask you to reason about

s

and

a

without an explicit formula. Here, acceleration is the slope of the velocity graph, and the particle is at rest where the velocity graph crosses the horizontal axis.

A frequent misconception: students confuse "velocity is decreasing" with "speed is decreasing." Velocity decreasing means

a(t)<0

; speed decreasing means

|v|

is falling, which requires

v

and

a

to have opposite signs. Keep these separate and label every conclusion with its justification, since free-response scoring rewards explicit reasoning tied to the sign of

v

and

a

Key terms

Position $s(t)$ .: A function giving the particle's location on a coordinate axis at time $t$ .
Velocity $v(t)$ .: The rate of change of position, $v(t)=s'(t)$ ; its sign gives the direction of motion.
Speed.: The absolute value of velocity, $|v(t)|$ ; always nonnegative, measures how fast without direction.
Acceleration $a(t)$ .: The rate of change of velocity, $a(t)=v'(t)=s''(t)$ ; describes how velocity is changing.
Displacement.: Net change in position over an interval, $s(b)-s(a)$ ; can be positive, negative, or zero.
Total distance.: Sum of the magnitudes of all movements over an interval, accounting for direction reversals.
Change of direction.: Occurs where $v(t)$ changes sign, not merely where $v(t)=0$ .
At rest.: The condition when $v(t)=0$ ; the particle is momentarily not moving.

Worked example

A particle moves along a line with position

s(t)=t^3-6t^2+9t

for

t\ge 0

, where

s

is in meters and

t

in seconds. (a) Find

v(t)

and

a(t)

. (b) When is the particle at rest? (c) On what intervals does it move right? (d) At

t=1

, is it speeding up or slowing down?

Start by differentiating.

v(t)=s'(t)=3t^2-12t+9

, and

a(t)=v'(t)=6t-12

.

For part (b), set

v(t)=0

3t^2-12t+9=3(t^2-4t+3)=3(t-1)(t-3)=0

, so

t=1

and

t=3

. The particle is at rest at

t=1

s and

t=3

s.

For part (c), build a sign chart for

v(t)=3(t-1)(t-3)

t\ge 0

. Test points: at

t=0

v=3(-1)(-3)=9>0

; at

t=2

v=3(1)(-1)=-3<0

; at

t=4

v=3(3)(1)=9>0

. So the particle moves right on

0\le t<1

and

t>3

, and moves left on

1<t<3

.

For part (d), evaluate at

t=1

. Since

t=1

is a turning point, check just how the signs behave near it, but the question asks at

t=1

exactly where

v(1)=0

. Instead evaluate the trend: use a point like

t=0.5

if needed. At

t=1

v(1)=0

, so speed is momentarily zero — the particle is neither speeding up nor slowing down at that instant; it is at rest and reversing direction. To illustrate the speeding-up test elsewhere, take

t=2

v(2)=-3

and

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

... use

t=2.5

v(2.5)=3(1.5)(-0.5)=-2.25<0

a(2.5)=6(2.5)-12=3>0

; opposite signs, so the particle is slowing down at

t=2.5

Practice questions

A particle moves along a line with velocity

v(t)=t^2-4t+3

for

t\ge 0

. At

t=4

, is the particle speeding up or slowing down?

Speeding up, because $v(4)>0$ and $a(4)>0$
Slowing down, because $v(4)>0$ and $a(4)<0$
Speeding up, because $v(4)<0$ and $a(4)<0$
Slowing down, because $v(4)<0$ and $a(4)>0$

Answer: Speeding up, because $v(4)>0$ and $a(4)>0$

Compute

v(4)=16-16+3=3>0

. Then

a(t)=v'(t)=2t-4

, so

a(4)=8-4=4>0

. Since velocity and acceleration share the same sign, their product is positive and the speed is increasing — the particle is speeding up.

A particle has position

s(t)=2t^3-9t^2+12t

for

0\le t\le 3

. Find the total distance traveled by the particle over this interval, and explain why it differs from the displacement.

Answer: Total distance = 5 meters; displacement = 3 meters.

First

v(t)=6t^2-18t+12=6(t-1)(t-2)

, zero at

t=1

and

t=2

. Evaluate position:

s(0)=0

s(1)=2-9+12=5

s(2)=16-36+24=4

s(3)=54-81+36=9

. From

t=0

1

the particle moves from 0 to 5 (distance 5); from 1 to 2 it moves 5 to 4 (distance 1); from 2 to 3 it moves 4 to 9 (distance 5). Total distance

=5+1+5=11

meters. Displacement is

s(3)-s(0)=9-0=9

meters. They differ because the particle reverses direction at

t=1

, so backtracking adds to distance but not to net displacement. (Corrected totals: distance 11 m, displacement 9 m.)

The velocity of a particle is

v(t)=\sin t

for

0\le t\le 2\pi

. On what interval(s) is the particle moving left, and where does it change direction?

Answer: It moves left on $\pi<t<2\pi$ and changes direction at $t=\pi$ .

The particle moves left where

v(t)<0

. Since

\sin t<0

\pi<t<2\pi

, that is the leftward interval. Velocity is zero at

t=0,\pi,2\pi

. At

t=\pi

\sin t

changes from positive to negative, so the particle genuinely reverses direction there. At

t=0

and

t=2\pi

(endpoints),

\sin t

does not switch sign within the interval in a way that reverses interior motion, so

t=\pi

is the direction change.

FAQ

What is the difference between velocity and speed?: Velocity $v(t)$ is signed and tells you both how fast and in which direction the particle moves. Speed is its absolute value $|v(t)|$ and is never negative — it strips away direction. A velocity of $-5$ m/s means moving left at a speed of 5 m/s.
How do I know when a particle changes direction?: A particle changes direction only where velocity $v(t)$ changes sign. Find where $v(t)=0$ or is undefined, then check the sign of $v$ on each side. If it switches from positive to negative or vice versa, the direction reverses. If $v$ touches zero without switching signs, no reversal occurs.
Does negative acceleration always mean slowing down?: No. Slowing down depends on velocity and acceleration having opposite signs. If the particle already moves left ( $v<0$ ) and acceleration is also negative, both signs match and it speeds up. Always compare the sign of $v$ with the sign of $a$ ; equivalently, check whether $v\cdot a$ is positive (speeding up) or negative (slowing down).
How is total distance different from displacement?: Displacement is the net change in position, $s(b)-s(a)$ , and can be negative. Total distance adds the absolute change in position over each leg between turning points, so it is always nonnegative and counts backtracking. When a particle reverses direction, total distance exceeds the magnitude of displacement.

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