Straight-Line Motion: Connecting Position, Velocity, and Acceleration

AP Calculus · Contextual Applications of Differentiation

Straight-Line Motion: Position, Velocity, Acceleration

One function and two derivatives. The marks are in the signs, and in telling speed from velocity.

key idea$s\xrightarrow{\ '\ }v\xrightarrow{\ '\ }a;\qquad \text{speed}=\lvert v\rvert$

Setting up the model

Position, velocity and acceleration are one function and its first two derivatives. The exam questions are almost never about differentiating — they are about reading signs and distinguishing velocity from speed.

Reading a motion problem

$s(t)$ is position — a signed coordinate, not a distance. It can be negative.
$v(t)=s'(t)$ is velocity — signed. Its sign gives the direction of travel.
$a(t)=v'(t)=s''(t)$ is acceleration — the rate of change of velocity.
Speed is $\lvert v(t)\rvert$ — never negative, and it is not $\lvert a\rvert$'s partner.
The particle is at rest when $v=0$, and changes direction when $v$ changes sign.

position $s(t)$

velocity $v(t)$

$s(t)=t^{3}-6t^{2}+9t$ and its velocity. The particle is at rest at $t=1$ and $t=3$, where $v$ crosses zero — and those are exactly the turning points of $s$.

Question asked	What to examine	Condition
When is the particle at rest?	$v(t)$	$v=0$
When does it change direction?	$v(t)$	$v$ changes sign
When is it moving left?	$v(t)$	$v\lt 0$
When is it speeding up?	$v$ and $a$	same sign
When is it slowing down?	$v$ and $a$	opposite signs
Total distance travelled	$\lvert v\rvert$	split at every $v=0$

Rows 4 and 5 are the ones students get wrong. Speeding up is not the same as positive acceleration — it is about the two signs agreeing.

Worked examples

EX 1

At rest and direction changes

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

For $t\ge0$, find when the particle is at rest and when it changes direction.

Differentiate

$v(t)=3t^{2}-12t+9=3\left(t^{2}-4t+3\right)=3(t-1)(t-3)$.

Solve $v=0$

$t=1$ and $t=3$ — the particle is at rest at those instants.

Test the sign in each interval

$v(0)=9\gt 0$; $v(2)=-3\lt 0$; $v(4)=9\gt 0$.

Identify direction changes

$v$ goes $+\to-$ at $t=1$ and $-\to+$ at $t=3$. Both are direction changes.

Note the distinction

At rest requires only $v=0$; changing direction requires a sign change. They coincide here, but not always.

EX 2

Speeding up or slowing down?

$v$ and $a$ signs

At $t=2$, is the particle speeding up or slowing down?

Find the velocity

$v(2)=3(1)(-1)=-3$, so $v\lt 0$ — moving left.

Find the acceleration

$a(t)=v'(t)=6t-12$, so $a(2)=0$.

Apply the rule

With $a=0$ the particle is momentarily neither speeding up nor slowing down.

Check a nearby instant

At $t=2.5$: $v=3(1.5)(-0.5)=-2.25$ and $a=3\gt 0$. Opposite signs, so it is slowing down.

State the rule plainly

Same signs ⇒ speeding up. Opposite signs ⇒ slowing down. Positive acceleration alone tells you nothing.

EX 3

Total distance versus displacement

$[0,4]$

Find the displacement and the total distance travelled on $[0,4]$.

Displacement is the net change

$s(4)-s(0)=(64-96+36)-0=4$ metres.

Find where the direction changes

$v=0$ at $t=1$ and $t=3$, so split $[0,4]$ into $[0,1]$, $[1,3]$, $[3,4]$.

Compute each leg

$s(0)=0$, $s(1)=4$, $s(3)=0$, $s(4)=4$. Legs: $+4$, $-4$, $+4$.

Add the magnitudes

Total distance $=\lvert4\rvert+\lvert-4\rvert+\lvert4\rvert=12$ metres.

Compare

Displacement $4$ m, distance $12$ m. Failing to split at $t=1$ and $t=3$ gives $4$ for both — the standard error.

EX 4

Reading it all from the graph

graph → conclusions

Using the velocity graph above, describe the motion on $[0,4]$ without any algebra.

Where $v$ is above the axis

On $[0,1)$ and $(3,4]$ the particle moves in the positive direction.

Where $v$ is below

On $(1,3)$ it moves in the negative direction.

Where $v$ touches zero

At $t=1$ and $t=3$ — at rest, and each is a direction change since $v$ crosses.

Read acceleration as a slope

$a$ is the slope of the velocity graph: negative until the vertex at $t=2$, positive after.

Combine for speeding up

On $(1,2)$: $v\lt0$ and $a\lt0$ — same signs, speeding up. On $(2,3)$: $v\lt0$, $a\gt0$ — slowing down.

Units & interpretation

Quantity	Symbol	Units	Sign means
position	$s(t)$	metres	which side of the origin
velocity	$v(t)$	m/s	direction of travel
speed	$\lvert v(t)\rvert$	m/s	— never negative
acceleration	$a(t)$	m/s$^{2}$	which way velocity is changing
displacement	$s(b)-s(a)$	metres	net change, can be $0$
total distance	sum of $\lvert$leg$\rvert$	metres	never negative

Speed and velocity share units but not signs. Displacement and distance share units but not meaning.

Watch out

Speeding up ≠ positive acceleration

It means $v$ and $a$ have the same sign. A particle moving left with negative acceleration is speeding up.

Split before totalling distance

Total distance requires breaking the interval at every $v=0$ where the sign changes. Skipping that returns the displacement instead.

$v=0$ does not always mean a direction change

The velocity must change sign. A velocity that touches zero and returns the same way — like $v=(t-2)^{2}$ — is a pause, not a reversal.

Speed is $\lvert v\rvert$, not $\lvert a\rvert$

Acceleration measures how velocity changes. It has no direct relationship to how fast the particle is going.

Course map. Topic 4.2 · LO CHA-3.B — calculate rates of change in applied contexts involving motion (EK CHA-3.B.1).