Unit 8Displacement vs. Total Distance

Concept 4.100Built for May 2027 AP Exam

Displacement vs. Total Distance

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

Displacement over a time interval is the integral of velocity, $\int_a^b v(t)\,dt$ , and it is signed. Total distance travelled is the integral of speed, $\int_a^b |v(t)|\,dt$ , and it is never negative. The two numbers agree exactly when the particle never reverses on the interval. The moment $v$ changes sign, the backward stretch cancels part of the forward stretch inside the signed integral, so displacement ends up smaller than distance. The mechanical fix is to split the interval at every zero of $v$ and integrate each piece separately, then add the absolute values. Released FRQs almost always ask both quantities in the same part, which means the grader is watching whether you produced two different numbers.

On the AP Exam

AP Calculus AB tests this skill through the representations and reasoning described below.

What you’ll see: A velocity function or a velocity table on a stated interval, followed by “find the displacement” and “find the total distance travelled.” Some years phrase it as “how far does the particle travel” without the word distance.
What you’ll do: Integrate $v$ for displacement. Find the zeros of $v$ , split the interval, integrate $v$ on each piece, and add the absolute values for distance. If only a table is given, use a left, right, or midpoint sum on each monotone piece.
Memory hook: Velocity signs, speed doesn’t. Displacement keeps the sign; distance throws it away.
Use it on the exam: For MCQs: If the choices include a negative number for a distance question, it is a distractor built from the signed integral. For FRQs: Show the split points explicitly. “ $v(t) = 0$ at $t = 1$ and $t = 3$ ” is usually worth a point on its own, and every later line depends on it.

Concept Review

Two Integrals, One Function

Figure 1 — The same velocity curve yields a signed net area and an unsigned total area.

Both quantities start from the same graph and differ only in what happens to the sign:

$\text{displacement} = \int_a^b v(t)\,dt \qquad\qquad \text{distance} = \int_a^b |v(t)|\,dt$

The left integral adds signed areas, so a region below the axis subtracts. The right integral reflects every below-axis region above the axis first, then adds. That single absolute value is the entire difference between the two concepts.

Two bar displays comparing a signed displacement of four thirds with a total distance of four

Why the Numbers Diverge

Suppose a particle moves forward $5$ meters, back $2$ meters, then forward $1$ meter. Its displacement is $5 - 2 + 1 = 4$ meters. Its distance is $5 + 2 + 1 = 8$ meters. The odometer reads $8$ ; the position reads $x_0 + 4$ .

Nothing about the motion is unusual. Any reversal opens a gap between the two numbers, and the gap equals twice the total backward travel. That is a useful check: distance minus displacement is always even in the sense that it counts the retraced ground twice.

The Splitting Procedure

There is one reliable algorithm, and it never changes:

Solve $v(t) = 0$ on $(a, b)$ and keep only the solutions inside the interval.
Sort them and use them, together with $a$ and $b$ , to cut the interval into subintervals.
On each subinterval, decide the sign of $v$ from a single test point.
Integrate $v$ on each subinterval to get a signed piece.
Displacement is the plain sum of the pieces. Distance is the sum of their absolute values.

Step 3 is where most lost points live. A sign chosen from the wrong test point flips one piece and quietly changes the answer, while the arithmetic on that piece still looks correct.

Reading the Position Function Instead

If the position function $s(t)$ is given rather than $v(t)$ , both quantities are still available. Displacement is $s(b) - s(a)$ . Total distance is the sum of $|s(t_{i+1}) - s(t_i)|$ over the turning points, where the turning points are the zeros of $v = s'$ .

Given	Displacement	Total distance
velocity $v(t)$	$\int_a^b v(t)\,dt$	$\int_a^b \lvert v(t)\rvert\,dt$
position $s(t)$	$s(b) - s(a)$	sum of $\lvert s(t_{i+1}) - s(t_i)\rvert$ across turning points
speed only	not recoverable	$\int_a^b \text{speed}(t)\,dt$
acceleration $a(t)$ plus $v(a)$	integrate twice, then subtract	split at zeros of the reconstructed $v$

The third row matters. Speed alone destroys the direction information, so displacement cannot be rebuilt from it. Questions that give only speed are asking for distance, and a question that gives only speed but asks for displacement is unanswerable as stated.

Units and Wording

Displacement and distance share units, because both come from velocity times time. They differ in interpretation, not in dimension. Watch the nouns:

“How far has the particle travelled” means distance.
“What is the particle’s displacement” means the signed integral.
“Change in position” means displacement, and it can be negative.
“Total distance” never takes a sign, and a negative answer is automatically wrong.

Common Misconception Check

Myth 1: “Displacement and distance are the same because both are integrals of velocity.” Truth: Only the distance integral carries an absolute value, and that is exactly what removes the cancellation.
Myth 2: “If the particle ends where it started, the distance is zero.” Truth: The displacement is zero; the distance is the full path length, which is positive.
Myth 3: “You can compute distance by integrating $|v|$ without finding the zeros.” Truth: You can if your calculator handles the absolute value, but on a no-calculator part you must split, and the split points are themselves scored.

Topic-Specific Study Method

Decide Which Integral Before You Reach for the Antiderivative

Borrowed idea: Displacement asks where the particle ended up, and distance asks how far it walked. Are those ever the same number? So when do the two numbers agree? They agree exactly when the velocity never changes sign on the interval. The integral of velocity gives a signed total, so a backward stretch subtracts from the forward stretch. What does the integral of speed give instead? It gives the total distance, because the absolute value throws the sign away. Splitting the interval at every zero of velocity is the mechanical way to compute it. Each piece then contributes its own positive length. Write the split points down before you integrate anything.

Exam disguise	What you see on the page	What your brain must translate	Required execution
“total distance travelled”	a velocity and a time window	unsigned accumulation	split at zeros, integrate each piece, add absolute values
“displacement of the particle”	the same velocity and window	signed accumulation	integrate once, keep the sign
“change in position”	a velocity, sometimes a table	displacement in disguise	$s(b) - s(a)$ , or the signed integral
“how far from the starting point”	a velocity that reverses	distance from the origin now	integrate, then take the absolute value of the total
“particle is at rest when”	the zeros of the velocity	the split points themselves	solve $v(t) = 0$ , list only interior solutions

① Circle the noun in the question before writing any integral. “Distance” and “displacement” lead to different scoring lines.

② List the zeros of $v$ as a separate numbered line. That line is scored even when the integral is wrong.

③ Sanity-check the pair at the end: distance must be at least the absolute value of displacement, and they must be equal when there is no interior zero.

Training & Error Diagnosis

Core Archetype (Simplest Form): A particle moves along a line with velocity $v(t) = t^{2} - 4t + 3$ meters per second for $0 \le t \le 4$ . Find its displacement and the total distance travelled.

Standard Execution:

Zeros. $t^{2} - 4t + 3 = (t-1)(t-3) = 0$ gives $t = 1$ and $t = 3$ , both interior.

Pieces. $\int_0^1 v\,dt = \frac{4}{3}$ , $\int_1^3 v\,dt = -\frac{4}{3}$ , $\int_3^4 v\,dt = \frac{4}{3}$ .

Displacement. $\frac{4}{3} - \frac{4}{3} + \frac{4}{3} = \frac{4}{3}$ meters.

Distance. $\frac{4}{3} + \frac{4}{3} + \frac{4}{3} = 4$ meters.

Check. Distance exceeds displacement, as it must whenever the particle reverses.

Exam Variation (Table / Context Shift): A particle moves along the $x$ -axis with velocity given by the table below. Using a trapezoidal sum on each monotone piece, find the total distance travelled on $0 \le t \le 6$ .

$t$	0	1	3	6
$v(t)$	2	$-1$	$-3$	4

Solution: The velocity changes sign between $t = 1$ and $t = 3$ , and again between $t = 3$ and $t = 6$ , so the table alone does not pin the zeros. The standard treatment on a table part is to split at the tabulated sign change and use the trapezoidal sum on each block. On $[0,1]$ the trapezoid is $\frac{2 + (-1)}{2} \cdot 1 = 0.5$ . On $[1,3]$ it is $\frac{(-1) + (-3)}{2} \cdot 2 = -4$ . On $[3,6]$ it is $\frac{(-3) + 4}{2} \cdot 3 = 1.5$ . Adding absolute values gives $0.5 + 4 + 1.5 = 6$ meters. The signed sum would give $-2$ , which is the displacement estimate.

Common Traps & Failure Signals:

Absolute value applied at the end (Signal: $\left|\int_a^b v\right|$ written for a distance question): the absolute value belongs inside, on each piece.
Interior zero missed (Signal: one integral where the velocity clearly reverses): solve $v(t) = 0$ first, always.
Distance reported as negative (Signal: a minus sign on a length): distance is a sum of absolute values.

6D Cognitive Error Diagnosis (Where Your Thinking Broke):

Sign Kept: You reported the signed integral as the distance.
- Zero Missed: You integrated the velocity in one stroke across a reversal, so the backward travel cancelled the forward travel and the absolute value was applied only to the final total instead of to each piece.
- Path Shortened: You treated the trip as one-way because the graph looked smooth, and a smooth graph hides a reversal that the algebra would have found in one line.

Past 10-Year AP Exam Appearance

Testing Frequency: Most Years; Five of the nine released exams since 2017 ask for a distance travelled or a displacement in a motion context, and the two quantities usually appear in the same part
10-Year Total Count: 5 of 9 Exams · 6 FRQ Parts; One year asks the pair twice, once as displacement and once as distance, which is why the part count exceeds the exam count
FRQ Years Appeared: 2019 (Q2c), 2021 (Q2d), 2023 (Q2c), 2023 (Q2d), 2024 (Q2d), 2026 (Q5C)

Practice Examples

MCQ 1

Distance With a Reversal

A particle moves along a line with velocity $v(t) = t - 2$ for $0 \le t \le 3$ . The total distance travelled is

MCQ 2

Displacement From a Table

The velocity of a particle is given by the table below. Using a trapezoidal sum, the particle’s displacement on $0 \le t \le 4$ is approximately

$t$	0	2	4
$v(t)$	3	1	$-5$

MCQ 3

Which Integral Answers the Question

A particle has velocity $v(t) = \sin t$ on $0 \le t \le 2\pi$ . Which expression gives the total distance travelled?

FRQ 4

A Full Motion Analysis

A particle moves along the $x$ -axis with velocity $v(t) = t^{2} - 4t + 3$ for $0 \le t \le 4$ , measured in meters per second. At $t = 0$ the particle is at $x = 2$ .

(a) Find all times in $0 < t < 4$ at which the particle is at rest.

(b) Find the displacement of the particle on $0 \le t \le 4$ .

(c) Find the total distance travelled by the particle on $0 \le t \le 4$ .

(d) Find the position of the particle at $t = 4$ and explain why it differs from the total distance.

Cheatsheet

Core Reference

	Action	Failure if skipped
1	read the noun: displacement or distance	the wrong integral gets built and every later line is wasted
2	solve $v(t) = 0$ for interior zeros	the reversal is invisible and the pieces are wrong
3	integrate $v$ on each subinterval	the signed pieces are unavailable
4	add signed pieces for displacement	the answer is an unsigned length
5	add absolute pieces for distance	the answer can come out negative
6	check distance $\ge \lvert$ displacement $\rvert$	a sign error survives to the final line

Sentences That Score

TOPIC REVIEW • NEXT STEP IN YOUR LEARNING LOOP

Ready to check your understanding?

Strengthen this topic with AP-style practice, or continue to next topic.

Start Topic Practice Next Topic: 8.5 Net Change vs. Total Change

Unit 8Displacement vs. Total Distance

Concept 4.100Built for May 2027 AP Exam

Displacement vs. Total Distance

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

On the AP Exam

AP Calculus AB tests this skill through the representations and reasoning described below.

What you’ll see: A velocity function or a velocity table on a stated interval, followed by “find the displacement” and “find the total distance travelled.” Some years phrase it as “how far does the particle travel” without the word distance.
What you’ll do: Integrate $v$ for displacement. Find the zeros of $v$ , split the interval, integrate $v$ on each piece, and add the absolute values for distance. If only a table is given, use a left, right, or midpoint sum on each monotone piece.
Memory hook: Velocity signs, speed doesn’t. Displacement keeps the sign; distance throws it away.
Use it on the exam: For MCQs: If the choices include a negative number for a distance question, it is a distractor built from the signed integral. For FRQs: Show the split points explicitly. “ $v(t) = 0$ at $t = 1$ and $t = 3$ ” is usually worth a point on its own, and every later line depends on it.

Concept Review

Two Integrals, One Function

Figure 1 — The same velocity curve yields a signed net area and an unsigned total area.

Both quantities start from the same graph and differ only in what happens to the sign:

$\text{displacement} = \int_a^b v(t)\,dt \qquad\qquad \text{distance} = \int_a^b |v(t)|\,dt$

Two bar displays comparing a signed displacement of four thirds with a total distance of four

Why the Numbers Diverge

The Splitting Procedure

There is one reliable algorithm, and it never changes:

Solve $v(t) = 0$ on $(a, b)$ and keep only the solutions inside the interval.
Sort them and use them, together with $a$ and $b$ , to cut the interval into subintervals.
On each subinterval, decide the sign of $v$ from a single test point.
Integrate $v$ on each subinterval to get a signed piece.
Displacement is the plain sum of the pieces. Distance is the sum of their absolute values.

Step 3 is where most lost points live. A sign chosen from the wrong test point flips one piece and quietly changes the answer, while the arithmetic on that piece still looks correct.

Reading the Position Function Instead

Given	Displacement	Total distance
velocity $v(t)$	$\int_a^b v(t)\,dt$	$\int_a^b \lvert v(t)\rvert\,dt$
position $s(t)$	$s(b) - s(a)$	sum of $\lvert s(t_{i+1}) - s(t_i)\rvert$ across turning points
speed only	not recoverable	$\int_a^b \text{speed}(t)\,dt$
acceleration $a(t)$ plus $v(a)$	integrate twice, then subtract	split at zeros of the reconstructed $v$

Units and Wording

Displacement and distance share units, because both come from velocity times time. They differ in interpretation, not in dimension. Watch the nouns:

“How far has the particle travelled” means distance.
“What is the particle’s displacement” means the signed integral.
“Change in position” means displacement, and it can be negative.
“Total distance” never takes a sign, and a negative answer is automatically wrong.

Common Misconception Check

Myth 1: “Displacement and distance are the same because both are integrals of velocity.” Truth: Only the distance integral carries an absolute value, and that is exactly what removes the cancellation.
Myth 2: “If the particle ends where it started, the distance is zero.” Truth: The displacement is zero; the distance is the full path length, which is positive.
Myth 3: “You can compute distance by integrating $|v|$ without finding the zeros.” Truth: You can if your calculator handles the absolute value, but on a no-calculator part you must split, and the split points are themselves scored.

Topic-Specific Study Method

Decide Which Integral Before You Reach for the Antiderivative

Exam disguise	What you see on the page	What your brain must translate	Required execution
“total distance travelled”	a velocity and a time window	unsigned accumulation	split at zeros, integrate each piece, add absolute values
“displacement of the particle”	the same velocity and window	signed accumulation	integrate once, keep the sign
“change in position”	a velocity, sometimes a table	displacement in disguise	$s(b) - s(a)$ , or the signed integral
“how far from the starting point”	a velocity that reverses	distance from the origin now	integrate, then take the absolute value of the total
“particle is at rest when”	the zeros of the velocity	the split points themselves	solve $v(t) = 0$ , list only interior solutions

① Circle the noun in the question before writing any integral. “Distance” and “displacement” lead to different scoring lines.

② List the zeros of $v$ as a separate numbered line. That line is scored even when the integral is wrong.

③ Sanity-check the pair at the end: distance must be at least the absolute value of displacement, and they must be equal when there is no interior zero.

Training & Error Diagnosis

Core Archetype (Simplest Form): A particle moves along a line with velocity $v(t) = t^{2} - 4t + 3$ meters per second for $0 \le t \le 4$ . Find its displacement and the total distance travelled.

Standard Execution:

Zeros. $t^{2} - 4t + 3 = (t-1)(t-3) = 0$ gives $t = 1$ and $t = 3$ , both interior.

Pieces. $\int_0^1 v\,dt = \frac{4}{3}$ , $\int_1^3 v\,dt = -\frac{4}{3}$ , $\int_3^4 v\,dt = \frac{4}{3}$ .

Displacement. $\frac{4}{3} - \frac{4}{3} + \frac{4}{3} = \frac{4}{3}$ meters.

Distance. $\frac{4}{3} + \frac{4}{3} + \frac{4}{3} = 4$ meters.

Check. Distance exceeds displacement, as it must whenever the particle reverses.

$t$	0	1	3	6
$v(t)$	2	$-1$	$-3$	4

Solution: The velocity changes sign between $t = 1$ and $t = 3$ , and again between $t = 3$ and $t = 6$ , so the table alone does not pin the zeros. The standard treatment on a table part is to split at the tabulated sign change and use the trapezoidal sum on each block. On $[0,1]$ the trapezoid is $\frac{2 + (-1)}{2} \cdot 1 = 0.5$ . On $[1,3]$ it is $\frac{(-1) + (-3)}{2} \cdot 2 = -4$ . On $[3,6]$ it is $\frac{(-3) + 4}{2} \cdot 3 = 1.5$ . Adding absolute values gives $0.5 + 4 + 1.5 = 6$ meters. The signed sum would give $-2$ , which is the displacement estimate.

Common Traps & Failure Signals:

Absolute value applied at the end (Signal: $\left|\int_a^b v\right|$ written for a distance question): the absolute value belongs inside, on each piece.
Interior zero missed (Signal: one integral where the velocity clearly reverses): solve $v(t) = 0$ first, always.
Distance reported as negative (Signal: a minus sign on a length): distance is a sum of absolute values.

6D Cognitive Error Diagnosis (Where Your Thinking Broke):

Sign Kept: You reported the signed integral as the distance.
- Zero Missed: You integrated the velocity in one stroke across a reversal, so the backward travel cancelled the forward travel and the absolute value was applied only to the final total instead of to each piece.
- Path Shortened: You treated the trip as one-way because the graph looked smooth, and a smooth graph hides a reversal that the algebra would have found in one line.

Past 10-Year AP Exam Appearance

Testing Frequency: Most Years; Five of the nine released exams since 2017 ask for a distance travelled or a displacement in a motion context, and the two quantities usually appear in the same part
10-Year Total Count: 5 of 9 Exams · 6 FRQ Parts; One year asks the pair twice, once as displacement and once as distance, which is why the part count exceeds the exam count
FRQ Years Appeared: 2019 (Q2c), 2021 (Q2d), 2023 (Q2c), 2023 (Q2d), 2024 (Q2d), 2026 (Q5C)

Practice Examples

MCQ 1

Distance With a Reversal

A particle moves along a line with velocity $v(t) = t - 2$ for $0 \le t \le 3$ . The total distance travelled is

MCQ 2

Displacement From a Table

The velocity of a particle is given by the table below. Using a trapezoidal sum, the particle’s displacement on $0 \le t \le 4$ is approximately

$t$	0	2	4
$v(t)$	3	1	$-5$

MCQ 3

Which Integral Answers the Question

A particle has velocity $v(t) = \sin t$ on $0 \le t \le 2\pi$ . Which expression gives the total distance travelled?

FRQ 4

A Full Motion Analysis

A particle moves along the $x$ -axis with velocity $v(t) = t^{2} - 4t + 3$ for $0 \le t \le 4$ , measured in meters per second. At $t = 0$ the particle is at $x = 2$ .

(a) Find all times in $0 < t < 4$ at which the particle is at rest.

(b) Find the displacement of the particle on $0 \le t \le 4$ .

(c) Find the total distance travelled by the particle on $0 \le t \le 4$ .

(d) Find the position of the particle at $t = 4$ and explain why it differs from the total distance.

Cheatsheet

Core Reference

	Action	Failure if skipped
1	read the noun: displacement or distance	the wrong integral gets built and every later line is wasted
2	solve $v(t) = 0$ for interior zeros	the reversal is invisible and the pieces are wrong
3	integrate $v$ on each subinterval	the signed pieces are unavailable
4	add signed pieces for displacement	the answer is an unsigned length
5	add absolute pieces for distance	the answer can come out negative
6	check distance $\ge \lvert$ displacement $\rvert$	a sign error survives to the final line

Sentences That Score

TOPIC REVIEW • NEXT STEP IN YOUR LEARNING LOOP

Ready to check your understanding?

Strengthen this topic with AP-style practice, or continue to next topic.

Start Topic Practice Next Topic: 8.5 Net Change vs. Total Change

Unit 8Displacement vs. Total Distance

Concept 4.100Built for May 2027 AP Exam

Displacement vs. Total Distance

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

On the AP Exam

AP Calculus AB tests this skill through the representations and reasoning described below.

What you’ll see: A velocity function or a velocity table on a stated interval, followed by “find the displacement” and “find the total distance travelled.” Some years phrase it as “how far does the particle travel” without the word distance.
What you’ll do: Integrate $v$ for displacement. Find the zeros of $v$ , split the interval, integrate $v$ on each piece, and add the absolute values for distance. If only a table is given, use a left, right, or midpoint sum on each monotone piece.
Memory hook: Velocity signs, speed doesn’t. Displacement keeps the sign; distance throws it away.
Use it on the exam: For MCQs: If the choices include a negative number for a distance question, it is a distractor built from the signed integral. For FRQs: Show the split points explicitly. “ $v(t) = 0$ at $t = 1$ and $t = 3$ ” is usually worth a point on its own, and every later line depends on it.

Concept Review

Two Integrals, One Function

Figure 1 — The same velocity curve yields a signed net area and an unsigned total area.

Both quantities start from the same graph and differ only in what happens to the sign:

$\text{displacement} = \int_a^b v(t)\,dt \qquad\qquad \text{distance} = \int_a^b |v(t)|\,dt$

Two bar displays comparing a signed displacement of four thirds with a total distance of four

Why the Numbers Diverge

The Splitting Procedure

There is one reliable algorithm, and it never changes:

Solve $v(t) = 0$ on $(a, b)$ and keep only the solutions inside the interval.
Sort them and use them, together with $a$ and $b$ , to cut the interval into subintervals.
On each subinterval, decide the sign of $v$ from a single test point.
Integrate $v$ on each subinterval to get a signed piece.
Displacement is the plain sum of the pieces. Distance is the sum of their absolute values.

Step 3 is where most lost points live. A sign chosen from the wrong test point flips one piece and quietly changes the answer, while the arithmetic on that piece still looks correct.

Reading the Position Function Instead

Given	Displacement	Total distance
velocity $v(t)$	$\int_a^b v(t)\,dt$	$\int_a^b \lvert v(t)\rvert\,dt$
position $s(t)$	$s(b) - s(a)$	sum of $\lvert s(t_{i+1}) - s(t_i)\rvert$ across turning points
speed only	not recoverable	$\int_a^b \text{speed}(t)\,dt$
acceleration $a(t)$ plus $v(a)$	integrate twice, then subtract	split at zeros of the reconstructed $v$

Units and Wording

Displacement and distance share units, because both come from velocity times time. They differ in interpretation, not in dimension. Watch the nouns:

“How far has the particle travelled” means distance.
“What is the particle’s displacement” means the signed integral.
“Change in position” means displacement, and it can be negative.
“Total distance” never takes a sign, and a negative answer is automatically wrong.

Common Misconception Check

Myth 1: “Displacement and distance are the same because both are integrals of velocity.” Truth: Only the distance integral carries an absolute value, and that is exactly what removes the cancellation.
Myth 2: “If the particle ends where it started, the distance is zero.” Truth: The displacement is zero; the distance is the full path length, which is positive.
Myth 3: “You can compute distance by integrating $|v|$ without finding the zeros.” Truth: You can if your calculator handles the absolute value, but on a no-calculator part you must split, and the split points are themselves scored.

Topic-Specific Study Method

Decide Which Integral Before You Reach for the Antiderivative

Exam disguise	What you see on the page	What your brain must translate	Required execution
“total distance travelled”	a velocity and a time window	unsigned accumulation	split at zeros, integrate each piece, add absolute values
“displacement of the particle”	the same velocity and window	signed accumulation	integrate once, keep the sign
“change in position”	a velocity, sometimes a table	displacement in disguise	$s(b) - s(a)$ , or the signed integral
“how far from the starting point”	a velocity that reverses	distance from the origin now	integrate, then take the absolute value of the total
“particle is at rest when”	the zeros of the velocity	the split points themselves	solve $v(t) = 0$ , list only interior solutions

① Circle the noun in the question before writing any integral. “Distance” and “displacement” lead to different scoring lines.

② List the zeros of $v$ as a separate numbered line. That line is scored even when the integral is wrong.

③ Sanity-check the pair at the end: distance must be at least the absolute value of displacement, and they must be equal when there is no interior zero.

Training & Error Diagnosis

Core Archetype (Simplest Form): A particle moves along a line with velocity $v(t) = t^{2} - 4t + 3$ meters per second for $0 \le t \le 4$ . Find its displacement and the total distance travelled.

Standard Execution:

Zeros. $t^{2} - 4t + 3 = (t-1)(t-3) = 0$ gives $t = 1$ and $t = 3$ , both interior.

Pieces. $\int_0^1 v\,dt = \frac{4}{3}$ , $\int_1^3 v\,dt = -\frac{4}{3}$ , $\int_3^4 v\,dt = \frac{4}{3}$ .

Displacement. $\frac{4}{3} - \frac{4}{3} + \frac{4}{3} = \frac{4}{3}$ meters.

Distance. $\frac{4}{3} + \frac{4}{3} + \frac{4}{3} = 4$ meters.

Check. Distance exceeds displacement, as it must whenever the particle reverses.

$t$	0	1	3	6
$v(t)$	2	$-1$	$-3$	4

Solution: The velocity changes sign between $t = 1$ and $t = 3$ , and again between $t = 3$ and $t = 6$ , so the table alone does not pin the zeros. The standard treatment on a table part is to split at the tabulated sign change and use the trapezoidal sum on each block. On $[0,1]$ the trapezoid is $\frac{2 + (-1)}{2} \cdot 1 = 0.5$ . On $[1,3]$ it is $\frac{(-1) + (-3)}{2} \cdot 2 = -4$ . On $[3,6]$ it is $\frac{(-3) + 4}{2} \cdot 3 = 1.5$ . Adding absolute values gives $0.5 + 4 + 1.5 = 6$ meters. The signed sum would give $-2$ , which is the displacement estimate.

Common Traps & Failure Signals:

Absolute value applied at the end (Signal: $\left|\int_a^b v\right|$ written for a distance question): the absolute value belongs inside, on each piece.
Interior zero missed (Signal: one integral where the velocity clearly reverses): solve $v(t) = 0$ first, always.
Distance reported as negative (Signal: a minus sign on a length): distance is a sum of absolute values.

6D Cognitive Error Diagnosis (Where Your Thinking Broke):

Sign Kept: You reported the signed integral as the distance.
- Zero Missed: You integrated the velocity in one stroke across a reversal, so the backward travel cancelled the forward travel and the absolute value was applied only to the final total instead of to each piece.
- Path Shortened: You treated the trip as one-way because the graph looked smooth, and a smooth graph hides a reversal that the algebra would have found in one line.

Past 10-Year AP Exam Appearance

Testing Frequency: Most Years; Five of the nine released exams since 2017 ask for a distance travelled or a displacement in a motion context, and the two quantities usually appear in the same part
10-Year Total Count: 5 of 9 Exams · 6 FRQ Parts; One year asks the pair twice, once as displacement and once as distance, which is why the part count exceeds the exam count
FRQ Years Appeared: 2019 (Q2c), 2021 (Q2d), 2023 (Q2c), 2023 (Q2d), 2024 (Q2d), 2026 (Q5C)

Practice Examples

MCQ 1

Distance With a Reversal

A particle moves along a line with velocity $v(t) = t - 2$ for $0 \le t \le 3$ . The total distance travelled is

MCQ 2

Displacement From a Table

The velocity of a particle is given by the table below. Using a trapezoidal sum, the particle’s displacement on $0 \le t \le 4$ is approximately

$t$	0	2	4
$v(t)$	3	1	$-5$

MCQ 3

Which Integral Answers the Question

A particle has velocity $v(t) = \sin t$ on $0 \le t \le 2\pi$ . Which expression gives the total distance travelled?

FRQ 4

A Full Motion Analysis

A particle moves along the $x$ -axis with velocity $v(t) = t^{2} - 4t + 3$ for $0 \le t \le 4$ , measured in meters per second. At $t = 0$ the particle is at $x = 2$ .

(a) Find all times in $0 < t < 4$ at which the particle is at rest.

(b) Find the displacement of the particle on $0 \le t \le 4$ .

(c) Find the total distance travelled by the particle on $0 \le t \le 4$ .

(d) Find the position of the particle at $t = 4$ and explain why it differs from the total distance.

Cheatsheet

Core Reference

	Action	Failure if skipped
1	read the noun: displacement or distance	the wrong integral gets built and every later line is wasted
2	solve $v(t) = 0$ for interior zeros	the reversal is invisible and the pieces are wrong
3	integrate $v$ on each subinterval	the signed pieces are unavailable
4	add signed pieces for displacement	the answer is an unsigned length
5	add absolute pieces for distance	the answer can come out negative
6	check distance $\ge \lvert$ displacement $\rvert$	a sign error survives to the final line

Sentences That Score

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