Unit 8

v(t) = \int a(t)dt

Concept 4.131Built for May 2027 AP Exam

$v(t) = \int a(t)dt$

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

Acceleration is the derivative of velocity, so velocity is an accumulation of acceleration: $v(t) = v(0) + \int_0^t a(\tau)\,d\tau$ . The integral measures the change in velocity, not the change in speed, and those two differ whenever the particle reverses. Two starting values therefore control the whole chain. The initial velocity anchors $v$ , and the initial position anchors $s$ , and each integration introduces exactly one new constant. On the released exams this direction is rare in free response, because the questions usually hand you $v$ and ask for $s$ . The idea still decides the multiple-choice items and every part that asks whether the particle is speeding up or slowing down, since that answer is a statement about the signs of $v$ and $a$ together.

On the AP Exam

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

What you’ll see: Either a free-response part that supplies $a(t)$ and an initial velocity, or, far more often, a part that asks whether the speed is increasing at a stated instant after giving $v$ explicitly.
What you’ll do: For the accumulation, integrate $a$ and add $v(0)$ . For the speeding question, compare the signs of $v$ and $a$ rather than looking at either alone.
Memory hook: The sign of $a$ is not the answer. Speeding up happens when $v$ and $a$ agree, and slowing down happens when they disagree.
Use it on the exam: For MCQs: A negative acceleration with a negative velocity means the particle is speeding up in the negative direction, which is the item most students miss. For FRQs: Write “ $v$ and $a$ have opposite signs, so the speed is decreasing” as a sentence. The reason is the scored part, not the arithmetic.

Concept Review

One More Integration

Figure 1 — The shaded area under $a(t)$ is added to the starting velocity to produce the velocity graph.

Position, velocity, and acceleration form a chain in which each step down is a derivative and each step up is an integral:

$s \;\xrightarrow{\ \frac{d}{dt}\ }\; v \;\xrightarrow{\ \frac{d}{dt}\ }\; a \qquad\text{and}\qquad a \;\xrightarrow{\ \int\ }\; v \;\xrightarrow{\ \int\ }\; s$

Going up costs one constant per step:

$v(t) = v(0) + \int_0^t a(\tau)\,d\tau, \qquad s(t) = s(0) + \int_0^t v(\tau)\,d\tau$

A question that asks for both needs both constants, and a question that supplies only one of them is telling you which quantity is out of reach.

An acceleration graph whose shaded area is added to an initial velocity to produce a velocity graph

Change in Velocity Versus Change in Speed

Take $v$ going from $-4$ to $+1$ . The velocity changed by $+5$ , while the speed changed by $1 - 4 = -3$ . The particle actually slowed down, even though the velocity increased. Any item that asks for a change in speed therefore requires the sign check first.

Speeding Up and Slowing Down

The rule is a product of signs, and it is the single most tested idea in this family of questions.

Sign of $v$	Sign of $a$	What happens
$+$	$+$	speed increasing, moving right
$+$	$-$	speed decreasing, moving right
$-$	$-$	speed increasing, moving left
$-$	$+$	speed decreasing, moving left
$0$	nonzero	instantaneously at rest, and the direction reverses or hesitates

The last row is why a rest point is not automatically a turning point. A particle can stop for an instant and continue in the same direction, which happens when $v$ touches zero without crossing.

Two Constants, Two Anchors

A question that starts from $a$ and asks for $s$ needs two pieces of history: the initial velocity and the initial position. Each released problem that does this states both, and each one scores the constants separately.

A useful check on the completed chain: differentiate what you wrote and confirm you recover the given function. If $s(t)$ came from integrating $v$ , then $s'(t)$ must equal $v(t)$ for every $t$ , and that check costs one line.

Reading the Direction of Travel From $a$

The sign of $a$ tells you which way the velocity is heading, not which way the particle is heading. A particle moving right with negative acceleration is still moving right; it is merely losing speed. Only a zero of velocity can reverse the direction, and only a change of sign at that zero actually does reverse it.

Common Misconception Check

Myth 1: “Negative acceleration means the particle moves left.” Truth: Acceleration describes how velocity changes, not where the particle goes.
Myth 2: “ $\int a\,dt$ gives the change in speed.” Truth: It gives the change in velocity; the two differ whenever $v$ changes sign.
Myth 3: “The initial velocity can be ignored if the question starts at $t = 0$ .” Truth: Starting at zero makes the limits easy, not the constant unnecessary.

Topic-Specific Study Method

Anchor the Chain, Then Read the Signs

Borrowed idea: Acceleration accumulates into velocity, and velocity accumulates into position. Each of those two integrations introduces its own constant of integration. The chain therefore needs two anchors before it can say anything about location. The integral of acceleration is a signed change in velocity. That signed change is not the same quantity as a change in speed. Speeding up is a claim about the two signs at once, not about the magnitudes. Every released part of this type scores the sign sentence rather than the arithmetic. Write the comparison out on the page instead of assuming it in your head.

Exam disguise	What you see on the page	What your brain must translate	Required execution
“ $a(t) = 2t - 5$ and $v(0) = 6$ ”	an acceleration and an anchor	$v(t) = 6 + \int_0^t a$	integrate, add the anchor
“is the speed increasing at $t = 1$ ?”	a two-sign question	compare $v(1)$ with $a(1)$	state both signs, then conclude
“find the velocity when the particle first stops”	a zero of velocity	solve $v(t) = 0$	solve, then substitute
“ $v$ is negative and $a$ is negative”	matching signs	speed increasing leftward	say “increasing” with the direction
“position from acceleration”	two integrations	two constants needed	ask for both anchors

① Compute $v$ before judging the speed, because the judgment needs a number and its sign, not a formula alone.

② Write the sign comparison in words: “ $v < 0$ and $a > 0$ , so the speed is decreasing.”

③ When two integrations are required, name both constants in the first line so neither gets lost.

Training & Error Diagnosis

Core Archetype (Simplest Form): A particle has acceleration $a(t) = 6t$ and initial velocity $v(0) = -2$ . Find $v(2)$ .

Standard Execution:

Integrate. $\int_0^2 6t\,dt = \left[3t^{2}\right]_0^2 = 12$ .

Add the anchor. $v(2) = -2 + 12 = 10$ .

Check. A positive constant acceleration pushes the velocity upward, and it crosses zero at $t = \sqrt{1/3}$ , so a positive value at $t = 2$ is expected.

Exam Variation (Context / Parameter Shift): A particle has $a(t) = 2t - 5$ and $v(0) = 6$ for $0 \le t \le 4$ . Find the intervals on which the speed is increasing.

Solution: Integrating gives $v(t) = 6 + t^{2} - 5t = (t - 2)(t - 3)$ , so the particle rests at $t = 2$ and $t = 3$ . The acceleration $a(t) = 2t - 5$ is negative before $t = 2.5$ and positive after. On $(0, 2)$ the velocity is positive and the acceleration is negative, so the speed decreases. On $(2, 2.5)$ both are negative, so the speed increases. On $(2.5, 3)$ the velocity is negative and the acceleration is positive, so the speed decreases again. On $(3, 4]$ both are positive, so the speed increases. The answer is $(2, 2.5)$ and $(3, 4]$ .

Common Traps & Failure Signals:

Anchor dropped (Signal: $v$ proportional to the integral of $a$ with no constant): the initial velocity must appear.
Speed and velocity merged (Signal: “the speed increased by $5$ ” for $v$ going from $-4$ to $1$ ): the speed change is $-3$ .
Sign compared with itself (Signal: “ $a < 0$ , so the speed decreases”): the comparison needs both signs.

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

Anchor Dropped: You integrated the acceleration without the initial velocity.
- One Sign Only: You decided the speed was decreasing because the acceleration was negative, which is a conclusion about velocity rather than about speed and reverses the answer half the time.
- Magnitude Mixed In: You treated the change in velocity as if it were the change in speed, which silently assumes the particle never reversed direction during the interval.

Past 10-Year AP Exam Appearance

Testing Frequency: MCQ Only; No released free-response part since 2017 asks a student to build $v$ from $a$; the free responses that mention acceleration ask for it by differentiation instead, and the accumulation direction lives in the multiple-choice section
10-Year Total Count: 0 of 9 Exams · 0 Parts; Seven parts do mention acceleration, and every one of them differentiates $v$ to get it or hands it over as given
FRQ Years Appeared: None in Released FRQs

Practice Examples

MCQ 1

Integrating an Acceleration

A particle has $a(t) = 6t$ and $v(0) = -2$ . Its velocity at $t = 2$ is

MCQ 2

A Full Interval

A particle has $a(t) = 2 - t$ and $v(0) = 3$ . Its velocity at $t = 4$ is

MCQ 3

Judging Without Computing

A particle has $a(t) < 0$ for all $t$ in $[0, 4]$ and $v(0) = 5$ . Which statement must be true?

FRQ 4

From Acceleration to Speed

A particle moves along the $x$ -axis with acceleration $a(t) = 2t - 5$ for $0 \le t \le 4$ , in meters per second squared. At time $t = 0$ its velocity is $6$ meters per second.

(a) Find $v(t)$ and the value of $v(4)$ .

(b) Find every time in $(0, 4)$ at which the particle is at rest, and determine the intervals on which the speed is increasing.

(c) A student claims that the particle must be slowing down at $t = 1$ because the acceleration is negative there. Explain why the claim is false.

Cheatsheet

Core Reference

	Action	Failure if skipped
1	write $v(t) = v(0) + \int a$	the constant is missing and $v$ is wrong everywhere
2	evaluate the velocity at the stated time	the sign comparison has no input
3	compute $a$ at the same time	the second sign is missing
4	compare the two signs in words	the justification point is lost
5	for position, integrate again and add $s(0)$	only one of two anchors is used
6	differentiate the result as a check	an algebraic slip survives to the answer

Sentences That Score

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Unit 8

v(t) = \int a(t)dt

Concept 4.131Built for May 2027 AP Exam

$v(t) = \int a(t)dt$

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: Either a free-response part that supplies $a(t)$ and an initial velocity, or, far more often, a part that asks whether the speed is increasing at a stated instant after giving $v$ explicitly.
What you’ll do: For the accumulation, integrate $a$ and add $v(0)$ . For the speeding question, compare the signs of $v$ and $a$ rather than looking at either alone.
Memory hook: The sign of $a$ is not the answer. Speeding up happens when $v$ and $a$ agree, and slowing down happens when they disagree.
Use it on the exam: For MCQs: A negative acceleration with a negative velocity means the particle is speeding up in the negative direction, which is the item most students miss. For FRQs: Write “ $v$ and $a$ have opposite signs, so the speed is decreasing” as a sentence. The reason is the scored part, not the arithmetic.

Concept Review

One More Integration

Figure 1 — The shaded area under $a(t)$ is added to the starting velocity to produce the velocity graph.

Position, velocity, and acceleration form a chain in which each step down is a derivative and each step up is an integral:

$s \;\xrightarrow{\ \frac{d}{dt}\ }\; v \;\xrightarrow{\ \frac{d}{dt}\ }\; a \qquad\text{and}\qquad a \;\xrightarrow{\ \int\ }\; v \;\xrightarrow{\ \int\ }\; s$

Going up costs one constant per step:

$v(t) = v(0) + \int_0^t a(\tau)\,d\tau, \qquad s(t) = s(0) + \int_0^t v(\tau)\,d\tau$

A question that asks for both needs both constants, and a question that supplies only one of them is telling you which quantity is out of reach.

An acceleration graph whose shaded area is added to an initial velocity to produce a velocity graph

Change in Velocity Versus Change in Speed

Speeding Up and Slowing Down

The rule is a product of signs, and it is the single most tested idea in this family of questions.

Sign of $v$	Sign of $a$	What happens
$+$	$+$	speed increasing, moving right
$+$	$-$	speed decreasing, moving right
$-$	$-$	speed increasing, moving left
$-$	$+$	speed decreasing, moving left
$0$	nonzero	instantaneously at rest, and the direction reverses or hesitates

The last row is why a rest point is not automatically a turning point. A particle can stop for an instant and continue in the same direction, which happens when $v$ touches zero without crossing.

Two Constants, Two Anchors

Reading the Direction of Travel From $a$

Common Misconception Check

Myth 1: “Negative acceleration means the particle moves left.” Truth: Acceleration describes how velocity changes, not where the particle goes.
Myth 2: “ $\int a\,dt$ gives the change in speed.” Truth: It gives the change in velocity; the two differ whenever $v$ changes sign.
Myth 3: “The initial velocity can be ignored if the question starts at $t = 0$ .” Truth: Starting at zero makes the limits easy, not the constant unnecessary.

Topic-Specific Study Method

Anchor the Chain, Then Read the Signs

Exam disguise	What you see on the page	What your brain must translate	Required execution
“ $a(t) = 2t - 5$ and $v(0) = 6$ ”	an acceleration and an anchor	$v(t) = 6 + \int_0^t a$	integrate, add the anchor
“is the speed increasing at $t = 1$ ?”	a two-sign question	compare $v(1)$ with $a(1)$	state both signs, then conclude
“find the velocity when the particle first stops”	a zero of velocity	solve $v(t) = 0$	solve, then substitute
“ $v$ is negative and $a$ is negative”	matching signs	speed increasing leftward	say “increasing” with the direction
“position from acceleration”	two integrations	two constants needed	ask for both anchors

① Compute $v$ before judging the speed, because the judgment needs a number and its sign, not a formula alone.

② Write the sign comparison in words: “ $v < 0$ and $a > 0$ , so the speed is decreasing.”

③ When two integrations are required, name both constants in the first line so neither gets lost.

Training & Error Diagnosis

Core Archetype (Simplest Form): A particle has acceleration $a(t) = 6t$ and initial velocity $v(0) = -2$ . Find $v(2)$ .

Standard Execution:

Integrate. $\int_0^2 6t\,dt = \left[3t^{2}\right]_0^2 = 12$ .

Add the anchor. $v(2) = -2 + 12 = 10$ .

Check. A positive constant acceleration pushes the velocity upward, and it crosses zero at $t = \sqrt{1/3}$ , so a positive value at $t = 2$ is expected.

Exam Variation (Context / Parameter Shift): A particle has $a(t) = 2t - 5$ and $v(0) = 6$ for $0 \le t \le 4$ . Find the intervals on which the speed is increasing.

Solution: Integrating gives $v(t) = 6 + t^{2} - 5t = (t - 2)(t - 3)$ , so the particle rests at $t = 2$ and $t = 3$ . The acceleration $a(t) = 2t - 5$ is negative before $t = 2.5$ and positive after. On $(0, 2)$ the velocity is positive and the acceleration is negative, so the speed decreases. On $(2, 2.5)$ both are negative, so the speed increases. On $(2.5, 3)$ the velocity is negative and the acceleration is positive, so the speed decreases again. On $(3, 4]$ both are positive, so the speed increases. The answer is $(2, 2.5)$ and $(3, 4]$ .

Common Traps & Failure Signals:

Anchor dropped (Signal: $v$ proportional to the integral of $a$ with no constant): the initial velocity must appear.
Speed and velocity merged (Signal: “the speed increased by $5$ ” for $v$ going from $-4$ to $1$ ): the speed change is $-3$ .
Sign compared with itself (Signal: “ $a < 0$ , so the speed decreases”): the comparison needs both signs.

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

Anchor Dropped: You integrated the acceleration without the initial velocity.
- One Sign Only: You decided the speed was decreasing because the acceleration was negative, which is a conclusion about velocity rather than about speed and reverses the answer half the time.
- Magnitude Mixed In: You treated the change in velocity as if it were the change in speed, which silently assumes the particle never reversed direction during the interval.

Past 10-Year AP Exam Appearance

Testing Frequency: MCQ Only; No released free-response part since 2017 asks a student to build $v$ from $a$; the free responses that mention acceleration ask for it by differentiation instead, and the accumulation direction lives in the multiple-choice section
10-Year Total Count: 0 of 9 Exams · 0 Parts; Seven parts do mention acceleration, and every one of them differentiates $v$ to get it or hands it over as given
FRQ Years Appeared: None in Released FRQs

Practice Examples

MCQ 1

Integrating an Acceleration

A particle has $a(t) = 6t$ and $v(0) = -2$ . Its velocity at $t = 2$ is

MCQ 2

A Full Interval

A particle has $a(t) = 2 - t$ and $v(0) = 3$ . Its velocity at $t = 4$ is

MCQ 3

Judging Without Computing

A particle has $a(t) < 0$ for all $t$ in $[0, 4]$ and $v(0) = 5$ . Which statement must be true?

FRQ 4

From Acceleration to Speed

A particle moves along the $x$ -axis with acceleration $a(t) = 2t - 5$ for $0 \le t \le 4$ , in meters per second squared. At time $t = 0$ its velocity is $6$ meters per second.

(a) Find $v(t)$ and the value of $v(4)$ .

(b) Find every time in $(0, 4)$ at which the particle is at rest, and determine the intervals on which the speed is increasing.

(c) A student claims that the particle must be slowing down at $t = 1$ because the acceleration is negative there. Explain why the claim is false.

Cheatsheet

Core Reference

	Action	Failure if skipped
1	write $v(t) = v(0) + \int a$	the constant is missing and $v$ is wrong everywhere
2	evaluate the velocity at the stated time	the sign comparison has no input
3	compute $a$ at the same time	the second sign is missing
4	compare the two signs in words	the justification point is lost
5	for position, integrate again and add $s(0)$	only one of two anchors is used
6	differentiate the result as a check	an algebraic slip survives to the answer

Sentences That Score

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Start Topic Practice Next Topic: 8.4 Displacement vs. Total Distance

Unit 8

v(t) = \int a(t)dt

Concept 4.131Built for May 2027 AP Exam

$v(t) = \int a(t)dt$

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: Either a free-response part that supplies $a(t)$ and an initial velocity, or, far more often, a part that asks whether the speed is increasing at a stated instant after giving $v$ explicitly.
What you’ll do: For the accumulation, integrate $a$ and add $v(0)$ . For the speeding question, compare the signs of $v$ and $a$ rather than looking at either alone.
Memory hook: The sign of $a$ is not the answer. Speeding up happens when $v$ and $a$ agree, and slowing down happens when they disagree.
Use it on the exam: For MCQs: A negative acceleration with a negative velocity means the particle is speeding up in the negative direction, which is the item most students miss. For FRQs: Write “ $v$ and $a$ have opposite signs, so the speed is decreasing” as a sentence. The reason is the scored part, not the arithmetic.

Concept Review

One More Integration

Figure 1 — The shaded area under $a(t)$ is added to the starting velocity to produce the velocity graph.

Position, velocity, and acceleration form a chain in which each step down is a derivative and each step up is an integral:

$s \;\xrightarrow{\ \frac{d}{dt}\ }\; v \;\xrightarrow{\ \frac{d}{dt}\ }\; a \qquad\text{and}\qquad a \;\xrightarrow{\ \int\ }\; v \;\xrightarrow{\ \int\ }\; s$

Going up costs one constant per step:

$v(t) = v(0) + \int_0^t a(\tau)\,d\tau, \qquad s(t) = s(0) + \int_0^t v(\tau)\,d\tau$

A question that asks for both needs both constants, and a question that supplies only one of them is telling you which quantity is out of reach.

An acceleration graph whose shaded area is added to an initial velocity to produce a velocity graph

Change in Velocity Versus Change in Speed

Speeding Up and Slowing Down

The rule is a product of signs, and it is the single most tested idea in this family of questions.

Sign of $v$	Sign of $a$	What happens
$+$	$+$	speed increasing, moving right
$+$	$-$	speed decreasing, moving right
$-$	$-$	speed increasing, moving left
$-$	$+$	speed decreasing, moving left
$0$	nonzero	instantaneously at rest, and the direction reverses or hesitates

The last row is why a rest point is not automatically a turning point. A particle can stop for an instant and continue in the same direction, which happens when $v$ touches zero without crossing.

Two Constants, Two Anchors

Reading the Direction of Travel From $a$

Common Misconception Check

Myth 1: “Negative acceleration means the particle moves left.” Truth: Acceleration describes how velocity changes, not where the particle goes.
Myth 2: “ $\int a\,dt$ gives the change in speed.” Truth: It gives the change in velocity; the two differ whenever $v$ changes sign.
Myth 3: “The initial velocity can be ignored if the question starts at $t = 0$ .” Truth: Starting at zero makes the limits easy, not the constant unnecessary.

Topic-Specific Study Method

Anchor the Chain, Then Read the Signs

Exam disguise	What you see on the page	What your brain must translate	Required execution
“ $a(t) = 2t - 5$ and $v(0) = 6$ ”	an acceleration and an anchor	$v(t) = 6 + \int_0^t a$	integrate, add the anchor
“is the speed increasing at $t = 1$ ?”	a two-sign question	compare $v(1)$ with $a(1)$	state both signs, then conclude
“find the velocity when the particle first stops”	a zero of velocity	solve $v(t) = 0$	solve, then substitute
“ $v$ is negative and $a$ is negative”	matching signs	speed increasing leftward	say “increasing” with the direction
“position from acceleration”	two integrations	two constants needed	ask for both anchors

① Compute $v$ before judging the speed, because the judgment needs a number and its sign, not a formula alone.

② Write the sign comparison in words: “ $v < 0$ and $a > 0$ , so the speed is decreasing.”

③ When two integrations are required, name both constants in the first line so neither gets lost.

Training & Error Diagnosis

Core Archetype (Simplest Form): A particle has acceleration $a(t) = 6t$ and initial velocity $v(0) = -2$ . Find $v(2)$ .

Standard Execution:

Integrate. $\int_0^2 6t\,dt = \left[3t^{2}\right]_0^2 = 12$ .

Add the anchor. $v(2) = -2 + 12 = 10$ .

Check. A positive constant acceleration pushes the velocity upward, and it crosses zero at $t = \sqrt{1/3}$ , so a positive value at $t = 2$ is expected.

Exam Variation (Context / Parameter Shift): A particle has $a(t) = 2t - 5$ and $v(0) = 6$ for $0 \le t \le 4$ . Find the intervals on which the speed is increasing.

Solution: Integrating gives $v(t) = 6 + t^{2} - 5t = (t - 2)(t - 3)$ , so the particle rests at $t = 2$ and $t = 3$ . The acceleration $a(t) = 2t - 5$ is negative before $t = 2.5$ and positive after. On $(0, 2)$ the velocity is positive and the acceleration is negative, so the speed decreases. On $(2, 2.5)$ both are negative, so the speed increases. On $(2.5, 3)$ the velocity is negative and the acceleration is positive, so the speed decreases again. On $(3, 4]$ both are positive, so the speed increases. The answer is $(2, 2.5)$ and $(3, 4]$ .

Common Traps & Failure Signals:

Anchor dropped (Signal: $v$ proportional to the integral of $a$ with no constant): the initial velocity must appear.
Speed and velocity merged (Signal: “the speed increased by $5$ ” for $v$ going from $-4$ to $1$ ): the speed change is $-3$ .
Sign compared with itself (Signal: “ $a < 0$ , so the speed decreases”): the comparison needs both signs.

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

Anchor Dropped: You integrated the acceleration without the initial velocity.
- One Sign Only: You decided the speed was decreasing because the acceleration was negative, which is a conclusion about velocity rather than about speed and reverses the answer half the time.
- Magnitude Mixed In: You treated the change in velocity as if it were the change in speed, which silently assumes the particle never reversed direction during the interval.

Past 10-Year AP Exam Appearance

Testing Frequency: MCQ Only; No released free-response part since 2017 asks a student to build $v$ from $a$; the free responses that mention acceleration ask for it by differentiation instead, and the accumulation direction lives in the multiple-choice section
10-Year Total Count: 0 of 9 Exams · 0 Parts; Seven parts do mention acceleration, and every one of them differentiates $v$ to get it or hands it over as given
FRQ Years Appeared: None in Released FRQs

Practice Examples

MCQ 1

Integrating an Acceleration

A particle has $a(t) = 6t$ and $v(0) = -2$ . Its velocity at $t = 2$ is

MCQ 2

A Full Interval

A particle has $a(t) = 2 - t$ and $v(0) = 3$ . Its velocity at $t = 4$ is

MCQ 3

Judging Without Computing

A particle has $a(t) < 0$ for all $t$ in $[0, 4]$ and $v(0) = 5$ . Which statement must be true?

FRQ 4

From Acceleration to Speed

A particle moves along the $x$ -axis with acceleration $a(t) = 2t - 5$ for $0 \le t \le 4$ , in meters per second squared. At time $t = 0$ its velocity is $6$ meters per second.

(a) Find $v(t)$ and the value of $v(4)$ .

(b) Find every time in $(0, 4)$ at which the particle is at rest, and determine the intervals on which the speed is increasing.

(c) A student claims that the particle must be slowing down at $t = 1$ because the acceleration is negative there. Explain why the claim is false.

Cheatsheet

Core Reference

	Action	Failure if skipped
1	write $v(t) = v(0) + \int a$	the constant is missing and $v$ is wrong everywhere
2	evaluate the velocity at the stated time	the sign comparison has no input
3	compute $a$ at the same time	the second sign is missing
4	compare the two signs in words	the justification point is lost
5	for position, integrate again and add $s(0)$	only one of two anchors is used
6	differentiate the result as a check	an algebraic slip survives to the answer

Sentences That Score

TOPIC REVIEW • NEXT STEP IN YOUR LEARNING LOOP

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Start Topic Practice Next Topic: 8.4 Displacement vs. Total Distance