Velocity vs Acceleration: What Is the Difference?

Velocity is the rate at which position changes, in m/s. Acceleration is the rate at which velocity changes, in m/s squared. They are two levels apart, so knowing one tells you nothing about the other: an object can have zero velocity and large acceleration, or huge velocity and none at all.

AP Physics: Unit 1 (topics 1.2 Displacement, Velocity, and Acceleration, 1.3 Representing Motion). AP Physics 1 Topic 1.2 defines both quantities under LO 1.2.B: EK 1.2.B.2 gives average velocity as displacement divided by the interval, EK 1.2.B.3 gives average acceleration as the change in velocity divided by the interval, EK 1.2.B.4 states that an object is accelerating if the magnitude and/or direction of its velocity are changing, and EK 1.2.B.5 relates a small-interval average to the instantaneous value. Topic 1.3 supplies the graph rules: EK 1.3.A.4.i makes instantaneous velocity the slope of a tangent to a position-time graph, EK 1.3.A.4.ii makes instantaneous acceleration the slope of a tangent to a velocity-time graph, EK 1.3.A.4.iii makes displacement the area under a velocity-time graph, and EK 1.3.A.4.iv makes the change in velocity the area under an acceleration-time graph. EK 1.3.A.2 lists the three constant-acceleration equations and EK 1.3.A.3 gives the near-Earth vertical acceleration as downward, constant and approximately 10 m/s squared. The Topic 1.3 boundary statement excludes quantitative analyzis of nonuniform acceleration while requiring qualitative analyzis, graph sketching and discussion of it, and a second boundary statement covers the value of g. The word deceleration appears in none of the four Course and Exam Descriptions. Unit 1 carries 10 to 15 percent of the AP Physics 1 multiple-choice section.

One is a rate of change of the other

They sit one step apart on the same ladder, and every difference between them comes from that.

Velocity is the rate of change of position. EK 1.2.B.2 gives the average version, $\vec{v}_{avg} = \frac{\Delta \vec{x}}{\Delta t}$ , and EK 1.3.A.4.i gives the instantaneous version as the slope of a line tangent to a point on a graph of position as a function of time. Units of m/s.

Acceleration is the rate of change of velocity. EK 1.2.B.3 gives $\vec{a}_{avg} = \frac{\Delta \vec{v}}{\Delta t}$ , and EK 1.3.A.4.ii gives the instantaneous version as the slope of a line tangent to a point on a graph of velocity as a function of time. Units of m/s squared.

Because acceleration is built out of a change in velocity, the current value of the velocity is not an input to it. That is the whole source of the trouble. An object with a velocity of zero can have any acceleration you like, and an object screaming along at 300 m/s can have none. Nothing about the size of one constrains the size of the other, and nothing about the sign of one constrains the sign of the other.

EK 1.2.B.4 is the definition worth memorizing, because it is broader than it first looks: an object is accelerating if the magnitude and/or direction of the object's velocity are changing. The and/or is doing real work. Change the direction while holding the speed fixed and the object is still accelerating.

The glossary defines each on its own at velocity and acceleration. This page is about the two places the pair actually costs marks: the signs, and the cases where one is zero and the other is not.

Velocity vs acceleration, side by side

	Velocity	Acceleration
What changes	Position	Velocity
Symbol	$\vec{v}$	$\vec{a}$
SI unit	m/s	m/s $^2$
Average form (CED)	$\vec{v}_{avg} = \frac{\Delta \vec{x}}{\Delta t}$ , EK 1.2.B.2	$\vec{a}_{avg} = \frac{\Delta \vec{v}}{\Delta t}$ , EK 1.2.B.3
Read off which graph, as a slope	Position against time	Velocity against time
Read off which graph, as an area	Velocity against time gives displacement	Acceleration against time gives change in velocity
Scalar partner	Speed, which is $\lvert \vec{v} \rvert$	None with its own name. The course uses the magnitude of the acceleration
Can be zero while the other is large	Yes, at the top of a throw	Yes, at constant velocity
What a minus sign means	Motion in the negative direction	Change of velocity toward the negative direction
What causes it	Nothing. It is a state	A nonzero net force, per EK 2.5.A.2
Does a constant value imply the other is zero	No. Constant velocity does imply zero acceleration	No. Constant acceleration does not imply constant velocity
Changes when direction changes at fixed speed	Yes	Yes, and that is EK 1.2.B.4

The last three rows are where the exam lives. A constant velocity forces the acceleration to zero, and that is a genuine implication. The reverse implication does not exist, and reading the table backwards is how students conclude that a steady acceleration means a steady velocity.

The four sign combinations, and the rule that actually works

A reliable way to lose a mark in kinematics is to read a minus sign in front of an acceleration as "slowing down". It does not mean that, and here is the complete case list. Pick a positive direction first, because none of these rows means anything until you have.

Velocity	Acceleration	Speed is	Why	An example, taking east as positive
Positive	Positive	Increasing	Same signs, so the change adds to what is there	A car heading east and pressing the accelerator
Positive	Negative	Decreasing	Opposite signs, so the change eats into it	A car heading east and braking
Negative	Negative	Increasing	Same signs, so the magnitude grows	A car heading west and pressing the accelerator
Negative	Positive	Decreasing	Opposite signs, so the magnitude shrinks	A car heading west and braking

Rows one and three both speed up. Rows two and four both slow down. The sign of the acceleration on its own appears in one speeding-up row and one slowing-down row, so it predicts nothing.

The rule: compare the two signs. Same sign, speeding up. Opposite signs, slowing down. In one dimension that is exact and complete.

Two consequences worth keeping.

A falling object has $v$ and $a$ both negative if you called up positive, so it is speeding up, and a negative acceleration accompanied it the whole way. A ball thrown upward has positive $v$ and negative $a$ on the way up, and the acceleration never changed. One value of $a$ , two behaviors, and the velocity is what switched.

And the labels are not physical. Choose down as positive for the same throw and every sign flips: the rise is negative velocity with positive acceleration, still slowing, and the fall is positive with positive, still speeding up. The signs are a bookkeeping choice; the same-or-opposite test is what survives the choice. The first worked example runs the same throw both ways to show it.

One piece of language to avoid entirely. Deceleration is not an AP term, it appears in none of the four Course and Exam Descriptions, and it hides exactly the information this table is about. Write that the object is slowing down, or that the acceleration is opposite to the velocity.

Zero one, nonzero the other: four cases that all appear on exams

Velocity	Acceleration	What it looks like	Where you meet it
Zero	Zero	At rest and staying there	A book on a table. Newton's first law, EK 2.4.A.3
Zero	Nonzero	Momentarily stopped, about to move	The top of a throw. A mass at maximum displacement in simple harmonic motion. A ball at the instant it reverses off a wall
Nonzero	Zero	Constant velocity, straight line, steady speed	A puck sliding on frictionless ice. Terminal velocity
Nonzero	Nonzero, perpendicular to $\vec{v}$	Speed unchanged, direction turning	Uniform circular motion. The horizontal motion of a projectile is the opposite case, with acceleration parallel to nothing horizontal at all

Row two is the one that gets asked. At the highest point of a ball's flight the velocity is exactly zero for an instant, and students conclude that the acceleration is zero too, or that gravity has switched off. Neither is true. EK 1.3.A.3 says the vertical acceleration caused by the force of gravity near Earth's surface is downward and constant, and constant means it did not pause at the top. If the acceleration really were zero there, the ball would stay there, because a velocity of zero with an acceleration of zero is row one.

Row four is the reason EK 1.2.B.4 says magnitude and/or direction. A car going round a bend at a fixed speedometer reading has a velocity that is changing every instant, because its direction is changing, so it is accelerating. Its speed is constant and its velocity is not, and the two words are not interchangeable, which is the subject of speed vs velocity.

Row three carries the first law: EK 2.4.A.3 says that if the net force exerted on a system is zero, the velocity of that system will remain constant. Note that constant velocity is the condition, not zero velocity, so a puck coasting at 6 m/s and a puck sitting still are the same case.

Reading both off a graph, which is where AP Physics 1 tests it

The CED devotes four consecutive essential knowledge statements to this in Topic 1.3, and between them they give you a complete grid.

Graph	The slope at a point gives	The area under it gives
Position against time	Instantaneous velocity (EK 1.3.A.4.i)	Nothing the course uses
Velocity against time	Instantaneous acceleration (EK 1.3.A.4.ii)	Displacement (EK 1.3.A.4.iii)
Acceleration against time	Nothing the course asks for	Change in velocity (EK 1.3.A.4.iv)

Three habits follow from that grid, and each one is worth a mark somewhere.

Never read a value where you should read a slope. A position-time graph high above the axis says the object is far away, not that it is fast. Fast is steep. A common multiple-choice distractor puts a high, flat position graph next to a low, steep one and asks which object is moving faster.

On a velocity-time graph, both quantities are visible at once, and that is why the exam prefers it. The height of the line is the velocity, the steepness is the acceleration, and the sign test from the section above becomes a picture: the graph is above the axis and rising, or below the axis and falling, when the object is speeding up. Above the axis and falling, or below the axis and rising, means slowing down. A line crossing the axis is the moment of reversal.

Area below the axis is negative displacement. EK 1.3.A.4.iii defines it as the area bounded by the function and the horizontal axis over the appropriate interval, so the parts below the axis subtract. That is the difference between displacement and distance, worked in distance vs displacement and demonstrated numerically in the second example below.

One scope limit. The Topic 1.3 boundary statement says AP Physics 1 does not expect students to quantitatively analyze nonuniform acceleration, but does expect them to be able to qualitatively analyze it, sketch appropriate graphs of it, and discuss situations in which acceleration is nonuniform. So a curved velocity-time graph is fair game to describe and to sketch, and not to integrate.

Where the confusion costs a mark

Reading a negative acceleration as slowing down. It means slowing down only if the velocity is positive. Compare the signs.
Saying zero velocity means zero acceleration. At the top of a throw the velocity is zero and the acceleration is the full 9.8 m/s squared downward. The two are unrelated numbers.
Saying the acceleration must point the way the object is moving. It points the way the net force points, per EK 2.5.A.2, and that direction is set by the forces rather than by the motion. A braking car accelerates backward while travelling forward.
Using the word deceleration. It is not an AP term and it conceals the sign, which is usually the thing being tested.
Reading velocity off the height of a position-time graph. Velocity is the slope, per EK 1.3.A.4.i.
Reading acceleration off the height of a velocity-time graph. Acceleration is the slope, per EK 1.3.A.4.ii.
Assuming constant speed means no acceleration. Only if the direction is also constant. EK 1.2.B.4 counts a change of direction as acceleration, which is the entire basis of circular motion.
Applying the kinematic equations where the acceleration is not constant. EK 1.3.A.2 offers the three equations for constant acceleration specifically. A curved velocity-time graph disqualifies them.
Changing the positive direction halfway through. Every sign in every equation depends on the choice, so declare it once at the top and keep it, even when a later part of the question makes a different choice look natural.
Mixing up the average and instantaneous versions when the acceleration is not constant. EK 1.2.B.5 says only that calculating an average over a very small time interval yields a value very close to the instantaneous one, which is an approximation and not an identity.

What the CED and the equation sheet give you

Topic 1.2, Displacement, Velocity, and Acceleration, is where both definitions live, under LO 1.2.B, describe the average velocity and acceleration of an object. Its five essential knowledge statements are the ones quoted throughout this page: EK 1.2.B.1 says averages are calculated considering the initial and final states of an object over an interval of time, 1.2.B.2 and 1.2.B.3 give the two definitions, 1.2.B.4 sets the condition for accelerating, and 1.2.B.5 relates a small-interval average to the instantaneous value. Topic 1.3, Representing Motion, adds the graph rules and the three kinematic equations.

The AP Physics 1 equation sheet prints the three constant-acceleration equations, which connect the two quantities directly:

v_x = v_{x0} + a_x t

x = x_0 + v_{x0}t + \tfrac{1}{2}a_x t^2

v_x^2 = v_{x0}^2 + 2a_x(x - x_0)

Read the first one as the sentence the sign table is made of: the velocity you end with is the velocity you started with plus the acceleration times the time. If $a_x$ carries the same sign as $v_{x0}$ that sum grows in magnitude, and if it carries the opposite sign it shrinks, passes through zero and reverses. That is the whole of the four-row table in one line of algebra. The procedure for using all three is in kinematic equations, and the kinematics calculator checks the arithmetic.

The defining ratios themselves, $\vec{v}_{avg} = \Delta \vec{x}/\Delta t$ and $\vec{a}_{avg} = \Delta \vec{v}/\Delta t$ , are listed in the CED as relevant equations under EK 1.2.B.2 and EK 1.2.B.3, and they are definitions rather than printed lines. The full sheet is here.

One value to watch. EK 1.3.A.3 states that near the surface of Earth the vertical acceleration caused by the force of gravity is downward, constant, and has a measured value approximately equal to $a_g = g \approx 10\ \text{m/s}^2$ , and the Topic 1.3 boundary statement says that value will be used wherever a numerical quantity is required for $g$ , while adding that students will not be penalized for correctly using the more precise commonly accepted values of $g = 9.81\ \text{m/s}^2$ or $g = 9.8\ \text{m/s}^2$ . The Table of Information prints 9.8. Every number on this page uses 9.8; is g 9.8 or 10 works out what the choice does to your answers.

Unit 1 carries 10 to 15 percent of the AP Physics 1 multiple-choice section. Topic 1.2 Displacement, Velocity, and Acceleration and Topic 1.3 Representing Motion carry the full framing.

One throw, both sign conventions: the labels flip and the physics does not

A ball is thrown straight up at 14.7 m/s and caught at the height it was released from. Ignore air resistance and use $g = 9.8\ \text{m/s}^2$ . Find the velocity and the acceleration at $t = 0.5\ \text{s}$ , at the highest point, and at $t = 2.0\ \text{s}$ , first with up as positive and then with down as positive, and state in each case whether the ball is speeding up or slowing down.

Convention A: up is positive. Then $v_0 = +14.7\ \text{m/s}$ and $a = -9.8\ \text{m/s}^2$ , constant for the whole flight per EK 1.3.A.3.
At $t = 0.5\ \text{s}$ : $v = v_0 + at = 14.7 + (-9.8)(0.5) = 14.7 - 4.9 = +9.8\ \text{m/s}$ . The signs are opposite, so the ball is slowing down, and the speed has fallen from 14.7 to 9.8 m/s.
Highest point: set $v = 0$ in $v = v_0 + at$ , giving $t = 14.7/9.8 = 1.5\ \text{s}$ . There $v = 0$ and $a = -9.8\ \text{m/s}^2$ still. The acceleration did not pause, which is row two of the zero table.
Maximum height, as a check: $\Delta y = v_0^2/(2g) = (14.7)^2/(2 \times 9.8) = 216.09/19.6 = 11.025\ \text{m}$ .
At $t = 2.0\ \text{s}$ : $v = 14.7 - 19.6 = -4.9\ \text{m/s}$ . Now both signs are negative, so the ball is speeding up, with a speed of 4.9 m/s. The acceleration is the same $-9.8\ \text{m/s}^2$ it has been throughout.
Convention B: down is positive. Nothing physical changes, so now $v_0 = -14.7\ \text{m/s}$ and $a = +9.8\ \text{m/s}^2$ .
At $t = 0.5\ \text{s}$ : $v = -14.7 + 4.9 = -9.8\ \text{m/s}$ . Opposite signs again, so still slowing down, and the speed is still 9.8 m/s.
At $t = 2.0\ \text{s}$ : $v = -14.7 + 19.6 = +4.9\ \text{m/s}$ . Both signs positive, so still speeding up, and the speed is still 4.9 m/s.
Compare the two conventions. Every sign reversed and every speed and every verdict stayed identical. In convention A the ball speeds up while the acceleration is negative; in convention B it speeds up while the acceleration is positive. The sign of $a$ alone predicted nothing in either case, and the same-or-opposite comparison predicted everything in both.

Up positive: $+9.8\ \text{m/s}$ slowing, then $0$ with $a = -9.8\ \text{m/s}^2$ , then $-4.9\ \text{m/s}$ speeding up. Down positive: $-9.8\ \text{m/s}$ slowing, then $0$ with $a = +9.8\ \text{m/s}^2$ , then $+4.9\ \text{m/s}$ speeding up. Maximum height 11.025 m either way. The acceleration was one unchanging number for the entire flight, and the ball slowed, stopped and sped up under it, which is the clearest possible demonstration that the acceleration does not tell you what the speed is doing until you look at the velocity as well.

A velocity-time graph where the acceleration is positive throughout and the object still slows down first

A cart moves along a straight track. Its velocity is $-8.0\ \text{m/s}$ at $t = 0$ and rises linearly to $+4.0\ \text{m/s}$ at $t = 4.0\ \text{s}$ . Find the acceleration, the instant the cart reverses, the velocity and speed at $t = 1.0\ \text{s}$ , the displacement over the four seconds, and the distance travelled. Say when the cart is speeding up and when it is slowing down.

Declare the convention: positive is the direction the cart eventually ends up moving, and the graph of velocity against time is a straight line, so the acceleration is constant.
Acceleration from the slope, per EK 1.3.A.4.ii: $a = \Delta v/\Delta t = (4.0 - (-8.0))/(4.0 - 0) = 12.0/4.0 = +3.0\ \text{m/s}^2$ , positive for the entire interval.
Reversal instant, where the line crosses the axis: $0 = -8.0 + 3.0t$ , so $t = 8.0/3.0 = 2.67\ \text{s}$ .
At $t = 1.0\ \text{s}$ : $v = -8.0 + (3.0)(1.0) = -5.0\ \text{m/s}$ , so the speed is 5.0 m/s, down from 8.0 m/s. Velocity negative and acceleration positive means opposite signs, so the cart is slowing down even though the acceleration is positive.
Verdicts across the interval. From $t = 0$ to $t = 2.67\ \text{s}$ the velocity is negative and the acceleration positive, so the cart slows from 8.0 m/s to a stop. From $t = 2.67\ \text{s}$ to $t = 4.0\ \text{s}$ both are positive, so it speeds up from rest to 4.0 m/s.
Displacement from the area under the graph, per EK 1.3.A.4.iii. The area below the axis is a triangle of base 2.67 s and height 8.0 m/s: $\tfrac{1}{2}(2.667)(8.0) = 10.67\ \text{m}$ , counted as negative. The area above the axis is a triangle of base $4.0 - 2.667 = 1.333\ \text{s}$ and height 4.0 m/s: $\tfrac{1}{2}(1.333)(4.0) = 2.67\ \text{m}$ , counted as positive.
Net displacement: $-10.67 + 2.67 = -8.0\ \text{m}$ .
Check with the kinematic equation: $x - x_0 = v_0 t + \tfrac{1}{2}at^2 = (-8.0)(4.0) + \tfrac{1}{2}(3.0)(4.0)^2 = -32 + 24 = -8.0\ \text{m}$ . The two routes agree exactly.
Distance travelled is the total path length, so the two areas add rather than subtract: $10.67 + 2.67 = 13.3\ \text{m}$ .

Acceleration +3.0 m/s squared throughout, reversal at 2.67 s, velocity at 1.0 s is -5.0 m/s with speed 5.0 m/s, displacement -8.0 m, distance 13.3 m. One constant positive acceleration produced 2.67 s of slowing down followed by 1.33 s of speeding up, so no property of the acceleration on its own could have predicted either. Note also that the object finished 8.0 m in the negative direction while its acceleration pointed positive for the whole four seconds.

Constant speed, changing velocity: a car on a circular track

A car drives around a circular track of radius 55 m at a constant speed of 22 m/s. Find its acceleration at any instant, then find the magnitude of its change in velocity over a quarter of a lap and the magnitude of its average acceleration over that quarter, and explain why the two acceleration figures differ.

Declare the convention: speeds and magnitudes are positive numbers, and directions are described in words, since the motion is two-dimensional.
Is the car accelerating at all? Yes. EK 1.2.B.4 says an object is accelerating if the magnitude and/or direction of its velocity is changing, and the direction is changing continuously here even though the speedometer never moves.
Instantaneous acceleration, from the centripetal relation on the equation sheet: $a_c = v^2/r = (22\ \text{m/s})^2/(55\ \text{m}) = 484/55 = 8.8\ \text{m/s}^2$ , directed toward the center of the circle at every instant.
Change in velocity over a quarter lap. The velocity is 22 m/s in one direction at the start and 22 m/s at right angles to it at the end, so the two vectors and their difference form a right triangle: $\lvert \Delta \vec{v} \rvert = \sqrt{22^2 + 22^2} = 22\sqrt{2} = 31.1\ \text{m/s}$ .
Time for a quarter lap: the quarter arc has length $\tfrac{1}{4}(2\pi r) = \tfrac{1}{2}\pi(55\ \text{m}) = 86.4\ \text{m}$ , so $t = 86.4/22 = 3.93\ \text{s}$ .
Average acceleration over the quarter, from EK 1.2.B.3: $\lvert \vec{a}_{avg} \rvert = \lvert \Delta \vec{v} \rvert/\Delta t = (31.1\ \text{m/s})/(3.93\ \text{s}) = 7.92\ \text{m/s}^2$ .
Explain the gap between 7.92 and 8.8. The instantaneous acceleration has a constant magnitude but a direction that swings through 90 degrees over the quarter lap, so the individual contributions partly point in different directions and their vector sum is shorter than their total length. Average acceleration uses only the endpoints, exactly as EK 1.2.B.1 says averages are calculated from the initial and final states.
Check the ratio for sense: $7.92/8.8 = 0.900$ , which is $2\sqrt{2}/\pi$ . Take a smaller and smaller arc instead of a quarter and that factor approaches 1, which is EK 1.2.B.5 in action, since an average over a very small interval approaches the instantaneous value.

Instantaneous acceleration 8.8 m/s squared toward the center, change in velocity over a quarter lap 31.1 m/s, average acceleration over that quarter 7.92 m/s squared. The car's speed is constant at 22 m/s and its velocity is different at every instant, which is the cleanest counterexample to the belief that constant speed means no acceleration. It also shows that an average acceleration and an instantaneous one are not interchangeable whenever the direction is changing.

Frequently asked questions

What is the difference between velocity and acceleration?

Velocity is the rate at which position changes, measured in meters per second, and acceleration is the rate at which velocity changes, measured in meters per second squared. They are one step apart, which means the value of one places no restriction on the value of the other. A ball at the top of its flight has zero velocity and an acceleration of 9.8 m/s squared downward. A puck sliding on frictionless ice has a large velocity and zero acceleration. AP Physics 1 EK 1.2.B.4 sets the condition for accelerating broadly: an object is accelerating if the magnitude and/or direction of its velocity are changing, so turning at a steady speed counts.

Does negative acceleration mean slowing down?

No. It means the acceleration points in the direction you chose to call negative, and nothing more. Whether the object is speeding up or slowing down depends on how that sign compares with the sign of the velocity. If the two signs match, the speed is increasing; if they are opposite, the speed is decreasing. A ball falling with a velocity of minus 3 m/s and an acceleration of minus 9.8 m/s squared is speeding up, because both are negative. A car moving east at 20 m/s with an acceleration of minus 4 m/s squared is slowing down, because the signs differ. Choose your positive direction, then compare.

Can an object have zero velocity and nonzero acceleration?

Yes, and it is the standard exam case. At the highest point of a throw the ball's velocity passes through zero for one instant while its acceleration is the full 9.8 m/s squared downward. AP Physics 1 EK 1.3.A.3 says the vertical acceleration caused by the force of gravity near Earth's surface is downward and constant, and constant leaves no room for a pause at the top. The reasoning check is simple: if the acceleration really were zero there, the velocity would stay at zero and the ball would hang in the air. Other examples are a mass at maximum displacement in simple harmonic motion and a ball at the instant it reverses off a wall.

Can an object be accelerating while its speed stays constant?

Yes, whenever its direction is changing. AP Physics 1 EK 1.2.B.4 defines accelerating as the magnitude and/or direction of the object's velocity changing, so a change of direction alone is enough. A car going round a bend at a steady 22 m/s on a 55 m radius track has an acceleration of 8.8 m/s squared pointed at the center of the circle, and the speedometer reads the same number the whole way round. This is the whole basis of uniform circular motion, and it is why speed and velocity are not interchangeable words: the speed is constant and the velocity is different at every instant.

How do I tell velocity and acceleration apart on a graph?

By which graph you are looking at and whether you take a slope or an area. On a position-time graph the slope of the tangent is the instantaneous velocity, which is AP Physics 1 EK 1.3.A.4.i. On a velocity-time graph the slope of the tangent is the instantaneous acceleration, EK 1.3.A.4.ii, and the area under the curve is the displacement, EK 1.3.A.4.iii, with area below the axis counting as negative. On an acceleration-time graph the area under the curve is the change in velocity, EK 1.3.A.4.iv. The error to guard against is reading a value off the height when the question wants a slope: a position graph high above the axis means far away, not fast.

Does the acceleration point the same way the object is moving?

Not necessarily, and assuming it does is a reliable way to get a sign wrong. Acceleration points in the direction of the net force, which AP Physics 1 EK 2.5.A.2 states as the acceleration of a system's center of mass being in the same direction as the net force exerted on the system. A car braking while driving forward has a forward velocity and a backward acceleration. A ball thrown upward has an upward velocity and a downward acceleration for the whole of its rise. An object in uniform circular motion has an acceleration at right angles to its velocity at every instant.

Should I use the word deceleration?

It is better avoided in AP work. The word does not appear in any of the four Course and Exam Descriptions, and it hides the sign information that questions are usually testing. Saying an object decelerates leaves the reader to work out both the direction of motion and the direction of the acceleration, and it invites the assumption that a negative acceleration always means slowing. The clearer phrasings are that the object is slowing down, or that its acceleration is directed opposite to its velocity. Both survive a change of sign convention, which deceleration does not.