2.3 What Is Acceleration? Formula and Examples

Acceleration is where kinematics stops describing where you are and starts describing how your motion feels. If you can master acceleration, you can read any position graph, velocity graph, and force problem that follows, because every dynamics question in AP Physics, A-Level, IB HL, and Olympiad physics begins with one skill: turning a change in velocity into a slope, a derivative, and a sign. In this master guide we will rebuild the entire concept from the ground up exactly as presented in Fundamentals of Physics, so you can solve graph problems, derivative problems, g-unit problems, and sign trap problems without ever needing the textbook open next to you.

Key Insights

• Average acceleration is change in velocity divided by time interval.
• Instantaneous acceleration is the derivative of velocity and second derivative of position.
• On a velocity versus time graph, acceleration is the slope of the curve.
• Your body senses acceleration, not velocity, which explains g-forces.
• If velocity and acceleration have same sign, speed increases; opposite signs means speed decreases.

Acceleration

Acceleration is the rate at which velocity changes. Velocity tells you how fast position changes, acceleration tells you how fast velocity itself is changing. Like displacement and velocity, acceleration is a vector quantity for motion along an axis. Its algebraic sign tells you which direction along the axis it points.

When a particle's velocity changes from \(v_1\) at time \(t_1\) to \(v_2\) at time \(t_2\), we say the particle has undergone acceleration.

Average Acceleration

Average acceleration over a time interval is the ratio of the change in velocity to the time interval in which that change occurs.

\(a_{avg}\) = (\(v_1 - v_2\)) / (\(t_1 - t_2\)) = \(Δv\) / \(Δt\)

This is formally identical to the definition of average velocity, but with velocity playing the role of position. The units follow: velocity is length / time, so dividing again by time gives length / time / time or length / time². In SI, m/s / s = m/s².

The sign of \(a_{avg}\) is crucial. It does not mean speeding up or slowing down by itself. It means direction on the \(x\)-axis. If \(Δv\) is positive, \(a_{avg}\) points in the positive \(x\) direction. If \(Δv\) is negative, it points in the negative direction.

Quantity	Definition	Graphical Meaning	Units
Average Velocity	\(Δx\)/\(Δt\)	Slope of secant on \(x\) vs \(t\)	m/s
Average Acceleration	\(Δv\)/\(Δt\)	Slope of secant on \(v\) vs \(t\)	m/s2

Callout Box: Why \(Δv\) and not just \(v\)? Many students use \(v\)/\(t\) instead of \(Δv\)/\(Δt\).Acceleration cares about change. A particle moving at a constant 4 m/s has zero acceleration because its velocity is not changing, even though \(v\)/\(t\) is not zero.

Average acceleration as slope of secant line between t1 and t2 versus instantaneous acceleration as slope of tangent line at a point on a 3D velocity vs time curve

In this 3D view, the blue curve is \(v\)(\(t\)). The green dashed line connecting the points at \(t_1\) and \(t_2\) has slope \(Δv\) / \(Δt\) = \(a_{avg}\). As you bring \(t_2\) closer to \(t_1\), that secant line becomes the red tangent line. Its slope at that single instant is the instantaneous acceleration.

Instantaneous Acceleration

To know the acceleration at a single instant, we shrink the interval to zero. The instantaneous acceleration, or simply acceleration \(a\), is the first time derivative of velocity \(v(t)\) and the second time derivative of position \(x(t)\):

\(a\) = \(dv\) / \(dt\)

a = \(dv\) / \(dt\) = \(d\) / \(dt\) (\(dx\) / \(dt\)) = \(d^2x\) / \(dt^2\)

In words, acceleration at any instant is the rate at which velocity is changing at that instant. Graphically, acceleration at any point is the slope of the curve of \(v(t)\) at that point.

When velocity is constant, either at zero or at a non-zero value like 4 m/s, the \(v(t)\) curve is horizontal. Its slope is zero, so \(a\) = 0. When the \(v(t)\) curve has a positive slope, \(a(t)\) is positive. When the \(v(t)\) curve has a negative slope, \(a(t)\) is negative.

Three stacked realistic 3D graphs x(t) top, v(t) middle, a(t) bottom showing derivative connections

This stack shows the calculus chain. Differentiating position gives velocity. Differentiating velocity gives acceleration. Notice: where \(v(t)\) is flat, \(a(t)\) sits on zero. Where \(v(t)\) rises, \(a(t)\) is above zero. Where \(v(t)\) falls, \(a(t)\) is below zero. If the cab starts from rest, the \(v(t)\) slope is positive and \(a(t)\) is positive. When the cab brakes to a stop, the \(v(t)\) slope is negative and \(a(t)\) is negative. The slope during braking is often steeper because the cab stops in half the time it took to get up to speed, meaning the magnitude of deceleration is larger than the magnitude of the initial acceleration.

Pro Tip for Exams:If you are given \(x(t)\) as a polynomial, you never need to memorize \(v(t)\) or \(a(t)\). Take derivatives. If \(x\) = 4 -27\(t\) + \(t^3\) meters, then \(v\) = \(dx\) / \(dt\) = -27 + 3\(t^2\) m/s, and \(a\) = \(dv\) / \(dt\) = 6\(t\) m/s². The units stay attached to the function.

Exam Trap:A common error is to think \(a\) = 0 means \(v\) = 0. It means \(v\) is not changing. A car cruising at 90 km/h on a highway and an airplane cruising at 900 km/h both have near-zero acceleration even though their velocities are large.

Sensations

Your body does not have a speedometer. It has an accelerometer.

When the cab first accelerates upward, you feel as though you are pressed downward into the floor. When the cab is later braked to a stop, you feel as though you are stretched upward, lighter. In between, when velocity is constant, you feel nothing special, no sense of motion.

This is why you have no bodily awareness of moving at 90 km/h in a car or 900 km/h in an airplane when the velocity is steady. You become keenly aware the moment velocity changes quickly. Part of the thrill of an amusement park ride is due to the quick changes of velocity that you undergo. You pay for the accelerations, not for the speed.

A more extreme example is the rocket sled experiments conducted to test human tolerance to large accelerations. A sled is rapidly accelerated along a track and then very rapidly braked to a stop. The test subject experiences acceleration out of the page during speed-up and acceleration into the page during braking, with facial distortion visible due to inertia.

Callout Box: Accelerometer vs SpeedometerYour inner ear and your muscles sense net force, which is proportional to acceleration. You cannot sense constant velocity without looking outside. That is the principle of inertial frames that will return in Newton's First Law.

g Units

Large accelerations are sometimes expressed in terms of \(g\) units, with

1\(g\) = 9.8 m/s² (g unit)

We will discuss later that \(g\) is the magnitude of the acceleration of a falling object near Earth's surface.

This unit is convenient because it compares any acceleration to what you feel normally due to gravity.

Situation	Typical Acceleration	In g units	What you feel
Standing on Earth	9.8 m/s2	1g	Normal weight
Roller coaster peak turn	∼29 m/s2	3g	Pressed hard, 3 times heavier
Fighter jet sharp turn	∼49 m/s2	5g	Vision narrows, need g-suit
Rocket sled braking	98 m/s2	10g	Extreme, tested by Col. Stapp

On a roller coaster, you may experience brief accelerations up to 3g, which is 3 \times 9.8 m/s², or about 29 m/s², more than enough to justify the cost of the ride.

Realistic 3D visualization of g-units scale from 1g to extreme rocket sled acceleration

Signs

In common language, the sign of an acceleration has a nonscientific meaning: positive acceleration means the speed of an object is increasing, and negative acceleration means the speed is decreasing, the object is decelerating.

In physics, the sign of an acceleration indicates a direction, not whether the speed is increasing or decreasing.

Example: If a car with an initial velocity \(v\) = -25 m/s is braked to a stop in 5.0 s, then \(a_{avg}\) = +5.0 m/s². The acceleration is positive, but the car's speed has decreased. The reason is the difference in signs: the direction of the acceleration is opposite that of the velocity.

Here then is the proper way to interpret the signs:

Callout Box: The Golden Rule of SpeedIf the signs of the velocity and acceleration of a particle are the same, the speed of the particle increases. If the signs are opposite, the speed decreases.

Four quadrants of velocity and acceleration sign combinations showing when speed increases or decreases

Let's apply this to a classic checkpoint. A wombat moves along an \(x\) axis. What is the sign of its acceleration if it is moving (a) in the positive direction with increasing speed, (b) in the positive direction with decreasing speed, (c) in the negative direction with increasing speed, and (d) in the negative direction with decreasing speed?

(a) \(v\) positive, speed up → \(a\) must be positive, same sign.
(b) \(v\) positive, speed down → \(a\) negative, opposite sign.
(c) \(v\) negative, speed up → speed up means same sign, so \(a\) negative.
(d) \(v\) negative, slowing → opposite sign, so \(a\) positive.

Worked Sample: \(x\) = 4 -27\(t\) + \(t^3\)

This example cements calculus. Given position in meters and time in seconds.

(a) Find \(v(t)\) and \(a(t)\). Differentiate: \(v\) = \(dx\) / \(dt\) = -27 + 3\(t^2\), \(a\) = \(dv\) / \(dt\) = 6\(t\).

(b) Is there ever a time when \(v\) = 0? Set -27 + 3\(t^2\) = 0 → \(t^2\) = 9 → \(t\) = ±3 s. So velocity is zero both 3 s before and 3 s after the clock reads 0.

(c) Describe motion for \(t\) ≥ 0. At \(t\) = 0, particle is at \(x\) = +4 m and moving with \(v\) = -27 m/s, in the negative direction, with \(a\)(0) = 0 because velocity is momentarily not changing. For 0 < t < 3 s, \(v\) negative, \(a\) positive and growing, signs opposite → slowing down while still moving negative. At \(t\) = 3 s, \(v\) = 0, \(x\) = -50 m, the farthest left point, it stops momentarily. For \(t\) > 3 s, both \(v\) and \(a\) are positive, signs same → speeding up to the right. Both grow larger in magnitude progressively.

Practice Questions

MCQs (Multiple Choice Questions)

01. A particle has \(v_1\) = 2 m/s at \(t_1\) = 1 s and \(v_2\) = 8 m/s at \(t_2\) = 3 s. What is \(a_{avg}\)?

1. 2 m/s²
2. 3 m/s²
3. 6 m/s²
4. 8 m/s²

Correct: B
\(a_{avg}\) = (8-2)/(3-1) = 6/2 = 3 m/s².

02. On a \(v\) vs \(t\) graph, instantaneous acceleration at \(t\) = 2 s is:

1. Area under curve to \(t\)=2
2. Slope of tangent at \(t\)=2
3. Value of \(v\) at \(t\)=2
4. \(v\)/\(t\) at t=2

Correct: B
Slope = \(dv\) / \(dt\). Area gives displacement.

03. If \(x(t)\) = 5\(t^2\), what is \(a(t)\)?

1. 5\(t\)
2. 10\(t\)
3. 10 m/s²
4. 0

Correct: C
\(v\) = 10\(t\), \(a\) = 10

04. A car moves with \(v\) = -10 m/s and \(a\) = -2 m/s². Its speed is:

1. Increasing
2. Decreasing
3. Constant
4. Zero

Correct: A
Same sign → speed increases (more negative, faster in negative direction).

05. Unit of \(a\) that is incorrect:

1. m/s²
2. km/h²
3. m/s
4. ft/s²

Correct: C
m/s is velocity.

06. 2g is how many m/s²

1. 9.8
2. 19.6
3. 29.4
4. 98

Correct: B
2 × 9.8 = 19.6.

07. If \(v\) is constant at 4 m/s, acceleration is:

1. 4 m/s²
2. 0
3. 9.8
4. Depends on \(x\)

Correct: B
No change in \(v\).

08. When a cab slows down, \(v(t)\) slope is:

1. Positive
2. Negative
3. Zero
4. Infinite

Correct: B
Decreasing \(v\) → negative slope → negative \(a\).

09. A particle with \(v\) = -25 m/s braked to 0 in 5 s has \(a_{avg}\):

1. -5 m/s²
2. +5 m/s²
3. -25 m/s²
4. 0

Correct: B
(0 - (-25))/5 = +5 m/s².

10. Your body feels:

1. Velocity
2. Position
3. Acceleration
4. Jerk only

Correct: C
Accelerometer, not speedometer.

FRQs (Free Response Questions)

FRQ 1: (AP style - Graph Interpretation)
Given \(v(t)\) is a parabola \(v\) = 2 + 4\(t\) - \(t^2\) for 0 ≤ \(t\) ≤ 4 s.
(a) Find \(a(t)\).
(b) Find times when \(a\)=0.
(c) Describe speed behavior.

Rubric:

• (a) a = dv/dt = 4 -2t (2 pts)
• (b) Set 0 = 4-2t \rightarrow t=2 s (1 pt)
• (c) For \(t\)<2, \(a\)>0, if \(v\)>0 speed increases until \(t\)=2, after \(t\)=2, \(a\)<0 speed decreases, signs opposite after peak (2 pts)

FRQ 2: (A-Level - Units & g)
A roller coaster experiences 3.5g at bottom of loop. Calculate in m/s² and explain sensation.

Rubric:

• 3.5 × 9.8 = 34.3 m/s² (1 pt)
• Sensation: feels 3.5 times heavier, pressed into seat, blood pools (2 pts)
• Vector direction toward center (1 pt)

FRQ 3: (IB HL - Derivative)
\(x\) = 4 -27\(t\) + \(t^3\). Show \(v\)=0 at \(t\)=3 s and find \(x\) there. Is acceleration zero there?

Rubric:

• \(v\) = -27 + 3\(t^2\), set zero → \(t\) = 3 (2 pts)
• \(x\)(3) = 4-81+27 = -50 m (1 pt)
• \(a\) = 6\(t\) = 18 m/s² not zero, so reversing (1 pt)

FRQ 4: (College - Sign Rule)
A particle moves with \(v\) = -6 m/s, \(a\) = +3 m/s². Is it speeding up or slowing? What happens 4 s later if \(a\) constant?

Rubric:

• Opposite signs → slowing, speed decreasing to 0 (1 pt)
• Time to stop: 0 = -6 +3\(t\) → \(t\)=2 s (1 pt)
• After 2 s moves positive with increasing speed (2 pts)

FRQ 5: (USAPhO - Conceptual)
Explain why an accelerometer in a car reads zero at constant highway speed but a GPS speedometer reads 30 m/s. How would a passenger test which instrument is which without looking outside?

Rubric:

• Accelerometer measures \(dv\) / \(dt\), zero when \(v\) constant (1 pt)
• GPS measures \(dx\) / \(dt\) (1 pt)
• Passenger can only feel changes, not steady motion, so cannot tell speed without visual reference but can feel braking (2 pts)
• Reference to inertial frame (1 pt)

Frequently Asked Questions

What is the difference between average and instantaneous acceleration?

Average acceleration is \(Δv\) / \(Δt\) over a finite interval, the slope of a secant on a \(v\) vs \(t\) graph. Instantaneous acceleration is \(dv\) / \(dt\), the limit as \(Δt\) &to; 0, the slope of the tangent at one instant.

How do you find acceleration from a velocity-time graph?

Acceleration at any time equals the slope of the \(v\) vs \(t\) curve at that time. A steeper slope means larger magnitude. A horizontal tangent means zero acceleration.

What are the units of acceleration and what is 1g?

SI unit is meters per second squared, m/s². 1g = 9.8 m/s², the acceleration due to Earth's gravity. 3g is 29.4 m/s², typical for roller coasters.

Is negative acceleration always deceleration?

No. In physics, sign indicates direction on an axis. If velocity is negative and acceleration is negative, same sign, speed increases. Deceleration means speed decreases, which occurs when \(v\) and \(a\) have opposite signs.

How is acceleration related to position?

Acceleration is the second derivative of position: \(a\) = \(d^2x\) / \(dt^2\). First derivative gives velocity, second gives acceleration. You differentiate twice.

Why do we feel acceleration and not velocity?

Your body senses net force via inner ear and muscles. Force is proportional to acceleration. At constant velocity, net force is zero, so you feel weightless to motion. Changes in velocity produce forces you feel.

Can acceleration be zero when velocity is not zero?

Yes. Zero acceleration means velocity is constant, not necessarily zero. A car cruising at 20 m/s has \(a\) = 0 but \(v\) = 20 m/s.

What is the sign rule for speeding up vs slowing down?

If \(v\) and \(a\) have the same algebraic sign, speed increases. If they have opposite signs, speed decreases. This works for both positive and negative directions.

Exam Mapping

Concept	A-Level Physics (Edexcel/AQA)	AP Physics 1 & C	IB Physics HL	College Physics 1	USAPhO / IPhO
\(a_{avg}\) = \(Δv\) / \(Δt\)	Kinematics equations, gradient calculations	Unit 1 Kinematics, average vs instantaneous	Topic 2.1 Motion, slope analysis	Chapter 2 Motion in 1D, basic calculation	Basic tool for non-uniform motion analysis
\(a\) = \(dv\) / \(dt\) = \(d^2x\) / \(dt^2\)	Differentiation of \(x(t)\), calculus optional in A-Level Maths link	AP C: calculus application, derivative	HL calculus extension, \(a\) - \(t\) graphs	Calculus-based physics, derivative	Essential for variable acceleration problems
\(a\) as slope of \(v\) - \(t\)	Required practical: \(v\) - \(t\) graphs, area vs slope	Lab: interpreting motion graphs	IA: motion graphs, data analysis	Graph interpretation skills	Graph analysis under time pressure
\(g\) units	N/A but context for forces, \(g\) = 9.81	Context for apparent weight	Weight vs mass, \(g\)	Free fall intro, \(g\)	Human tolerance, rocket problems
Sign rule: speed up/down	Vector nature, negative acceleration misconceptions	Common MCQ trap	Conceptual question: speeding up/slowing	Conceptual check	Sign convention in 1D collisions

Conclusion

Acceleration ties together calculus, graphs, and physical sensation. You now have the full toolkit: average acceleration as a secant slope, instantaneous acceleration as a tangent slope and derivative \(a\) = \(dv\) / \(dt\) = \(d^2x\) / \(dt^2\), the meaning of zero and negative slopes, the \(g\)-unit scale for real-world forces, and the golden sign rule that replaces the vague idea of deceleration. When you see \(x(t)\) = 4 -27\(t\) + \(t^3\), you know to differentiate, to find where \(v\) = 0, to check where \(v\) and \(a\) share signs, and to describe four stages of motion from leftward slowing to rightward speeding. Carry this into constant acceleration problems and Newton's laws, and every exam graph becomes a slope problem you have already solved.

References

Fundamentals of Physics by Halliday, Resnick, and Walker, 12th Edition, Chapter 2.3 Acceleration - Primary source for definitions, derivations, \(g\)-units, sign interpretation, and sample problem \(x\) = 4 -27\(t\) + \(t^3\).