In everyday conversation, to accelerate means to speed up. The accelerator in a car can in fact cause it to speed up. The greater the acceleration, the greater the change in velocity over a given time. The formal definition of acceleration is consistent with these notions, but more inclusive.

Because acceleration is velocity in m/s divided by time in s, the SI units for acceleration are ${\text{m/s}}^{2}$, meters per second squared or meters per second per second, which literally means by how many meters per second the velocity changes every second.

Recall that velocity is a vector—it has both magnitude and direction. This means that a change in velocity can be a change in magnitude (or speed), but it can also be a change in *direction*. For example, if a car turns a corner at constant speed, it is accelerating because its direction is changing. The quicker you turn, the greater the acceleration. So there is an acceleration when velocity changes either in magnitude (an increase or decrease in speed) or in direction, or both.

Keep in mind that although acceleration is in the direction of the *change* in velocity, it is not always in the direction of *motion*. When an object slows down, its acceleration is opposite to the direction of its motion. This is known as deceleration.

A racehorse coming out of the gate accelerates from rest to a velocity of 15.0 m/s due west in 1.80 s. What is its average acceleration?

**Strategy**

First we draw a sketch and assign a coordinate system to the problem. This is a simple problem, but it always helps to visualize it. Notice that we assign east as positive and west as negative. Thus, in this case, we have negative velocity.

We can solve this problem by identifying $\mathrm{\Delta}v$ and $\mathrm{\Delta}t$ from the given information and then calculating the average acceleration directly from the equation $\stackrel{-}{a}=\frac{\mathrm{\Delta}v}{\mathrm{\Delta}t}=\frac{{v}_{\mathrm{f}}-{v}_{0}}{{t}_{\mathrm{f}}-{t}_{0}}$.

**Solution**

1. Identify the knowns. ${v}_{0}=0$, ${v}_{\mathrm{f}}=-\text{15}\text{.0 m/s}$ (the negative sign indicates direction toward the west), $\mathrm{\Delta}t=1\text{.80 s}$.

2. Find the change in velocity. Since the horse is going from zero to $-\text{15.0 m/s}$, its change in velocity equals its final velocity: $\mathrm{\Delta}v={v}_{\mathrm{f}}=-\text{15}\text{.0 m/s}$.

3. Plug in the known values ($\mathrm{\Delta}v$ and $\mathrm{\Delta}t$) and solve for the unknown $\stackrel{-}{a}$.

**Discussion**

The negative sign for acceleration indicates that acceleration is toward the west. An acceleration of $8\text{.33 m}{\text{/s}}^{2}$ due west means that the horse increases its velocity by 8.33 m/s due west each second, that is, 8.33 meters per second per second, which we write as $8\text{.33 m}{\text{/s}}^{2}$. This is truly an average acceleration, because the ride is not smooth. We shall see later that an acceleration of this magnitude would require the rider to hang on with a force nearly equal to his weight.

# Instantaneous Acceleration

Instantaneous acceleration $a$, or the *acceleration at a specific instant in time*, is obtained by the same process as discussed for instantaneous velocity in Time, Velocity, and Speed—that is, by considering an infinitesimally small interval of time. How do we find instantaneous acceleration using only algebra? The answer is that we choose an average acceleration that is representative of the motion. [link] shows graphs of instantaneous acceleration versus time for two very different motions. In [link](a), the acceleration varies slightly and the average over the entire interval is nearly the same as the instantaneous acceleration at any time. In this case, we should treat this motion as if it had a constant acceleration equal to the average (in this case about $1\text{.}\mathrm{8\; m}{\text{/s}}^{2}$). In [link](b), the acceleration varies drastically over time. In such situations it is best to consider smaller time intervals and choose an average acceleration for each. For example, we could consider motion over the time intervals from 0 to 1.0 s and from 1.0 to 3.0 s as separate motions with accelerations of $+3\text{.}\mathrm{0\; m}{\text{/s}}^{2}$ and $\text{\u20132}\text{.}\mathrm{0\; m}{\text{/s}}^{2}$, respectively.

The next several examples consider the motion of the subway train shown in [link]. In (a) the shuttle moves to the right, and in (b) it moves to the left. The examples are designed to further illustrate aspects of motion and to illustrate some of the reasoning that goes into solving problems.

What are the magnitude and sign of displacements for the motions of the subway train shown in parts (a) and (b) of [link]?

**Strategy**

A drawing with a coordinate system is already provided, so we don’t need to make a sketch, but we should analyze it to make sure we understand what it is showing. Pay particular attention to the coordinate system. To find displacement, we use the equation $\mathrm{\Delta}x={x}_{\mathrm{f}}-{x}_{0}$. This is straightforward since the initial and final positions are given.

**Solution**

1. Identify the knowns. In the figure we see that ${x}_{\mathrm{f}}=\text{6.70 km}$ and ${x}_{0}=\text{4.70 km}$ for part (a), and ${x\prime}_{\mathrm{f}}=3\text{.75 km}$ and ${x\prime}_{0}=5\text{.25 km}$ for part (b).

2. Solve for displacement in part (a).

3. Solve for displacement in part (b).

**Discussion**

The direction of the motion in (a) is to the right and therefore its displacement has a positive sign, whereas motion in (b) is to the left and thus has a negative sign.

What are the distances traveled for the motions shown in parts (a) and (b) of the subway train in [link]?

**Strategy**

To answer this question, think about the definitions of distance and distance traveled, and how they are related to displacement. Distance between two positions is defined to be the magnitude of displacement, which was found in [link]. Distance traveled is the total length of the path traveled between the two positions. (See Displacement.) In the case of the subway train shown in [link], the distance traveled is the same as the distance between the initial and final positions of the train.

**Solution**

1. The displacement for part (a) was +2.00 km. Therefore, the distance between the initial and final positions was 2.00 km, and the distance traveled was 2.00 km.

2. The displacement for part (b) was $\text{\u22121.5 km.}$ Therefore, the distance between the initial and final positions was 1.50 km, and the distance traveled was 1.50 km.

**Discussion**

Distance is a scalar. It has magnitude but no sign to indicate direction.

Suppose the train in [link](a) accelerates from rest to 30.0 km/h in the first 20.0 s of its motion. What is its average acceleration during that time interval?

**Strategy**

It is worth it at this point to make a simple sketch:

This problem involves three steps. First we must determine the change in velocity, then we must determine the change in time, and finally we use these values to calculate the acceleration.

**Solution**

1. Identify the knowns. ${v}_{0}=0$ (the trains starts at rest), ${v}_{\mathrm{f}}=\text{30}\text{.}\text{0 km/h}$, and $\mathrm{\Delta}t=\text{20}\text{.}\text{0 s}$.

2. Calculate $\mathrm{\Delta}v$. Since the train starts from rest, its change in velocity is $\mathrm{\Delta}v\text{=}\phantom{\rule{0.20em}{0ex}}\text{+}\text{30.0 km/h}$, where the plus sign means velocity to the right.

3. Plug in known values and solve for the unknown, $\stackrel{-}{a}$.

4. Since the units are mixed (we have both hours and seconds for time), we need to convert everything into SI units of meters and seconds. (See Physical Quantities and Units for more guidance.)

**Discussion**

The plus sign means that acceleration is to the right. This is reasonable because the train starts from rest and ends up with a velocity to the right (also positive). So acceleration is in the same direction as the *change* in velocity, as is always the case.

Now suppose that at the end of its trip, the train in [link](a) slows to a stop from a speed of 30.0 km/h in 8.00 s. What is its average acceleration while stopping?

**Strategy**

In this case, the train is decelerating and its acceleration is negative because it is toward the left. As in the previous example, we must find the change in velocity and the change in time and then solve for acceleration.

**Solution**

1. Identify the knowns. ${v}_{0}=\text{30}\text{.0 km/h}$, ${v}_{\mathrm{f}}=\mathrm{0\; km/h}$ (the train is stopped, so its velocity is 0), and $\mathrm{\Delta}t=\text{8.00 s}$.

2. Solve for the change in velocity, $\mathrm{\Delta}v$.

3. Plug in the knowns, $\mathrm{\Delta}v$ and $\mathrm{\Delta}t$, and solve for $\stackrel{-}{a}$.

4. Convert the units to meters and seconds.

**Discussion**

The minus sign indicates that acceleration is to the left. This sign is reasonable because the train initially has a positive velocity in this problem, and a negative acceleration would oppose the motion. Again, acceleration is in the same direction as the *change* in velocity, which is negative here. This acceleration can be called a deceleration because it has a direction opposite to the velocity.

The graphs of position, velocity, and acceleration vs. time for the trains in [link] and [link] are displayed in [link]. (We have taken the velocity to remain constant from 20 to 40 s, after which the train decelerates.)

What is the average velocity of the train in part b of [link], and shown again below, if it takes 5.00 min to make its trip?

**Strategy**

Average velocity is displacement divided by time. It will be negative here, since the train moves to the left and has a negative displacement.

**Solution**

1. Identify the knowns. ${x\prime}_{\mathrm{f}}=3\text{.75 km}$, ${x\prime}_{0}=\text{5.25 km}$, $\mathrm{\Delta}t=\text{5.00 min}$.

2. Determine displacement, $\mathrm{\Delta}x\prime $. We found $\mathrm{\Delta}x\prime $ to be $-\text{1.5 km}$ in [link].

3. Solve for average velocity.

4. Convert units.

**Discussion**

The negative velocity indicates motion to the left.

Finally, suppose the train in [link] slows to a stop from a velocity of 20.0 km/h in 10.0 s. What is its average acceleration?

**Strategy**

Once again, let’s draw a sketch:

As before, we must find the change in velocity and the change in time to calculate average acceleration.

**Solution**

1. Identify the knowns. ${v}_{0}=-\text{20 km/h}$, ${v}_{\mathrm{f}}=\mathrm{0\; km/h}$, $\mathrm{\Delta}t=\text{10}\text{.}\mathrm{0\; s}$.

2. Calculate $\mathrm{\Delta}v$. The change in velocity here is actually positive, since

3. Solve for $\stackrel{-}{a}$.

4. Convert units.

**Discussion**

The plus sign means that acceleration is to the right. This is reasonable because the train initially has a negative velocity (to the left) in this problem and a positive acceleration opposes the motion (and so it is to the right). Again, acceleration is in the same direction as the *change* in velocity, which is positive here. As in [link], this acceleration can be called a deceleration since it is in the direction opposite to the velocity.

# Sign and Direction

Perhaps the most important thing to note about these examples is the signs of the answers. In our chosen coordinate system, plus means the quantity is to the right and minus means it is to the left. This is easy to imagine for displacement and velocity. But it is a little less obvious for acceleration. Most people interpret negative acceleration as the slowing of an object. This was not the case in [link], where a positive acceleration slowed a negative velocity. The crucial distinction was that the acceleration was in the opposite direction from the velocity. In fact, a negative acceleration will *increase* a negative velocity. For example, the train moving to the left in [link] is sped up by an acceleration to the left. In that case, both $v$ and $a$ are negative. The plus and minus signs give the directions of the accelerations. If acceleration has the same sign as the change in velocity, the object is speeding up. If acceleration has the opposite sign of the change in velocity, the object is slowing down.

# Section Summary

- Acceleration is the rate at which velocity changes. In symbols, average acceleration $\stackrel{-}{a}$ is
$\stackrel{-}{a}=\frac{\mathrm{\Delta}v}{\mathrm{\Delta}t}=\frac{{v}_{\mathrm{f}}-{v}_{0}}{{t}_{\mathrm{f}}-{t}_{0}}\text{.}$
- The SI unit for acceleration is ${\text{m/s}}^{2}$.
- Acceleration is a vector, and thus has a both a magnitude and direction.
- Acceleration can be caused by either a change in the magnitude or the direction of the velocity.
- Instantaneous acceleration $a$ is the acceleration at a specific instant in time.
- Deceleration is an acceleration with a direction opposite to that of the velocity.

# Conceptual Questions

Is it possible for speed to be constant while acceleration is not zero? Give an example of such a situation.

Is it possible for velocity to be constant while acceleration is not zero? Explain.

Give an example in which velocity is zero yet acceleration is not.

If a subway train is moving to the left (has a negative velocity) and then comes to a stop, what is the direction of its acceleration? Is the acceleration positive or negative?

Plus and minus signs are used in one-dimensional motion to indicate direction. What is the sign of an acceleration that reduces the magnitude of a negative velocity? Of a positive velocity?

# Problems & Exercises

A cheetah can accelerate from rest to a speed of 30.0 m/s in 7.00 s. What is its acceleration?

$4\text{.}\text{29}\phantom{\rule{0.25em}{0ex}}{\text{m/s}}^{2}$

**Professional Application**

Dr. John Paul Stapp was U.S. Air Force officer who studied the effects of extreme deceleration on the human body. On December 10, 1954, Stapp rode a rocket sled, accelerating from rest to a top speed of 282 m/s (1015 km/h) in 5.00 s, and was brought jarringly back to rest in only 1.40 s! Calculate his (a) acceleration and (b) deceleration. Express each in multiples of $g$ $(9\text{.}\text{80 m}{\text{/s}}^{2})$ by taking its ratio to the acceleration of gravity.

A commuter backs her car out of her garage with an acceleration of $1\text{.}{\text{40 m/s}}^{2}$. (a) How long does it take her to reach a speed of 2.00 m/s? (b) If she then brakes to a stop in 0.800 s, what is her deceleration?

(a) $1\text{.}\text{43 s}$

(b) $-2\text{.}\text{50}\phantom{\rule{0.25em}{0ex}}{\text{m/s}}^{2}$

Assume that an intercontinental ballistic missile goes from rest to a suborbital speed of 6.50 km/s in 60.0 s (the actual speed and time are classified). What is its average acceleration in ${\text{m/s}}^{2}$ and in multiples of $g$ $(9\text{.}\text{80 m}{\text{/s}}^{2})?$

### Tập tin đính kèm

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- College Physics
- Preface
- Introduction: The Nature of Science and Physics
- Kinematics
- Introduction to One-Dimensional Kinematics
- Displacement
- Vectors, Scalars, and Coordinate Systems
- Time, Velocity, and Speed
- Acceleration
- Motion Equations for Constant Acceleration in One Dimension
- Problem-Solving Basics for One-Dimensional Kinematics
- Falling Objects
- Graphical Analysis of One-Dimensional Motion

- Two-Dimensional Kinematics
- Dynamics: Force and Newton's Laws of Motion
- Introduction to Dynamics: Newton’s Laws of Motion
- Development of Force Concept
- Newton’s First Law of Motion: Inertia
- Newton’s Second Law of Motion: Concept of a System
- Newton’s Third Law of Motion: Symmetry in Forces
- Normal, Tension, and Other Examples of Forces
- Problem-Solving Strategies
- Further Applications of Newton’s Laws of Motion
- Extended Topic: The Four Basic Forces—An Introduction

- Further Applications of Newton's Laws: Friction, Drag, and Elasticity
- Uniform Circular Motion and Gravitation
- Work, Energy, and Energy Resources
- Linear Momentum and Collisions
- Statics and Torque
- Rotational Motion and Angular Momentum
- Introduction to Rotational Motion and Angular Momentum
- Angular Acceleration
- Kinematics of Rotational Motion
- Dynamics of Rotational Motion: Rotational Inertia
- Rotational Kinetic Energy: Work and Energy Revisited
- Angular Momentum and Its Conservation
- Collisions of Extended Bodies in Two Dimensions
- Gyroscopic Effects: Vector Aspects of Angular Momentum

- Fluid Statics
- Fluid Dynamics and Its Biological and Medical Applications
- Introduction to Fluid Dynamics and Its Biological and Medical Applications
- Flow Rate and Its Relation to Velocity
- Bernoulli’s Equation
- The Most General Applications of Bernoulli’s Equation
- Viscosity and Laminar Flow; Poiseuille’s Law
- The Onset of Turbulence
- Motion of an Object in a Viscous Fluid
- Molecular Transport Phenomena: Diffusion, Osmosis, and Related Processes

- Temperature, Kinetic Theory, and the Gas Laws
- Heat and Heat Transfer Methods
- Thermodynamics
- Introduction to Thermodynamics
- The First Law of Thermodynamics
- The First Law of Thermodynamics and Some Simple Processes
- Introduction to the Second Law of Thermodynamics: Heat Engines and Their Efficiency
- Carnot’s Perfect Heat Engine: The Second Law of Thermodynamics Restated
- Applications of Thermodynamics: Heat Pumps and Refrigerators
- Entropy and the Second Law of Thermodynamics: Disorder and the Unavailability of Energy
- Statistical Interpretation of Entropy and the Second Law of Thermodynamics: The Underlying Explanation

- Oscillatory Motion and Waves
- Introduction to Oscillatory Motion and Waves
- Hooke’s Law: Stress and Strain Revisited
- Period and Frequency in Oscillations
- Simple Harmonic Motion: A Special Periodic Motion
- The Simple Pendulum
- Energy and the Simple Harmonic Oscillator
- Uniform Circular Motion and Simple Harmonic Motion
- Damped Harmonic Motion
- Forced Oscillations and Resonance
- Waves
- Superposition and Interference
- Energy in Waves: Intensity

- Physics of Hearing
- Electric Charge and Electric Field
- Introduction to Electric Charge and Electric Field
- Static Electricity and Charge: Conservation of Charge
- Conductors and Insulators
- Coulomb’s Law
- Electric Field: Concept of a Field Revisited
- Electric Field Lines: Multiple Charges
- Electric Forces in Biology
- Conductors and Electric Fields in Static Equilibrium
- Applications of Electrostatics

- Electric Potential and Electric Field
- Electric Current, Resistance, and Ohm's Law
- Circuits, Bioelectricity, and DC Instruments
- Magnetism
- Introduction to Magnetism
- Magnets
- Ferromagnets and Electromagnets
- Magnetic Fields and Magnetic Field Lines
- Magnetic Field Strength: Force on a Moving Charge in a Magnetic Field
- Force on a Moving Charge in a Magnetic Field: Examples and Applications
- The Hall Effect
- Magnetic Force on a Current-Carrying Conductor
- Torque on a Current Loop: Motors and Meters
- Magnetic Fields Produced by Currents: Ampere’s Law
- Magnetic Force between Two Parallel Conductors
- More Applications of Magnetism

- Electromagnetic Induction, AC Circuits, and Electrical Technologies
- Introduction to Electromagnetic Induction, AC Circuits and Electrical Technologies
- Induced Emf and Magnetic Flux
- Faraday’s Law of Induction: Lenz’s Law
- Motional Emf
- Eddy Currents and Magnetic Damping
- Electric Generators
- Back Emf
- Transformers
- Electrical Safety: Systems and Devices
- Inductance
- RL Circuits
- Reactance, Inductive and Capacitive
- RLC Series AC Circuits

- Electromagnetic Waves
- Geometric Optics
- Vision and Optical Instruments
- Wave Optics
- Introduction to Wave Optics
- The Wave Aspect of Light: Interference
- Huygens's Principle: Diffraction
- Young’s Double Slit Experiment
- Multiple Slit Diffraction
- Single Slit Diffraction
- Limits of Resolution: The Rayleigh Criterion
- Thin Film Interference
- Polarization
- *Extended Topic* Microscopy Enhanced by the Wave Characteristics of Light

- Special Relativity
- Introduction to Quantum Physics
- Atomic Physics
- Introduction to Atomic Physics
- Discovery of the Atom
- Discovery of the Parts of the Atom: Electrons and Nuclei
- Bohr’s Theory of the Hydrogen Atom
- X Rays: Atomic Origins and Applications
- Applications of Atomic Excitations and De-Excitations
- The Wave Nature of Matter Causes Quantization
- Patterns in Spectra Reveal More Quantization
- Quantum Numbers and Rules
- The Pauli Exclusion Principle

- Radioactivity and Nuclear Physics
- Medical Applications of Nuclear Physics
- Particle Physics
- Frontiers of Physics
- Atomic Masses
- Selected Radioactive Isotopes
- Useful Information
- Glossary of Key Symbols and Notation