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The Nature of Science and Physics

Kinematics

Introduction to One-Dimensional KinematicsDisplacementVectors, Scalars, and Coordinate SystemsTime, Velocity, and SpeedVideo: One-Dimensional KinematicsAccelerationMotion Equations for Constant Acceleration in One DimensionProblem-Solving Basics for One-Dimensional KinematicsFalling ObjectsGraphical Analysis of One-Dimensional Motion

Two-Dimensional Kinematics

Dynamics: Force and Newton's Laws of Motion

Introduction to Dynamics: Newton's Laws of MotionDevelopment of Force ConceptNewton's First Law of Motion: InertiaNewton's Second Law of Motion: Concept of a SystemNewton's Third Law of Motion: Symmetry in ForcesVideo: Newton's LawsNormal, Tension, and Other Examples of ForcesProblem-Solving StrategiesFurther Applications of Newton's Laws of MotionExtended Topic: The Four Basic Forces—An Introduction

Further Applications of Newton's Laws: Friction, Drag, and Elasticity

Uniform Circular Motion and Gravitation

Introduction to Uniform Circular Motion and GravitationRotation Angle and Angular VelocityCentripetal AccelerationCentripetal ForceFictitious Forces and Non-inertial Frames: The Coriolis ForceNewton's Universal Law of GravitationVideo: GravitationSatellites and Kepler's Laws: An Argument for Simplicity

Work, Energy, and Energy Resources

Introduction to Work, Energy, and Energy ResourcesWork: The Scientific DefinitionKinetic Energy and the Work-Energy TheoremGravitational Potential EnergyVideo: Potential and Kinetic EnergyConservative Forces and Potential EnergyNonconservative ForcesConservation of EnergyPowerWork, Energy, and Power in HumansWorld Energy Use

Linear Momentum and Collisions

Rotational Motion and Angular Momentum

Introduction to Rotational Motion and Angular MomentumAngular AccelerationKinematics of Rotational MotionVideo: Rotational MotionDynamics of Rotational Motion: Rotational InertiaRotational Kinetic Energy: Work and Energy RevisitedAngular Momentum and Its ConservationVideo: Angular MomentumCollisions of Extended Bodies in Two DimensionsGyroscopic Effects: Vector Aspects of Angular Momentum

Statics and Torque

Fluid Statics

Introduction to Fluid StaticsWhat Is a Fluid?DensityPressureVariation of Pressure with Depth in a FluidPascal's PrincipleGauge Pressure, Absolute Pressure, and Pressure MeasurementArchimedes' PrincipleVideo: BuoyancyCohesion and Adhesion in Liquids: Surface Tension and Capillary ActionPressures in the Body

Fluid Dynamics and Its Biological and Medical Applications

Introduction to Fluid Dynamics and Biological and Medical ApplicationsFlow Rate and Its Relation to VelocityBernoulli's EquationVideo: Fluid FlowThe Most General Applications of Bernoulli's EquationViscosity and Laminar Flow; Poiseuille's LawThe Onset of TurbulenceMotion of an Object in a Viscous FluidMolecular Transport Phenomena: Diffusion, Osmosis, and Related Processes

Temperature, Kinetic Theory, and the Gas Laws

Heat and Heat Transfer Methods

Thermodynamics

Introduction to ThermodynamicsThe First Law of ThermodynamicsThe First Law of Thermodynamics and Some Simple ProcessesIntroduction to the Second Law of Thermodynamics: Heat Engines and Their EfficiencyCarnot's Perfect Heat Engine: The Second Law of Thermodynamics RestatedApplications of Thermodynamics: Heat Pumps and RefrigeratorsEntropy and the Second Law of Thermodynamics: Disorder and the Unavailability of EnergyStatistical Interpretation of Entropy and the Second Law of Thermodynamics: The Underlying Explanation

Oscillatory Motion and Waves

Introduction to Oscillatory Motion and WavesHooke's Law: Stress and Strain RevisitedPeriod and Frequency in OscillationsSimple Harmonic Motion: A Special Periodic MotionVideo: Harmonic MotionThe Simple PendulumEnergy and the Simple Harmonic OscillatorUniform Circular Motion and Simple Harmonic MotionDamped Harmonic MotionForced Oscillations and ResonanceWavesSuperposition and InterferenceEnergy in Waves: Intensity

Physics of Hearing

- Define and distinguish between instantaneous acceleration, average acceleration, and deceleration.
- Calculate acceleration given initial time, initial velocity, final time, and final velocity.

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

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

Acceleration is a vector in the same direction as the *change* in velocity, Δ*v*. Since velocity is a vector, it can change either in magnitude or in direction. Acceleration is therefore a change in either speed or direction, or both.

Deceleration always refers to acceleration in the direction opposite to the direction of the velocity. Deceleration always reduces speed. Negative acceleration, however, is acceleration *in the negative direction in the chosen coordinate system*. Negative acceleration may or may not be deceleration, and deceleration may or may not be considered negative acceleration. For example, consider Figure 3.

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?

$\bar{a}=\frac{\Delta v}{\Delta t}=\frac{{v}_{f}-{v}_{0}}{{t}_{f}-{t}_{0}}\\$

.2. Find the change in velocity. Since the horse is going from zero to −15.0 m/s, its change in velocity equals its final velocity: Δ

3. Plug in the known values (Δ

$\bar{a}\\$

.$\bar{a}=\frac{\Delta v}{\Delta t}=\frac{-\text{15}\text{.0 m/s}}{1\text{.80 s}}=-8\text{.33 m}{\text{/s}}^{2}\\$

What are the magnitude and sign of displacements for the motions of the subway train shown in parts (a) and (b) of Figure 7?

**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 Δ*x *= *x*_{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*_{f }= 6.70 km and *x*_{0 }= 4.70 km for part (a), and *x*′_{f }= 3.75 km and *x*′_{0 }= 5.25 km for part (b).

2. Solve for displacement in part (a).

**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.

2. Solve for displacement in part (a).

$\Delta x={x}_{f}-{x}_{0}=6.70\text{ km}-4.70\text{ km} = \text{+}2.00\text{ km}\\$

3. Solve for displacement in part (b).$\Delta x′ ={x′}_{f}-{x′}_{0}=\text{3.75 km}-\text{5.25 km} = -\text{1.50 km}\\$

What are the distances traveled for the motions shown in parts (a) and (b) of the subway train in Figure 7?

**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 Example 1. 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 Figure 7, 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 −1.5 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.

2. The displacement for part (b) was −1.5 km. Therefore, the distance between the initial and final positions was 1.50 km, and the distance traveled was 1.50 km.

Suppose the train in Figure 7(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:

**Solution**

1. Identify the knowns. *v*_{0 }= 0 (the trains starts at rest), *v*_{f }= 30.0 km/h, and Δ*t *= 20.0 s.

2. Calculate Δ*v*. Since the train starts from rest, its change in velocity is

3. Plug in known values and solve for the unknown,

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.

2. Calculate Δ

$\Delta v\text{=}\text{+}\text{30.0 km/h}\\$

, where the plus sign means velocity to the right.3. Plug in known values and solve for the unknown,

$\bar{a}\\$

.$\bar{a}=\frac{\Delta v}{\Delta t}=\frac{+\text{30.0 km/h}}{\text{20}\text{.}0 s}\\$

$\bar{a}=\left(\frac{+\text{30 km/h}}{\text{20.0 s}}\right)\left(\frac{{\text{10}}^{3}\text{m}}{\text{1 km}}\right)\left(\frac{\text{1 h}}{\text{3600 s}}\right)=0\text{.}{\text{417 m/s}}^{2}\\$

Now suppose that at the end of its trip, the train in Figure 7(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**

**Solution**

1. Identify the knowns. *v*_{0 }= 30.0 km/h, *v*_{f }= 0 km/h (the train is stopped, so its velocity is 0), and Δ*t *= 8.00 s.

2. Solve for the change in velocity, Δ*v*.

*v* and Δ*t*, and solve for

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 Example 4 and Example 5 are displayed in Figure 10. (We have taken the velocity to remain constant from 20 to 40 s, after which the train decelerates.)2. Solve for the change in velocity, Δ

Δ*v* = *v*_{f } − *v*_{0 } = 0 − 30.0 km/h = − 30.0 km/h

3. Plug in the knowns, Δ$\bar{a}\\$

.$\bar{a}=\frac{\Delta v}{\Delta t}=\frac{-\text{30}\text{.}\text{0 km/h}}{8\text{.}\text{00 s}}\\$

$\bar{a}=\frac{\Delta v}{\Delta t}=\left(\frac{-\text{30.0 km/h}}{\text{8.00 s}}\right)\left(\frac{{\text{10}}^{3}\text{m}}{\text{1 km}}\right)\left(\frac{\text{1 h}}{\text{3600 s}}\right)={\text{-1.04 m/s}}^{2}\text{.}\\$

What is the average velocity of the train in part b of Example 2, 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*′_{f }= 3.75 km, *x*′_{0 }= 5.25 km, Δ*t *= 5.00 min.

2. Determine displacement, Δ*x*′. We found Δ*x*′ to be −1.5 km in Example 2.

3. Solve for average velocity.

4. Convert units.

**Discussion**

The negative velocity indicates motion to the left.

2. Determine displacement, Δ

3. Solve for average velocity.

$\bar{v}=\frac{\Delta x′}{\Delta t}=\frac{-\text{1.50 km}}{\text{5.00 min}}\\$

$\bar{v}=\frac{\Delta x′}{\Delta t}=\left(\frac{-1\text{.}\text{50 km}}{5\text{.}\text{00 min}}\right)\left(\frac{\text{60 min}}{1 h}\right)=-\text{18}\text{.0 km/h}\\$

Finally, suppose the train in Figure 2 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 }= −20 km/h, *v*_{f }= 0 km/h, Δ*t *= 10.0 s.

2. Calculate Δ*v*. The change in velocity here is actually positive, since

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 Example 5, this acceleration can be called a deceleration since it is in the direction opposite to the velocity.

2. Calculate Δ

\Delta v={v}\text{\textunderscore}{f}-{v}\text{\textunderscore}{0}=0-\left(-20 km/h\right)\text{=}\phantom{\rule{0.25}{0ex}}\text{+}\text{20 km/h}\\

3. Solve for $\bar{a}\\$

.$\bar{a}=\frac{\Delta v}{\Delta t}=\frac{+\text{20}\text{.0 km/h}}{\text{10}\text{.}0 s}\\$

$\bar{a}=\left(\frac{+\text{20}\text{.}\text{0 km/h}}{\text{10}\text{.}\text{0 s}}\right)\left(\frac{{\text{10}}^{3}m}{1 km}\right)\left(\frac{1 h}{\text{3600 s}}\right)\text{=}\text{+}0\text{.556 m}{\text{/s}}^{2}\\$

An airplane lands on a runway traveling east. Describe its acceleration.

**Solution**

If we take east to be positive, then the airplane has negative acceleration, as it is accelerating toward the west. It is also decelerating: its acceleration is opposite in direction to its velocity.

- Acceleration is the rate at which velocity changes. In symbols, average acceleration $\bar{a}\\$is

$\bar{a}=\frac{\Delta v}{\Delta t}=\frac{{v}_{f}-{v}_{0}}{{t}_{f}-{t}_{0}}\text{.}\\$ - 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.

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

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

4. 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?

5. 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?

2.

3. A commuter backs her car out of her garage with an acceleration of 1.40 m/s

4. 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 m/s

- acceleration:
- the rate of change in velocity; the change in velocity over time

- average acceleration:
- the change in velocity divided by the time over which it changes

- instantaneous acceleration:
- acceleration at a specific point in time

- deceleration:
- acceleration in the direction opposite to velocity; acceleration that results in a decrease in velocity

3. (a) 1.43 s (b) -2.50 m/s