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HomeStudy GuidesPhysics

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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

- Describe a straight-line graph in terms of its slope and
*y*-intercept. - Determine average velocity or instantaneous velocity from a graph of position vs. time.
- Determine average or instantaneous acceleration from a graph of velocity vs. time.
- Derive a graph of velocity vs. time from a graph of position vs. time.
- Derive a graph of acceleration vs. time from a graph of velocity vs. time.

$y=\text{mx}+b\\$

$\bar{v}\\$

and the intercept is displacement at time zero—that is, $y=\text{mx}+b\\$

gives$x=\bar{v}t+{x}_{0}\\$

$x={x}_{0}+\bar{v}t\\$

.
The slope of the graph of displacement *x* vs. time *t* is velocity *v*.

$\text{slope}=\frac{\Delta x}{\Delta t}=v\\$

Notice that this equation is the same as that derived algebraically from other motion equations in Motion Equations for Constant Acceleration in One Dimension.
Find the average velocity of the car whose position is graphed in Figure 2.

**Strategy**

The slope of a graph of *x* vs. *t* is average velocity, since slope equals rise over run. In this case, rise = change in displacement and run = change in time, so that

Since the slope is constant here, any two points on the graph can be used to find the slope. (Generally speaking, it is most accurate to use two widely separated points on the straight line. This is because any error in reading data from the graph is proportionally smaller if the interval is larger.)

**Solution**

1. Choose two points on the line. In this case, we choose the points labeled on the graph: (6.4 s, 2000 m) and (0.50 s, 525 m). (Note, however, that you could choose any two points.)

2. Substitute the*x* and *t* values of the chosen points into the equation. Remember in calculating change (Δ) we always use final value minus initial value.

yielding

**Discussion**

This is an impressively large land speed (900 km/h, or about 560 mi/h): much greater than the typical highway speed limit of 60 mi/h (27 m/s or 96 km/h), but considerably shy of the record of 343 m/s (1234 km/h or 766 mi/h) set in 1997.

$\text{slope}=\frac{\Delta x}{\Delta t}=\bar{v}\\$

.2. Substitute the

$\bar{v}=\frac{\Delta x}{\Delta t}=\frac{\text{2000 m}-\text{525 m}}{6\text{.}\text{4 s}-0\text{.}\text{50 s}}\\$

,$\bar{v}=\text{250 m/s}$

.The graph of displacement versus time in Figure 3(a) is a curve rather than a straight line. The slope of the curve becomes steeper as time progresses, showing that the velocity is increasing over time. The slope at any point on a displacement-versus-time graph is the instantaneous velocity at that point. It is found by drawing a straight line tangent to the curve at the point of interest and taking the slope of this straight line. Tangent lines are shown for two points in Figure 3(a). If this is done at every point on the curve and the values are plotted against time, then the graph of velocity versus time shown in Figure 3(b) is obtained. Furthermore, the slope of the graph of velocity versus time is acceleration, which is shown in Figure 3(c).

Calculate the velocity of the jet car at a time of 25 s by finding the slope of the *x* vs. *t* graph in the graph below.

**Strategy**

The slope of a curve at a point is equal to the slope of a straight line tangent to the curve at that point. This principle is illustrated in Figure 5, where Q is the point at *t*=25 s.

**Solution**

1. Find the tangent line to the curve at *t *= 25 s.

2. Determine the endpoints of the tangent. These correspond to a position of 1300 m at time 19 s and a position of 3120 m at time 32 s.

3. Plug these endpoints into the equation to solve for the slope,*v*.

Thus,

**Discussion**

This is the value given in this figure’s table for *v* at *t *= 25 s. The value of 140 m/s for *v*_{Q} is plotted in Figure 5. The entire graph of *v* vs. *t* can be obtained in this fashion.

2. Determine the endpoints of the tangent. These correspond to a position of 1300 m at time 19 s and a position of 3120 m at time 32 s.

3. Plug these endpoints into the equation to solve for the slope,

$\text{slope}={v}_{Q}=\frac{{\Delta x}_{Q}}{{\Delta t}_{Q}}=\frac{\left(\text{3120 m}-\text{1300 m}\right)}{\left(\text{32 s}-\text{19 s}\right)}\\$

${v}_{Q}=\frac{\text{1820 m}}{\text{13 s}}=\text{140 m/s.}\\$

The slope of a graph of velocity *v* vs. time *t* is acceleration *a**.*

$\text{slope}=\frac{\Delta v}{\Delta t}=a\\$

Additional general information can be obtained from Figure 5 and the expression for a straight line,

$y=\text{mx}+b\\$

.In this case, the vertical axis

$v={v}_{0}+\text{at}\\$

.
A general relationship for velocity, acceleration, and time has again been obtained from a graph. Notice that this equation was also derived algebraically from other motion equations in Motion Equations for Constant Acceleration in One Dimension.It is not accidental that the same equations are obtained by graphical analysis as by algebraic techniques. In fact, an important way to

Calculate the acceleration of the jet car at a time of 25 s by finding the slope of the *v* vs. *t* graph in Figure 6(b).

**Strategy**

The slope of the curve at *t *= 25 s is equal to the slope of the line tangent at that point, as illustrated in Figure 6(b).

**Solution**

Determine endpoints of the tangent line from the figure, and then plug them into the equation to solve for slope, *a*.

**Discussion**

Note that this value for *a* is consistent with the value plotted in Figure 6(c) at *t *= 25 s.

$\text{slope}=\frac{\Delta v}{\Delta t}=\frac{\left(\text{260 m/s}-\text{210 m/s}\right)}{\left(\text{51 s}-1.0 s\right)}\\$

$a=\frac{\text{50 m/s}}{\text{50 s}}=1\text{.}0 m{\text{/s}}^{2}\\$

.
(a) The ship moves at constant velocity and then begins to decelerate at a constant rate. At some point, its deceleration rate decreases. It maintains this lower deceleration rate until it stops moving.

(b) A graph of acceleration vs. time would show zero acceleration in the first leg, large and constant negative acceleration in the second leg, and constant negative acceleration.

(b) A graph of acceleration vs. time would show zero acceleration in the first leg, large and constant negative acceleration in the second leg, and constant negative acceleration.

- Graphs of motion can be used to analyze motion.
- Graphical solutions yield identical solutions to mathematical methods for deriving motion equations.
- The slope of a graph of displacement
*x*vs. time*t*is velocity*v*. - The slope of a graph of velocity
*v*vs.*t*graph is acceleration*a*. - Average velocity, instantaneous velocity, and acceleration can all be obtained by analyzing graphs.

1. (a) Explain how you can use the graph of position versus time in Figure 9 to describe the change in velocity over time. Identify: (b) the time (*t*_{a}, *t*_{b}, *t*_{c}, *t*_{d}, or *t*_{e}) at which the instantaneous velocity is greatest, (c) the time at which it is zero, and (d) the time at which it is negative.

2. (a) Sketch a graph of velocity versus time corresponding to the graph of displacement versus time given in Figure 10. (b) Identify the time or times (*t*_{a}, *t*_{b}, *t*_{c}, etc.) at which the instantaneous velocity is greatest. (c) At which times is it zero? (d) At which times is it negative?

2. (a) Sketch a graph of velocity versus time corresponding to the graph of displacement versus time given in Figure 10. (b) Identify the time or times (

5. Consider the velocity vs. time graph of a person in an elevator shown in Figure 13 Suppose the elevator is initially at rest. It then accelerates for 3 seconds, maintains that velocity for 15 seconds, then decelerates for 5 seconds until it stops. The acceleration for the entire trip is not constant so we cannot use the equations of motion from Motion Equations for Constant Acceleration in One Dimension for the complete trip. (We could, however, use them in the three individual sections where acceleration is a constant.) Sketch graphs of: (a) position vs. time and (b) acceleration vs. time for this trip.

Note: There is always uncertainty in numbers taken from graphs. If your answers differ from expected values, examine them to see if they are within data extraction uncertainties estimated by you.

1. a) By taking the slope of the curve in Figure 14, verify that the velocity of the jet car is 115 m/s at
2. Using approximate values, calculate the slope of the curve in Figure 16 to verify that the velocity at *t*=10.0 s is 0.208 m/s. Assume all values are known to 3 significant figures.

5. Construct the displacement graph for the subway shuttle train as shown in Figure 2 from Acceleration (shown again below). Your graph should show the position of the train, in kilometers, from t = 0 to 20 s. You will need to use the information on acceleration and velocity given in the examples for this figure.

6. (a) Take the slope of the curve in Figure 11 to find the jogger’s velocity at*t *= 2.5 s. (b) Repeat at 7.5 s. These values must be consistent with the graph in Figure 18.

3. Using approximate values, calculate the slope of the curve in Figure 16 to verify that the velocity at t = 30.0 s is 0.238 m/s. Assume all values are known to 3 significant figures.

4. By taking the slope of the curve in Figure 17, verify that the acceleration is 3.2 m/s^{2} at *t* = 10 s.

4. By taking the slope of the curve in Figure 17, verify that the acceleration is 3.2 m/s

6. (a) Take the slope of the curve in Figure 11 to find the jogger’s velocity at

7. A graph of *v*(*t*) is shown for a world-class track sprinter in a 100-m race. (See Figure 21). (a) What is his average velocity for the first 4 s? (b) What is his instantaneous velocity at *t *= 5 s? (c) What is his average acceleration between 0 and 4 s? (d) What is his time for the race?

8. Figure 22 shows the displacement graph for a particle for 5 s. Draw the corresponding velocity and acceleration graphs.

- independent variable:
- the variable that the dependent variable is measured with respect to; usually plotted along the
*x*-axis

- dependent variable:
- the variable that is being measured; usually plotted along the
*y*-axis

- slope:
- the difference in
*y*-value (the rise) divided by the difference in*x*-value (the run) of two points on a straight line

- y-intercept:
- the
*y*-value when*x*= 0, or when the graph crosses the*y*-axis

3.

$v=\frac{\left(\text{11.7}-6.95\right)\times {\text{10}}^{3}\text{m}}{\left(40\text{.}\text{0 - 20}.0\right)\text{s}}=\text{238 m/s}\\$

5.

7. (a) 6 m/s (b) 12 m/s (c) 3 m/s