Chapter 2

Copyright © 2015 Pearson Education Inc Modified SH 8/15.

PowerPoint® Lectures forUniversity Physics, Thirteenth Edition – Hugh D. Young and Roger A. Freedman

Chapter 2

Motion Along a Straight Line

Copyright © 2015 Pearson Education Inc. Modified SH 8/15

Goals for Chapter 2

• Describe straight-line motion in terms of velocity and acceleration

• Distinguish between average and instantaneous velocity and acceleration

• Interpret graphs

– position versus time, x(t) versus t

• (slope = velocity!)


Goals for Chapter 2

• Describe straight-line motion in terms of velocity and acceleration

• Distinguish between average and instantaneous velocity and acceleration

• Interpret graphs

– velocity versus time, v(t) versus t

• Slope = acceleration!


Goals for Chapter 2

• Understand straight-line motion with constant acceleration

• Examine freely falling bodies

• Analyze straight-line motion when the acceleration is not constant


Introduction

• Kinematics is the study of motion.

• Displacement, velocity and acceleration are important physical quantities.

• A bungee jumper speeds up during the first part of his fall and then slows to a halt.

Bryan Spencer

If picture is changed to 13e opener of a bungee jumper, then the picture won't match lecture text.


Displacement vs. Distance

• Displacement (blue line) = how far the object is from its starting point, regardless of path

• Distance traveled (dashed line) is measured along the actual path.



Q: You make a round trip to the store 1 mile away.

•What distance do you travel?

•What is your displacement?



Q: You walk 70 meters across the campus, hear a friend call from behind, and walk 30 meters back the way you came to meet her.

•What distance do you travel?

•What is your displacement?


Displacement is written:

•SIGN matters! Direction matters!

• It is a VECTOR!!

<= Positive displacement

Displacement is negative =>



Speed is how far an object travels in a given time interval (in any direction)

Speed vs. Velocity

Ex: Go 10 miles to Chabot in 30 minutes

Average speed = 10 mi / 0.5 hr = 20 mph


Velocity includes directional information:

Speed vs. Velocity

Ex: Go 20 miles on 880 Northbound to Chabot in 20 minutes

Average velocity = 20 mi / 0.333 hr = 60 mph NORTH

VECTOR!


Speed vs. Velocity




Velocity includes directional information:

Speed vs. Velocity



VECTOR!


Speed is a SCALAR

• 60 miles/hour, 88 ft/sec, 27 meters/sec

Velocity is a VECTOR

• 60 mph North

• 88 ft/sec East

• 27 m/s @ azimuth of 173 degrees

Speed vs. Velocity


• Position of runner as a function of time is plotted as moving along the x axis of a coordinate system.

• During a 3.00-s time interval, a runner’s position changes from x1 = 50.0 m to x2 = 30.5 m

• What was the runner’s average velocity?

Example of Average Velocity


During a 3.00-sec interval, runner’s position changes from x1 = 50.0 m to x2 = 30.5 m What was the runner’s average velocity?

Vavg = (30.5 - 50.0) meters/3.00 sec

= -6.5 m/s in the x direction.

The answer must have value1,

units2, & DIRECTION3

Example of Average Velocity

Note! x = FINAL – INITIAL

position


During a 3.00-s time interval, the runner’s position changes from x1 = 50.0 m to x2 = 30.5 m. What was the runner’s average speed?

Savg = |30.5-50.0| meters/3.00 sec = 6.5 m/s

The answer must have value & units

Example of Average SPEED


Negative velocity???

• Average x-velocity is negative during a time interval if particle moves in negative x-direction for that time interval.


Displacement, time, and average velocity

• A racing car starts from rest, and after 1 second is 19 meters from the starting line.

• After the next 3 seconds, the car is 277 meters from the starting line.

• What was its average velocity in those 3 seconds?


Displacement, time, and average velocity—Figure 2.1

Q A racing car starts from rest, and after 1 second is 19 meters from the starting line. After the next 3 seconds, the car is 277 meters from the starting line. What was its average velocity in those 3 seconds?

Solution Method:

1. What do you know? What do you need to find? What are the units? What might be a reasonable estimate?

2. DRAW it! Visualize what is happening. Create a coordinate system, label the drawing with everything.

3. Find what you need from what you know



A racing car starts from rest, and after 1 second is 19 meters from the starting line. After the next 3 seconds, the car is 277 meters from the starting line. What was its average velocity in those 3 seconds?

• “starts from rest” = initial velocity = 0

• car moves along straight (say along an x-axis)

– has coordinate x = 0 at t=0 seconds

– has coordinate x=+19 meters at t =1 second

– Has coordinate x=+277 meters at t = 1+3 = 4 seconds.



Q A racing car starts from rest, and after 1 second is 19 meters from the starting line. After the next 3 seconds, the car is 277 meters from the starting line. What was its average velocity in those 3 seconds?


A position-time graph—Figure 2.3

• A position-time graph (an “x-t” graph) shows the particle’s position x as a function of time t.

• Average x-velocity is related to the slope of an x-t graph.


Instantaneous speed is the average speed in the limit as the time interval becomes infinitesimally short.

Ideally, a speedometer would measure instantaneous speed; in fact, it measures average speed, but over a very short time interval.

Note: It doesn’t measure direction!

Instantaneous Speed


Instantaneous velocity is the average velocity in the limit as the time interval becomes infinitesimally short.

Velocity is a vector; you must include direction!

V = 27 m/s west…

Instantaneous Speed


Instantaneous velocity

• The instantaneous velocity is the velocity at a specific instant of time or specific point along the path and is given by vx = dx/dt.


Instantaneous velocity

• The instantaneous velocity is the velocity at a specific instant of time or specific point along the path and is given by vx = dx/dt.

• The average speed is not the magnitude of the average velocity!


A jet engine moves along an experimental track (the x axis) as shown.

Its position as a function of time is given by the equation

x = At2 + B

where A = 2.10 m/s2 and B = 2.80 m.

Instantaneous Velocity Example


A jet engine’s position as a function of time is x = At2 + B, where A = 2.10 m/s2 and B = 2.80 m.

(a)Determine the displacement of the engine during the time interval from t1 = 3.00 s to t2 = 5.00 s.

(b)Determine the average velocity during this time interval.

(c)Determine the magnitude of the instantaneous velocity at t = 5.00 s.




(a)Determine the displacement of the engine during the time interval from t1 = 3.00 s to t2 = 5.00 s.

@ t = 3.00 s x1 = 21.7

@ t = 5.00 s x2 = 55.3

x2 – x1 = 33.6 meters in +x direction




b) Determine the average velocity during this time interval.

Vavg = 33.6 m/ 2.00 sec

= 16.8 m/s

in the + x direction




(c) Determine the magnitude of the instantaneous velocity at t = 5.00 s

|v| = dx/dt @ t = 5.00 seconds

= 21.0 m/s



Acceleration = the rate of change of velocity.

Units: meters/sec/sec or m/s^2 or m/s2 or ft/s2

Since velocity is a vector, acceleration is ALSO a vector, so direction is crucial…

A = 2.10 m/s2 in the +x direction

Acceleration


A car accelerates along a straight road from rest to 90 km/h in 5.0 s.

What is the magnitude of its average acceleration?

KEY WORDS:“straight road” = assume constant acceleration

“from rest” = starts at 0 km/h

Acceleration Example


A car accelerates along a straight road from rest to 90 km/h in 5.0 s What is the magnitude of its average acceleration?

|a| = (90 km/hr – 0 km/hr)/5.0 sec = 18 km/h/sec along road

better – convert to more reasonable units

90 km/hr = 90 x 103 m/hr x 1hr/3600 s = 25 m/s

So|a| = 5.0 m/s2 (note – magnitude only is requested)



Acceleration

Acceleration = the rate of change of velocity.


(a) If the velocity of an object is zero, does it mean that the acceleration is zero?

(b) If the acceleration is zero, does it mean that the velocity is zero?

Think of some examples.

Acceleration vs. Velocity?


An automobile is moving to the right along a straight highway. Then the driver puts on the brakes.

If the initial velocity (when the driver hits the brakes) is v1 = 15.0 m/s, and it takes 5.0 s to slow down to v2 = 5.0 m/s, what was the car’s average acceleration?



An automobile is moving to the right along a straight highway. Then the driver puts on the brakes. If the initial velocity (when the driver hits the brakes) is v1 = 15.0 m/s, and it takes 5.0 s to slow down to v2 = 5.0 m/s, what was the car’s average acceleration?



A semantic difference between negative acceleration and deceleration:

“Negative” acceleration is acceleration in the negative direction (defined by coordinate system).

“Deceleration” occurs when the acceleration is opposite in direction to the velocity.



Finding velocity on an x-t graph

• At any point on an x-t graph, the instantaneous x-velocity is equal to the slope of the tangent to the curve at that point.


Motion diagrams

• A motion diagram shows position of a particle at various instants, and arrows represent its velocity at each instant.


Average acceleration• Acceleration describes the rate of change of velocity with time.

• The average x-acceleration is aav-x = vx/t.


Instantaneous acceleration

• The instantaneous acceleration is ax = dvx/dt.

• Consider an accelerating racing car:


Average Acceleration

Q A racing car starts from rest, and after 1 second is 19 meters from the starting line.

After the next 3 seconds, the car is 277 meters from the starting line.

What was its average acceleration in the first second?

What was its average acceleration in the first 4 seconds?


Finding acceleration on a vx-t graph

• Use Velocity vs. Time (x - t) graph to find instantaneous acceleration & average acceleration.


A vx-t graph and a motion diagram

• Figure 2.13 shows the vx-t graph and the motion diagram for a particle.


An x-t graph and a motion diagram

• Figure 2.14 shows the x-t graph and the motion diagram for a particle.


Motion with constant acceleration—Figures 2.15 and 2.17

• For a particle with constant acceleration, the velocity changes at the same rate throughout the motion.


Acceleration given x(t)

A particle is moving in a straight line with its position is given by x = (2.10 m/s2)t2 + (2.80 m).

Calculate (a) its average acceleration during the interval from t1 = 3.00 s to t2 = 5.00 s, & (b) its instantaneous acceleration as a function of time.



Calculate (a) its average acceleration during the interval from t1 = 3.00 s to t2 = 5.00 s

V = dx/dt = (4.2 m/s) t

V1 = 12.6 m/s

V2 = 21 m/s

v/t = 8.4 m/s/2.0 s = 4.2 m/s2




Calculate (b) its instantaneous acceleration as a function of time.



Graph shows Velocity as a function of time for two cars accelerating from 0 to 100 km/h in a time of 10.0 s Compare (a) the average acceleration; (b) instantaneous acceleration; and (c) total distance traveled for the two cars.

Analyzing acceleration


Velocity as a function of time for two cars accelerating from 0 to 100 km/h in a time of 10.0 s Compare (a) the average acceleration; (b) instantaneous acceleration; and (c) total distance traveled for the two cars.

Same final speed in time =>

Same average acceleration

But Car A accelerates faster…

Analyzing acceleration


Constant Acceleration Equations

FIVE key variables:

xdisplacement vinitial , vfinal , acceleration time

FIVE key equations:

x = ½ (vi+vf)t

x = vit + ½ at2

x = vft – ½ at2

vf = vi + at

vf2 = vi

2 + 2ax


The equations of motion with constant acceleration

vx v0x axt

xx0v0xt12

axt2

vx2 v0x

2 2ax x x0

x x0 v0x vx

2

t

• Initial velocity, final velocity, acceleration, time

• Displacement (x – x0), initial velocity, time, acceleration

• Initial velocity, final velocity, acceleration, displacement

• Displacement, initial velocity, final velocity, time

• Displacement (x – x0), final velocity, time, acceleration

Equation of Motion Variables Present

x = x0 + vxt – 1/2axt2


The equations of motion with constant acceleration

vx v0x axt

xx0v0xt12

axt2

vx2 v0x

2 2ax x x0

x x0 v0x vx

2

t





Equation of Motion Find 3 of 4, solve for 4th!





• Displacement (x – x0), final velocity, time, acceleration

x = x0 + vxt – 1/2axt2


A motorcycle with constant acceleration

• What is position and velocity at t = 2.0 sec?

• Where is he when speed = 25 m/s?


Two bodies with different accelerations (Ex. 2.5)

• A motorist traveling at a constant 15m/s passes school crossing where speed limit is 20 mph.

• Just as motorist passes the sign a police officer stopped on a motorcycle starts off in pursuit with acceleration 3.0 m/s2.



• WHEN does officer catch up to the car?

• How FAST is officer going at that time?

• How FAR has each travelled?



• Two different initial/final velocities and accelerations

• Two different equations of position in time x(t)

– Xm(t) for the motorist gives his position as f(time)

– Xp(t) for the police officer gives *his* position

• But…time & displacement are linked!



• Two different initial/final velocities and accelerations

• TIME and displacement are linked!


Freely falling bodies

• Free fall is the motion of an object under the influence of only gravity.

• In the figure, a strobe light flashes with equal time intervals between flashes.

• The velocity change is the same in each time interval, so the acceleration is constant.


Freely Falling Objects

In the absence of air resistance,

all objects fall with the same acceleration,

although this may be tricky to tell by testing in an environment where there is air resistance.


The acceleration due to gravity at the Earth’s surface is approximately 9.80 m/s2.

At a given location on the Earth and in the absence of air resistance, all objects fall with the same constant acceleration.



Example: Falling from a tower.

Suppose that a ball is dropped (v0 = 0) from a tower 70.0 m high. How far will it have fallen after a time t1 = 1.00 s, t2 = 2.00 s, and t3 = 3.00 s? Ignore air resistance.



Example: Thrown down from a tower

Suppose a ball is thrown downward with an initial velocity of 3.00 m/s, instead of being dropped.

(a) What then would be its position after 1.00 s and 2.00 s?

(b) What is its speed after 1.00 s and 2.00 s?

Compare with the speeds of a dropped ball.


Example: Ball thrown up!

A person throws a ball upward into the air with an initial velocity of 15.0 m/s

Calculate (a) how high it goes, & (b) how long the ball is in the air before it comes back to the hand. Ignore air resistance.



Up-and-down motion in free fall

• An object is in free fall even when it is moving upward.


Is the acceleration zero at the highest point?—Figure 2.25

• The vertical velocity, but not the acceleration, is zero at the highest point.


Ball thrown upward (cont.)

Consider again a ball thrown upward, & calculate

(a) how much time it takes for the ball to reach the maximum height,

(b) the velocity of the ball when it returns to the thrower’s hand (point C).


Give examples to show the error in these two common misconceptions: (1) that acceleration and velocity are always in the same direction

(2) that an object thrown upward has zero acceleration at the highest point.



The quadratic formula

For a ball thrown upward at an initial speed of 15.0 m/s, calculate at what time t the ball passes a point 8.00 m above the person’s hand.


Ball thrown at edge of cliff

• A ball is thrown upward at a speed of 15.0 m/s by a person on the edge of a cliff, so that the ball can fall to the base of the cliff 50.0 m below.

• (a) Ignoring air resistance, how long does it take the ball to reach the base of the cliff?

• (b) What is the total distance traveled by the ball?



A ball is thrown upward at a speed of 15.0 m/s by a person on the edge of a cliff, so that the ball can fall to the base of the cliff 50.0 m below. Step 1:

DRAW IT! Establish your coordinate system!




• Step 2:What do you know?

• V0 = +15 m/s

• a = - 9.8 m/s/s

• y = - 50 m




• Step 3:What do you need to find? Units?

• “How long does it take the ball to reach the base of the cliff?”

• TIME!




• Step 4:What equation relates what you need to what you know?

• y = v0* t + ½ at2

• -50 = +15t - 4.9 t2

• Solve the quadratic to find…

• t = 5.07 sec OR – 2.01 sec ???


Variable Acceleration; Integral Calculus

Deriving the kinematic equations through integration:

For constant acceleration,



Then:

For constant acceleration,



Example: Integrating a time-varying acceleration.

An experimental vehicle starts from rest (v0 = 0) at t = 0 and accelerates at a rate given by a = (7.00 m/s3)t. What is

(a) its velocity and

(b) its displacement 2.00 s later?


Graphical Analysis and Numerical Integration

The total displacement of an object can be described as the area under the v-t curve:


Graphical Analysis and Numerical Integration

Similarly, the velocity may be written as the area under the a-t curve.

However, if the velocity or acceleration is not integrable, or is known only graphically, numerical integration may be used instead.


Example: Numerical integration

An object starts from rest at t = 0 and accelerates at a rate a(t) = (8.00 m/s4)t2. Determine its velocity after 2.00 s using numerical methods.


Velocity and position by integration

• The acceleration of a car is not always constant.

• The motion may be integrated over many small time intervals to give .00 0

and t t

x ox x xv v a dt x x v dt


Motion with changing acceleration (Ex. 2.9)

• Look at the example where acceleration DOES vary in time. ax = 2.0 m/s2 – (0.10 m/s3)t

• You cannot use the set equations - you must INTEGRATE

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Chapter 2

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