chapter 2: motion in one dimension how to solve any physics problem: 1. diagram or draw a picture 2....

Chapter 2:Motion in One Dimension

How to solve any physics problem:

1. Diagram or draw a picture2. Units and variables labeled3. Formulas written and ready to use4. Algebra shown with numbers and units5. Solution boxed with correct units and sig. figs.

There are three ways to describe motion.

1. Displacement

2. Velocity

3. Acceleration

Displacement and Velocity

First off, to describe motion, we must specify position.

But to specify position, we need to relate it to another known location

You want to be here

You are here

From this concept, we can create a number line.

For motion in one dimension, it is convenient to use the x-axis.

0

Displacement

Displacement of an object - the change of position of the object

NOTE: displacement is not always equal to the distance traveled.

0

Displacement (x)

x = xf – xi Where x = displacement

xf = final position xi = initial position

Example

What is the displacement if a car moves from an initial position of 10 m to a final position of 80 m?

x =80 m – 10 m = 70 m

ExampleWhat is the displacement if a car moves from an initial position of 80 m to a final position of 20 m?

x =20 m – 80 m = -60 m

Note: the negative sign indicates the direction of displacement.

Distance and DisplacementDistance and Displacement

1. Your displacement is 0, distance traveled is 8 mi

2. Your displacement is 8 mi, distance traveled is 0

3. Your displacement is 0, distance traveled is 0

4. Your displacement is 8 mi, distance traveled is 8 mi

Every morning you drive 4 miles from home to NPHS and then come back home at night using the same route. What is the best statement describing your daily trip? (Assume you do not move far while at NPHS)

VELOCITY – change in displacement over time. Average velocity :

if

ifavg tt

xx

t

xv

)( avgv

Units for velocity

•The SI unit for velocity is m/s; however, a velocity can be given in other units such as:

•Km/sec or km/hr or mi/hr

•When calculating the velocity, be sure all your units match. If they don’t, you need to convert.

Example: It takes 6.0 hours to drive to San Francisco from Newbury Park if driving at 65 mi/hr. How far is the trip?

Newbury Park

San Fransiscot = 6.0 hrs

v = 65 mi/hrs

x = ?

t

xv

vtx

hrshr

mi0.665

milesx 390

Example: The United States is 4,300 km wide. How long (in hours) would it take to drive across the country if someone were to drive at a steady 22 m/s the whole way?

t = ?

v = 22 m/s

x = 4,300 km

t

xv

v

xt

sm

m

/22

000,300,4

sec195455t

km

m

1

1000 m000,300,4

min60

1

sec60

min1 hr hrs54

Speed vs. Velocity

Speed is a SCALAR meaning it is only a number, direction doesn’t matter. Speed = (total distance)/(time)

Velocity is a VECTOR meaning it has a number (magnitude) and direction. Direction matters. Velocity = (displacement)/(time) http://www.youtube.com/watch?v=GKQ

dkS0qG3g

Acceleration

Acceleration – rate of change in velocity over time.

if

ifavg tt

vv

t

va

onaccelerati average is where avga

Units of Acceleration

a

avg

v t

m / ss

ms

1s

ms 2

Example: A car is initially coasting up hill at 8.5 m/s. 4.75 seconds later, the car is now rolling backwards at 2.6 m/s down the same hill. What is the car’s acceleration during this time period?

+ Vo - V f

t = 4.75 sec

Vf = - 2.6 m/s

Vo = 8.5 m/s

t

vva of

75.4

5.86.2

s

sm

75.4

/1.11

2/3.2 sma

Another way to describe motion is to graph displacement and velocity as a function of time it

GLX Time!Plug motion sensor into one of the ports at the top (line the groove up)

A position vs time graph should automatically open

The “Play” button lets you start and stop data taking.

There is no need to erase your graph. Just hit “Play” again to take more data and overwrite your old graph.

Posi

tion

(m

)

Time (s)

(1)Create this graph on your GLX using your motion sensor.

(2) Describe your motion in your notes.

In this case, displacement stays constant with time…there is no movement.

Posi

tion

(m

)

Time (sec)



Here the object has a low constant velocity in the beginning, and then it changes to a higher constant velocity.

Posi

tion

(m

)

Time (s)



This time the object travels away from the starting point at constant velocity, stops for some amount of time and then travels back to its starting point at constant velocity.

Posi

tion

(m

)

Time (s)

Velocity is not constant. Velocity (slope) is increasing at a constant rate.



Good luck on this one!

SPEED vs. VELOCITY Both describe how fast the position is changing with

respect to time.

Speed is a SCALAR quantity. It indicates an amount (magnitude), but not direction.

Velocity is a VECTOR quantity. It indicates both an amount (magnitude) and a direction.

SLOPE REVIEW

12

12xx

yy

run

risemslope

SLOPE REVIEW

Posi

tion

(m

)

Time (s)

Rise

Run

Run

Riseslope

t

x v

Velocity is represented by the SLOPE of the curve on a displacement vs. time graph.

In this class, a positive (+) slope indicates a forward direction and a negative (-) slope indicates a backwards direction (return).

VELOCITY

Use the position graph to answer the following:

b. What is the object’s velocity from 15 – 25 seconds?

a. What is the object’s velocity from 10 – 15 seconds?

c. What is the object’s velocity from 0 – 40 seconds?

0 m/s (object is at rest)

run

riseslopev

ss

mm

1525

6020

= -4 m/s

run

riseslopev

ss

mm

040

040

= -1 m/s

Posi

tion

(m

)

Time (s)

Consider this trip…

Posi

tion

(m

)

Time (s)

INSTANTANEOUS VELOCITY

How fast is the car going at this instant in time?

Posi

tion

(m

)

Time (s)


The slope of a curve at a given point, is equal to the slope of a tangent line at that point.

Posi

tion

(m

)

Time (s)


The slope of this line represents instantaneous velocity at the indicated point.

Describing MotionDescribing Motion

1. Speeds up all the time

2. Slows down all the time

3. Speeds up part of the time

4. Slows down part of the time

5. Speeds up & then slows down

6. Slows down & then speeds up

Hint: What is the meaning of the slope of the

x(t) graph?

The x(t) graph describes a 1-D motion of a train. What must be true about this motion?

Position vs. Time Position vs. Time GraphsGraphs

The x(t) graph displays motions of two trains A and B on parallel tracks. Which statement is true?

1. At t1 both trains have the same velocity

2. At t1 both trains have the same speed

3. At t1 both trains have the same acceleration

4. Both trains have the same velocity sometime before t1

5. The trains never have the same velocity

B

t

x A

t1

End Position Graphs

Motion can be described with a velocity vs. time graph.

Time (s)

Vel

ocity

(m

/s)

constant velocity consta

nt acc

eleration

Velocity vs. Time 1

Time (s)

Vel

ocity

(m

/s)

0

+

-

+ velocity

0 velocity

- velocity

Velocity vs. Time 2

Vel

ocity

(m

/s)

Time (s)

Rise = v

Run = t

t

vSlope aonaccelerati

The Slope of Velocity vs. Time Graphs

Use the velocity graph to answer the following:

b. What is the object’s acceleration from 4 – 7 seconds?

a. What is the object’s velocity from 4 – 7 seconds?

c. What is the object’s acceleration from 2 – 4 seconds?

3 m/s

run

riseslopea

ss

smsm

24

/1/3

= 1 m/s2

0 m/s2 (constant velocity)

Consider a trip…

A train travels at 5 miles per hour for 1 hour. What is its displacement after 1 hour?

t

xv

tvx

hr) mi/hr)(1 5(xmi 5x

Displacement can also be determined by finding the area under the curve of a velocity vs. time graph.

5 mi/hr

1 hour

Vel

ocity

(m

/s)

Time (hrs)

Using a graph

5 mi/hr

1 hour

Area = l x w = 1 hour x 5 mi/hr = 5 miles.

Find the “area under the curve” (AUC).V

eloc

ity (

m/s

)

Time (s)

Displacement vs. Time graph:Slope = velocity

Velocity vs. Time graph:

Slope = acceleration

AUC = displacement

Graphical analysis summary

A blue car moving at a constant speed of 10 m/s passes a red car that is at rest. This occurs at a stoplight the moment that the light turns green.

The red car accelerates from rest at 4 m/s/s for three seconds and then maintains a constant speed for 9 s. The blue car maintains a constant speed of 10 m/s for the entire 12 seconds.

Draw a “v vs. t” graph for:

The answer

QUICK QUIZ 2.3Parts (a), (b), and (c) of the figure below represent three graphs of the velocities of different objects moving in straight-line paths as functions of time. The possible accelerations of each object as functions of time are shown in parts (d), (e), and (f). Match each velocity-time graph with the acceleration-time graph that best describes the motion.

One way to look at the motion of an object is by using a motion map.

Imagine a toy car traveling along a piece of paper and dropping a dot of ink at a given time interval (say 1 drop every second). It could produce a trail that looks like this:

distance

Same time intervalBetween dots

Position and directionof object at a given

instant in timeTime starts

at zero

distance




at zero

If the car produced the following motion map, how long did it take the car to travel the length of the paper?

5 sec1 sec 2 sec 3 sec 4 sec 6 sec 8 sec7 sec

8 seconds total

distance




at zero

Describe the motion of the car? Stopped, constant velocity, accelerating or

decelerating?

How do you know?

Graphing Examples

a. Draw the motion map for the following:

•Object accelerates for 3 seconds.

•Then travels at a constant velocity for 2 seconds

•Then decelerates for 3 seconds

•Stops for 2 seconds

•Then returns to the start in 4 seconds at a constant velocity.

b. Sketch the velocity graph for the above motion.

time

velo

city

v

vo

ttime

Velocity with Constant Acceleration

time t

vf

vovelo

city Rise = v

= vf - vo

Run = t = (t – 0) = t

slopevt

v

f v

o

ta

vfv

oat

(p. 35)

time t

vf

vivelo

city

Find the AUC…

Break the area into two parts…

x = vit + ½ t(vf-vi)

Area = (l x w) + (½b x h)

time t

vf

vivelo

city

Find the AUC…

AUC = x = vot + ½ t(vf-vo)

x = ½ vot + ½ vft simplification

x = vot + ½ vft – ½ vot distribution

x = ½ (vo+vf)t p.(35)

1) vf = vo + at

x = ½ [vo + (at +vo)]t substitution

and 2) x = ½ (vo + vf)t

x = ½ (2vo + at)t combining terms

x = vot + ½ at2 (p. 36)

We now know…

By substitution of equation #1 into #2

Solve Eq #1 for t & substitute into Eq #2…

1) vf = vo + at

tvf v

o

a

2) x = ½ (vo + vf)t

(p. 36)

x (vov

f)

2(v

f v

o)

a

x vf

2 vo

2

2av

f

2 vo

2 2ax

1. a = acceleration

2s

m

time

velocity

2. x = displacement m

3. vf = final velocity s

m

4. vi = initial velocity

5. t = time sec

s

m

5 Parameters of Motion

To solve a constant acceleration problem, you must know, or be able to find, three of the five parameters.

Then use the following equations to solve for the other two:

vf = vo + at

x = ½(vo + vf)t

x = vot + ½a(t)2

vf2 = vo

2 + 2ax

Example…Example: A jet plane lands with a velocity of 100 m/sec and can slow down (-acceleration) at a maximum rate of –5.0 m/s2. Find (a) the time required for the plane to come to rest, and (b) the minimum size of the runway.

a) vi = +100m/svf= 0 m/sa = -5.0 m/s2

vf = vi + at

0m/s = 100m/s + (-5m/s2)t

t = 20s

vf2 = vi

2 + 2ax

0m/s = (100m/s)2 + 2(- 5m/s2) x

x = 1000 m

b) Solve for x

Example: A train is traveling down a straight track at 20 m/sec when the engineer applies the brakes, resulting in an acceleration of –1m/sec2 as long as the train is in motion. How far does the train travel in the first 6 seconds after the breaks are applied?

vi = 20 m/s

a = -1 m/sec2

t = 6 sec

x = ?

x = vit + ½ at2

x = (20m/s)(6s) + ½ (-1m/s2)(6s)2

x = 120m – 18m = 102 m

x = 100 m (sig. figs!)

Example: A racing car starting from rest accelerates at a rate of 5.00 m/s2. What is the velocity of the car after it has traveled 100. ft?

vi = 0 m/s

a = 5.0 m/sec2

x = 100 ft= 30.5 m

vf = ?

vf2 = vi

2 + 2ax

vf2 = 0 + 2(5m/s2)(30.5m)

vf = 17.5 m/s

22f /sm 305v

chapter 2: motion in one dimension how to solve any physics problem: 1. diagram or draw a picture 2....

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