One Dimensional Motion. Distance How far something has moved.

One Dimensional

Motion

Distance•How far

something has moved

Distance•Scalar quantity

Displacement•How far something

is from its starting position

Displacement

•A vector quantity

Time•The interval between two occurrences

Uniform Motion•Equal displacement

occurs during successive equal

time intervals

Uniform Motion•Velocity is

constant during uniform motion

t (s) 0 1 2 3 4 5 6P (m) 0 1 2 3 4 5 6

Position (P) vs Time Chart (t)

Distance vs Time GraphDistance vs Time

0

1

2

3

4

5

6

7

0 1 2 3 4 5 6 7

Time (s)

Dis

tan

ce (

m)

Distance

Slope•Slope = rise/run

•Slope = y/x

Slope•On a distance vs time graph:

•Slope = d/t

Slope•Slope = d/t

•Slope = velocity

Average Velocity•v = d/t

•v = d1 – d0

t1 – t0

Displacement

•d1 = d0 + vt1

•d = d0 + vt

t (s) 0 1 2 3 4 5 6P (m) 0 2 8 16 36 44 48

Position (P) vs Time Chart (t)

Distance vs Time GraphDistance vs Time

0

10

20

30

40

50

60

0 2 4 6 8

Time (s)

Dis

tan

ce (

m)

Distance

Distance vs Time GraphDictance vs Time

0

10

20

30

40

50

60

0 1 2 3 4 5 6

Time (s)

Dis

tan

ce (

m)

Dictance

Displacement vs Time

-30

-20

-10

0

10

20

30

40

0 10 20 30 40 50 60 70 80

Time (s)

Dis

pla

cem

en

t (m

)

Displacement

Accelerationa = v/t

a = v1 - v0

t1 - t0

Velocity

v = v0 + at

vf = vi + at

Displacement

d = d0 + v0t + ½ at2

Displacement

df =

di + vit + ½ at2

v2 = v0

2 + 2 a(d1 – d0)

v2 = v02 + 2ad

vf2 = vi

2 + 2ad

v = v0 + atd = d0 + v0t + ½ at2

v2 = v02 + 2ad

v = v0 + at

vf = vi + at

d =

d0 + v0t + ½ at2

df =

di + vit + ½ at2

df =

di + vit + ½ at2

d = vit + ½ at2

v2 = v02 + 2ad

vf2 = vi

2 + 2ad

vf = vi + at

d = vit + ½ at2

vf2 = vi

2+ 2ad

DrillA ball is dropped from

490 m. Calculate its:

vf & tair

A car starts 200.0 m west of town, and moves at 15 m/s east.1) write its best equation2) where will the car be at 10.0 s3) When will the car be in town

Determining Instantaneous Velocity

1) Graph the d/t data

2) Draw tangent to point of interest

3) Determine slope of tangent


0

5

10

15

20

0 1 2 3 4 5 6 7 8 9 10

Time (s)

Dis

pla

ce

me

nt

(m)

Velocity vs Time Graphs

Velocity vs Time

0

5

10

15

20

25

30

0 2 4 6 8 10Time (s)

Ve

loc

ity

(m/s

)

a = sloped = xy or vtd = area

Velocity vs Time

0102030405060708090

100

0 1 2 3 4 5 6 7 8 9 10Time (s)

Ve

loc

ity

(m/s

)

a = slope = y/x = v/t = 62/5 = 12.4 m/s2

Velocity vs Time

0102030405060708090

100

0 1 2 3 4 5 6 7 8 9 10Time (s)

Ve

loc

ity

(m/s

) d = area under curve

Define each of the following

•DistanceDisplacement

•Speed Velocity

•Acceleration


-15

-10

-5

0

5

10

15

0 5 10 15 20 25 30

Time (s)

Dis

pla

cem

en

t (m

)

A

B

C

D

Describe the motion for each series

Drill: •The velocity of a car is increased from 25 to 75 m/s west in 10.0 s. Calculate: a & d

vf = vi + at

d = vit + ½ at2

vf2 = vi

2+ 2ad


-30

-20

-10

0

10

20

30

40

0 10 20 30 40 50 60 70 80

Time (s)

Dis

pla

cem

en

t (m

)

Displacement

Describe the motion between each interval


0

2

4

6

8

10

0 1 2 3 4 5 6

Time (s)

Dis

pla

cem

en

t (m

)

A

B

C

D

Describe the motion of each series

Draw a position time graph for a person who walks

uniformly from the positive side of the origin back thru

the origin to the negative side. Repeat for the negative side.

Make the following conversions:

a) 10 m/s to km/hrb) 72 mph to m/s

1.6 km/mile

Draw a position time graph of a person who walks one

block briskly, waits at a traffic light, walks the next

block slowly, waits at another light, then runs the

last block.

A truck starts 400.0 m east of town, and moves at 12 m/s west Find the time & place where the car from the last problem & the truck will be at the same place

A car increases its velocity from 4.0 m/s to 36 m/s over 4.0 s.

Calculate: a & d

The same car slows from 36 m/s to 15 m/s

in 3.0 s. 1) Calculate the average

acceleration & dis

A car accelerates from 15 m/s to 25 m/s in

125 m. 1) Calculate its time &

acceleration

Drill:A car is coasting backwards at 3.0 m/s when its engine starts. After 2.5 s the car is

going 4.5 m/s. Calculate a & d

Motion Variables:•vi

•vf

•a• t•d

Make a chart

like the one to

the right

vi

vf

a

t

d

A car going 4.0 m/s accelerates at 3.0

m/s2 for 4.0 s. 1) Calculate: vf & d

A car slows from 44 m/s to 22 m/s

in 11 s.Calculate: a & d

Motion Affected by

Gravity

Gravity•A force of

attraction between two masses

Gravity•This force causes objects to accelerate towards each other

Gravity•The acceleration of gravity is relatively constant over the Earth’s surface

Acceleration of Gravity (ag or g)

9.81 m/s2

Down or (-)

GravityAny object in air will

have a vertical acceleration of

-9.81 m/s2

A ball is dropped from a 0.49 km cliff. The

acceleration of gravity is -9.8 m/s2.

Calculate: vmax & t

HW: A ball is thrown straight up at 19.6 m/s.

The acceleration of gravity is -9.8 m/s2. Calculate: hmax & tair

Drill: A ball is dropped from a ledge & lands

8.0 s later. Calculate: hledge & vmax

Homework•Problems: 27 – 30

•Page 103

Drill•A ball drops from 0.49 km.

•Calculate: tair & vmax

A man on the ground shoots a gun straight up & the bullet exits the barrel at

980 m/s. The acceleration of gravity is -9.8 m/s2.

Calculate its tair, vf, & hmax

A ball was dropped & landed at 70.0 m/s. The

acceleration of gravity is -9.8 m/s2.

1) Calculate: d & tair

A pumpkin was dropped from a plane & stayed in

air for 10.0 s. The acceleration of gravity is

-9.81 m/s2. Calculate: h & vmax.

Homework•Problems: 31 - 33

•Page 106

A cannon on a 2500 m cliff is fired straight up & the

ball exits the barrel at 0.98 km/s. The acceleration of

gravity is -9.8 m/s2. Calculate its tair, vf, & hmax

A ball is shot straight up to a height of 1.96 km. The acceleration of gravity is -9.8 m/s2.

Calculate: tair & vi

Drill: A car increases its velocity from 36 km/hr

to 72 km/hr in 5.0 s. Calculate: a & d

A car rolling backwards at 5.0 m/s accelerates at3.0 m/s2 for 4.0 s.

Calculate: vfinal & d

A car rolling backwards at 25.0 m/s accelerates at

5.0 m/s2 for 12.0 s.Calculate: vf & d


•Page 112 & 113

Drill: A ball is thrown straight up at 9800 cm/s.

Calculate: tair & hmax

A ball is thrown straight down at 25 m/s & stays

in air for 4.0 s. g = -9.8 m/s2

Calculate: initial height& vf of the ball.

A ball is thrown straight up to a height of 49 m.

g = -9.8 m/s2

Calculate: vi & tair


•Page 114

Drill: A ball is thrown straight sideways &

lands in 5.0 s. Calculate: initial

height & vmax down

A car rolling backwards at 5.0 m/s accelerates to 11

m/s forward in 4.0 s.Calculate: a & d during

that time.

A car going 36 km/hr slams on brakes, but still hits a tree at 6.0 km/hr after 1.0 s. Calculate: a & d during that time.


-10

0

10

20

30

40

0 2 4 6 8 10

Time (s)

Dis

plac

emen

t (m

)

Velocity vs Time

-20

0

20

40

0 2 4 6 8 10

Time (s)

Vel

ocit

y (m

/s)

Drill: A ball is thrown straight up

and hits the ground in 6.0 sec.

Calculate: hmax & vi

Test Date•Wednesday

10/25/06

Classwork•Work Problem Section B on pages 56 & 57 in the gray Physics Book.

Drill: A car goes from 36 km/hr to 54 km/hr in

5.0 s. Calculate: a & d during that time.

Matching Stuff•Formulas

•Definitions

•Units

Definitions• Motion Position

• Velocity Speed

• Distance Displacement

• Time Acceleration

Units• Displacement

• Time

• Velocity

• Acceleration

40

30

20

10

0

-10

-20

-30

-401 2 3 4 5 6 7 8 9 10

Dis

plac

emen

t (m

)Displacement vs Time graph

A ball is thrown straight up at 49 m/s. Calculate: tair & hmax

A car rolling backwards at 6.0 km/hr

accelerates to 30.0 km/hr forward in 6.0 s.

Calculate: a & d during that time.

A ball is thrown straight up to a

height of 490 m. Calculate: tair & vi

5

4

3

2

1

0

-1

-2

-3

-4

-5

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Vel

ocit

y (m

/s)

Time (s)

One Dimensional Motion. Distance How far something has moved.

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Transcript of One Dimensional Motion. Distance How far something has moved.