Physics Unit 1Physics Unit 1OneOne--Dimensional MotionDimensional Motion

11--D MotionD Motion

Motion in one direction.Motion in one direction.–– Dependant on the specific, chosen Dependant on the specific, chosen

frame of referenceframe of referenceA coordinate system for specifying the A coordinate system for specifying the precise location of objects in spaceprecise location of objects in space

–– Described by displacement, velocity, Described by displacement, velocity, and acceleration.and acceleration.

DisplacementDisplacementThe change in position of an objectThe change in position of an object

∆∆x = xx = xff –– xxii

displacement=final position displacement=final position –– initial positioninitial position

∆∆ (delta) means (delta) means changechange or or change inchange in

. . . Good News . . .. . . Good News . . .Check out figure 2Check out figure 2--2 on page 41.2 on page 41.–– What is the geckoWhat is the gecko’’s displacement?s displacement?

–– What would be the geckoWhat would be the gecko’’s displacement s displacement if we changed the frame of reference so if we changed the frame of reference so that 0 corresponded to the geckothat 0 corresponded to the gecko’’s s initial position?initial position?

Displacement vs. Distance DemoDisplacement vs. Distance Demo

DisplacementDisplacement–– What was the carWhat was the car’’s s

displacement?displacement?

DistanceDistance–– How far did the car How far did the car

travel?travel?

Displacement needs a magnitude (20 cm) and a direction (to the left).

Can they ever be negative?

Average VelocityAverage VelocityThe total displacement divided by the The total displacement divided by the time interval during which the time interval during which the displacement occurreddisplacement occurred

vvavgavg==∆∆xx==xxff--xxii∆∆t tt tff--ttii

average velocity=average velocity=change in positionchange in position==displacementdisplacementchange in time time intervalchange in time time interval

Velocity, the rate of change of position, can be positive Velocity, the rate of change of position, can be positive or negative depending on the displacement.or negative depending on the displacement.

Velocity ExampleVelocity ExampleMr. Burr travels from Greencastle to Middletown (50 km). Mr. Burr travels from Greencastle to Middletown (50 km). If his average speed is 50 km/h, what is the time it takes to If his average speed is 50 km/h, what is the time it takes to get to Middletown?get to Middletown?

Sample ProblemSample ProblemDuring a race on level ground Colin During a race on level ground Colin runs with an average velocity of 8.04 runs with an average velocity of 8.04 m/s to the east. What distance does m/s to the east. What distance does Colin cover in 243 s?Colin cover in 243 s?

Practice 2APractice 2A

Try: 2 & 4 on pg 44Try: 2 & 4 on pg 44–– 2 together, 4 on own2 together, 4 on own–– Answers follow:Answers follow:

–– 3.1 km to the south, 3.00 h3.1 km to the south, 3.00 h

Distance vs. Time GraphsDistance vs. Time Graphs

Distance

(m)

Time (s)

A B C D E

Graph the distance/time graphs on graph paper.Graph the distance/time graphs on graph paper.

00

20010162912889877265053241838221

Distance (m) Time (s)00

201018916814712610584634221

Distance (m) Time (s)

Graphing ContinuedGraphing ContinuedWhat is the shape of each graph?What is the shape of each graph?Calculate the slope of the Calculate the slope of the blackblack data.data.–– Calculate the velocity of the corresponding Calculate the velocity of the corresponding

object. How do they compare?object. How do they compare?–– What about the slope of the What about the slope of the orange orange data?data?

Calculate the velocity for every data point Calculate the velocity for every data point on each graph. Record this in two new on each graph. Record this in two new tables.tables.–– WhatWhat’’s happening to the velocity for the s happening to the velocity for the

blackblack graph? graph? –– WhatWhat’’s happening to the velocity for the s happening to the velocity for the

orange orange graph?graph?

Graphing ContinuedGraphing Continued

On a separate sheet of graph paper On a separate sheet of graph paper plot a velocity/time graph for both plot a velocity/time graph for both data sets.data sets.What is the shape of each graph?What is the shape of each graph?What is the slope of each graph?What is the slope of each graph?–– The slope is the acceleration of the The slope is the acceleration of the

objects.objects.

Unit 1 PhysicsGraphing Speed

d vs. t graphs

• Graph the motion • use the slope to interpret the relationship

between the variables (d & t)

Slope = riserun = =

ΔYΔX

dt = Speed

faster the speed

slower the speed

*steeper the slope →

*more gentle the slope →

d

t

d d d

t tt

When interpreting the motion answer:•As time goes by, distance (d ↑ or ↓)•The line shows relationship (constant or changing)

•Moving forward (d↑)

•Constant speed (straight)

•Moving backward (d↓)

•Moving forward (d↑)

•Constant speed (straight)

•Constant speed (straight)

•No motion (d stays same)

•0 m/s

•Changing speed (curved)•positive acceleration

d dd

t t t•Moving forward (d↑)

•Moving backward (d↓)

•Moving backward (d↓)

•changing speed (curved)

•changing speed (curved)

•changing speed (curved)

•(-) acceleration •(-) acceleration•(+) acceleration

MOTION

-

Time (s)

0 5 10 15 20 25 30 35 40 45 50 55

5

10

15

20

25

30

Dis

tance

(m

)

A

B C

D

E

H

G

F

AB•Move forward•Constant speed•ΔYΔX =10m

10s =1m/s

BC•At rest•Constant speed•0m/s

MOTION

-

Time (s)

0 5 10 15 20 25 30 35 40 45 50 55

5

10

15

20

25

30

Dis

tance

(m

)

A

B C

D

E

H

G

F

CD•Moving backward•Constant speed•ΔYΔX = 5m

5s =1m/s

DE•Moving forward•Changing speed•(+) acceleration

MOTION

-

Time (s)

0 5 10 15 20 25 30 35 40 45 50 55

5

10

15

20

25

30

Dis

tance

(m

)

A

B C

D

E

H

G

F

EF•Moving backward•Constant speed•ΔYΔX =25m

5s =5m/s

FG•Moving forward•Constant speed•ΔYΔX ==10m

5s2m/s

GH•Moving forward•Constant speed•ΔYΔX

= =12.5m12.5s

1m/s

MOTION

-

Time (s)

0 5 10 15 20 25 30 35 40 45 50 55

5

10

15

20

25

30

Dis

tance

(m

)

A

B C

D

E

H

G

F

Which line had the fastest speed?•EF•Steepest slopeDistance increases?

Distance decreases?•Moving forward

•Moving backward

Curved line?

Straight line?•Constant speed

•Changing speed

d vs. t graph shows?•Speed and direction

D v TD v T V v TV v T A v TA v T

AccelerationAccelerationJust like displacement over time gave Just like displacement over time gave velocity (rate of change of position), velocity (rate of change of position), velocity over time will give acceleration velocity over time will give acceleration (rate of change of velocity).(rate of change of velocity).

aaavgavg==∆∆vv==vvff--vvii∆∆t tt tff--ttii

average acceleration = average acceleration = change in velocitychange in velocitychange in timechange in time

Acceleration must have a direction as well, so it Acceleration must have a direction as well, so it can be positive or negative.can be positive or negative.

Sample ProblemSample ProblemThe last time Mr. Burr rode his motorcycle The last time Mr. Burr rode his motorcycle

he was thrown off and skidded on the he was thrown off and skidded on the blacktop for about 4 seconds. Before the blacktop for about 4 seconds. Before the accident he was traveling at 40 km/h. accident he was traveling at 40 km/h. What acceleration was I (he) subjected What acceleration was I (he) subjected to?to?

Practice 2BPractice 2B

Do 1 and 2 on page 49Do 1 and 2 on page 49–– 1 together, 2 on own1 together, 2 on own–– Answers followAnswers follow

–– 2.0 2.0 ss

HomeworkHomework

Practice 2A pg 44:Practice 2A pg 44:–– 1,31,3

Section Review pg 47:Section Review pg 47:–– 33--55

Practice 2B pg 49:Practice 2B pg 49:–– 33

Will the car speed up, slow down, retain Will the car speed up, slow down, retain a constant velocity or not move?a constant velocity or not move?

77

66

55

44

33

22

11

SituationSituation

0000

+ or + or --00

00+ or + or --

++--

--++

----

++++aavv

Motion with Constant AccelerationMotion with Constant AccelerationSome systems will have a variable Some systems will have a variable acceleration while some will have acceleration while some will have constant acceleration. constant acceleration. –– Free Fall (gravity)Free Fall (gravity)–– Kinematic EquationsKinematic Equations

Equations defining displacement and Equations defining displacement and velocity in terms of position, time, velocity velocity in terms of position, time, velocity and constant acceleration.and constant acceleration.

Displacement with Constant Displacement with Constant AccelerationAcceleration

Start with average velocity:Start with average velocity:

–– True with constant accelerationTrue with constant acceleration

–– Solve for Solve for ∆∆x=x=

Sample ProblemSample ProblemMrs. Burr was traveling at a velocity of 30 Mrs. Burr was traveling at a velocity of 30

m/s. As she rounded a turn she noticed m/s. As she rounded a turn she noticed a poor, defenseless turkey in the road. a poor, defenseless turkey in the road. She, fearing damage to her car despite She, fearing damage to her car despite the lure of free meat, applied her brakes the lure of free meat, applied her brakes and slowed to a stop in 3.6 s. How far and slowed to a stop in 3.6 s. How far did she travel while applying her brakes?did she travel while applying her brakes?

Sample Problem ContinuedSample Problem ContinuedGivens:Givens:

Unknowns:Unknowns:

Practice 2CPractice 2C

Pg 53:Pg 53:–– 2, 42, 4–– Answers followAnswers follow

–– 18.8 m and 9.1 s18.8 m and 9.1 s

Velocity with Constant AccelerationVelocity with Constant AccelerationFrom the equation for acceleration:From the equation for acceleration:

Solving for vSolving for vff ==

Another Displacement EquationAnother Displacement EquationSubstitute the equation vf =Substitute the equation vf =into our original equation for into our original equation for displacement:displacement:

∆∆x =x =

Sample ProblemSample ProblemCourtney fires an arrow from her bow. The Courtney fires an arrow from her bow. The

arrow undergoes a uniform acceleration arrow undergoes a uniform acceleration of 1.9 x 10of 1.9 x 1044 m/sm/s22 in 5.4 ms. What is the in 5.4 ms. What is the final speed of the arrow? Over what final speed of the arrow? Over what distance was the arrow accelerated?distance was the arrow accelerated?

Sample ContinuedSample ContinuedGivens:Givens:

Unknowns:Unknowns:

Practice 2DPractice 2D

Pg 55:Pg 55:–– 2,42,4–– Answers followAnswers follow

–– 19 m/s; 6.0 x 1019 m/s; 6.0 x 1011 m and 2.5 s; 32 mm and 2.5 s; 32 m

22ndnd Velocity EquationVelocity Equation

Velocity in terms of displacement Velocity in terms of displacement instead of time.instead of time.

vvff22=v=vii

22+2a+2a∆∆xx

II’’ll gladly show the derivation of this at ll gladly show the derivation of this at another time to anyone who is another time to anyone who is

interested.interested.

Sample ProblemSample ProblemAfter two days worth of notes and After two days worth of notes and

equations, Mr. Burrequations, Mr. Burr’’s class revolts and s class revolts and chases him. In order to escape, Mr. Burr chases him. In order to escape, Mr. Burr accelerates from rest at a rate of 0.500 accelerates from rest at a rate of 0.500 m/sm/s22. What would his velocity be after . What would his velocity be after traveling 8.58 m?traveling 8.58 m?

Sample ContinuedSample ContinuedGivens:Givens:

Unknowns:Unknowns:

Practice 2EPractice 2E

Pg 58:Pg 58:–– 2,4,62,4,6–– Answers followAnswers follow

+21 m/s, +16 m/s, +13 m/s; 87 m; 7.4 m+21 m/s, +16 m/s, +13 m/s; 87 m; 7.4 m

Table 2Table 2--4: Need to Know and Use4: Need to Know and UseForm to use with some Form to use with some initial velocityinitial velocity

∆∆x=x=½½(v(vii+v+vff))∆∆tt

vvff=v=vii+a(+a(∆∆t)t)

∆∆x=vx=vii((∆∆t)+t)+½½a(a(∆∆t)t)22

vvff22=v=vii

22+2a+2a∆∆xx

Form to use when Form to use when starting from reststarting from rest

∆∆x=x=½½(v(vff))∆∆tt

vvff=a(=a(∆∆t)t)

∆∆x=x=½½a(a(∆∆t)t)22

vvff22=2a=2a∆∆xx

Which Equations Do I Use, and Which Equations Do I Use, and When?When?

Overwhelmed?Overwhelmed?Strategy for solving physics problems:Strategy for solving physics problems:–– What am I given? (Givens)What am I given? (Givens)–– What am I looking for? (Unknowns)What am I looking for? (Unknowns)–– Is acceleration constant? (limits the Is acceleration constant? (limits the

equations one can use)equations one can use)–– Which equation(s) give me what I am looking Which equation(s) give me what I am looking

for by using what I already have? (puzzle!)for by using what I already have? (puzzle!)–– Plug and Chug!!Plug and Chug!!

Strategy PracticeStrategy Practice

Nathan accelerates his skateboard Nathan accelerates his skateboard uniformly along a straight path from uniformly along a straight path from rest to 12.5 m/s in 2.5 s.rest to 12.5 m/s in 2.5 s.–– What is NathanWhat is Nathan’’s acceleration?s acceleration?–– What is NathanWhat is Nathan’’s displacement during s displacement during

this time interval?this time interval?–– What is NathanWhat is Nathan’’s average velocity s average velocity

during this time interval?during this time interval?

What are You Given?What are You Given?vvii = 0 m/s= 0 m/svvff = 12.5 m/s= 12.5 m/s∆∆t = tt = tff--ttii = 2.5 m/s= 2.5 m/s

What are You Looking For?What are You Looking For?a = ?a = ?∆∆x = ?x = ?vvavgavg = ?= ?

Acceleration is Constant!!Acceleration is Constant!!

Part 1: a=?Part 1: a=?Equations?Equations?––aaavgavg==∆∆vv==vvff--vvii=a (uniform acceleration)=a (uniform acceleration)

∆∆t tt tff--ttii

––vvff=a(=a(∆∆t)t)

Pick one and solve.Pick one and solve.

Part 2: Part 2: ∆∆x = ?x = ?Equations?Equations?––∆∆x=x=½½(v(vff))∆∆tt––∆∆x=x=½½a(a(∆∆t)t)22

–– vvff22=2a=2a∆∆xx

Pick and solve.Pick and solve.

Part 3: vPart 3: vavgavg = ?= ?

Equations:Equations:

––vvavgavg==∆∆xx==xxff--xxii∆∆t tt tff--ttii

Only one equation, so PLUG AND Only one equation, so PLUG AND CHUG!!CHUG!!

Section ReviewSection Review

Pg 59Pg 59–– 1,5,61,5,6

Free FallFree Fall

Objects in free fall accelerate at a Objects in free fall accelerate at a constant rate, the acceleration due constant rate, the acceleration due to gravity.to gravity.

g = 9.81 m/sg = 9.81 m/s22

acts in only the yacts in only the y--axisaxissame equations apply, change x to ysame equations apply, change x to y

Quick LabQuick Lab

Pg 62 in bookPg 62 in book–– need meterstick need meterstick

and stopwatchand stopwatch

Time Time (s)(s)

∆∆y y (cm)(cm)

NameName

Sample ProblemSample ProblemA baseball is hit straight up in the air with A baseball is hit straight up in the air with

a velocity of 25 m/s. Create a table a velocity of 25 m/s. Create a table showing the ballshowing the ball’’s position s position (displacement), velocity, and acceleration (displacement), velocity, and acceleration for the first 5.00 s of its flight.for the first 5.00 s of its flight.Givens:Givens:–– vvii = 25.0 m/s= 25.0 m/s–– t = 1.00s t = 1.00s 5.00 s5.00 s–– g = g = --9.81 m/s9.81 m/s22

Unknowns:Unknowns:––∆∆y and vy and vff for each intervalfor each interval

tt yy vv aa(s)(s) (m)(m) (m/s)(m/s) (m/s(m/s22))

1.001.00 20.120.1 15.215.2 --9.819.812.002.00 30.430.4 5.45.4 --9.819.813.003.00 30.930.9 --4.44.4 --9.819.814.004.00 21.621.6 --14.214.2 --9.819.815.005.00 2.502.50 --24.024.0 --9.819.81

HomeworkHomework

Practice 2CPractice 2C–– 1,3,51,3,5

Practice 2DPractice 2D–– 1,31,3

Practice 2EPractice 2E–– 1,3,51,3,5

Practice 2FPractice 2F–– 1,3,51,3,5