Physics Unit 1Physics Unit 1

One-Dimensional MotionOne-Dimensional Motion

1-D Motion1-D 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.

DisplacementDisplacement The change in position of an objectThe change in position of an object

∆∆x = xx = xff – x – xii

displacement=final position – initial positiondisplacement=final position – initial position

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

. . . Good News . . .. . . Good News . . . Check out figure 2-2 on page 41.Check out figure 2-2 on page 41.

– What is the gecko’s displacement?What is the gecko’s displacement?

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

Displacement vs. Distance DemoDisplacement vs. Distance Demo

DisplacementDisplacement– What was the car’s What was the car’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 Velocity The 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==∆x∆x==xxff-x-xii

∆ ∆t tt tff-t-tii

average velocity=average velocity=change in positionchange in position==displacementdisplacement change in time time change in time time

intervalinterval

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 Mr. Burr travels from Greencastle to Middletown

(50 km). If the first half of the distance is (50 km). If the first half of the distance is driven at 100 km/h and the second half is driven at 100 km/h and the second half is driven at 50 km/h, what is the average speed?driven at 50 km/h, what is the average speed?

Sample ProblemSample Problem During 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: 2,4,6 on pg 44Try: 2,4,6 on pg 44– Answers follow:Answers follow:

– 3.1 km to the south, 3.00 h, 6.4 h and 77km/h 3.1 km to the south, 3.00 h, 6.4 h and 77km/h to the south.to the south.

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.

Time (s) Distance (m)

0 0

1 2

2 8

3 18

4 32

5 50

6 72

7 98

8 128

9 162

10 200

Time (s) Distance (m)

0 0

1 2

2 4

3 6

4 8

5 10

6 12

7 14

8 16

9 18

10 20

Graphing ContinuedGraphing Continued What 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 each data point.Calculate the velocity for each data point.

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.

D v TD v T V v T V v T A v T A v T

AccelerationAcceleration Just 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==∆v∆v==vvff-v-vii

∆ ∆t tt tff-t-tii

average acceleration = average acceleration = change in velocitychange in velocity change 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 2 and 4 on page 49Do 2 and 4 on page 49– Answers followAnswers follow

– 2.0 s and -3.5 x 102.0 s and -3.5 x 10-3-3 m/s m/s22

HomeworkHomework

Practice 2A pg 44:Practice 2A pg 44:– 1,3,51,3,5

Section Review pg 47:Section Review pg 47:– 2,3,52,3,5

Practice 2B pg 49:Practice 2B pg 49:– 1,3,51,3,5

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?

SituationSituation vv aa11 ++ ++

22 -- --

33 ++ --

44 -- ++

55 + or -+ or - 00

66 00 + or -+ or -

77 00 00

Motion with Constant AccelerationMotion with Constant Acceleration Some 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 ∆x=Solve for ∆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 a m/s. As she rounded a turn she noticed a poor, defenseless turkey in the road. 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 did and slowed to a stop in 3.6 s. How far did she travel while applying her brakes?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 Acceleration From the equation for acceleration:From the equation for acceleration:

Solving for vSolving for vff = =

Another Displacement EquationAnother Displacement Equation Substitute 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/s m/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 m m and 2.5 s; 32 m

22ndnd Velocity Equation Velocity Equation

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

vvff22=v=vii

22+2a∆x+2a∆x

I’ll gladly show the derivation of this at I’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. Burr’s class revolts and equations, Mr. Burr’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 Continued Givens: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 2-4: Need to Know and UseTable 2-4: Need to Know and Use Form to use with some Form to use with some

initial velocityinitial velocity

∆∆x=½(vx=½(vii+v+vff)∆t)∆t

vvff=v=vii+a(∆t)+a(∆t)

∆∆x=vx=vii(∆t)+½a(∆t)(∆t)+½a(∆t)22

vvff22=v=vii

22+2a∆x+2a∆x

Form to use when starting Form to use when starting from restfrom rest

∆∆x=½(vx=½(vff)∆t)∆t

vvff=a(∆t)=a(∆t)

∆∆x=½a(∆t)x=½a(∆t)22

vvff22=2a∆x=2a∆x

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 equations Is acceleration constant? (limits the equations

one can use)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 Nathan’s acceleration?What is Nathan’s acceleration?– What is Nathan’s displacement during What is Nathan’s displacement during

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

vvff = 12.5 m/s = 12.5 m/s

∆∆t = tt = tff-t-tii = 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==∆v∆v==vvff-v-vii=a (uniform acceleration)=a (uniform acceleration) ∆ ∆t tt tff-t-tii

– vvff=a(∆t)=a(∆t)

Pick one and solve.Pick one and solve.

Part 2: ∆x = ?Part 2: ∆x = ? Equations?Equations?

– ∆∆x=½(vx=½(vff)∆t)∆t

– ∆∆x=½a(∆t)x=½a(∆t)22

– vvff22=2a∆x=2a∆x

Pick and solve.Pick and solve.

Part 3: vPart 3: vavgavg = ? = ?

Equations:Equations:

– vvavgavg==∆x∆x==xxff-x-xii

∆ ∆t tt tff-t-tii 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 y-axisacts in only the y-axissame 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

NameName ∆∆y y (cm)(cm)

Time Time (s)(s)

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

velocity of 25 m/s. Create a table velocity of 25 m/s. Create a table showing the ball’s position showing the ball’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 s 5.00 s– g = -9.81 m/sg = -9.81 m/s22

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

tt y y v v a a(s)(s) (m)(m) (m/s)(m/s) (m/s(m/s22))

1.001.00 20.1 20.1 15.2 15.2 -9.81 -9.81

2.002.00 30.4 30.4 5.4 5.4 -9.81 -9.81

3.003.00 30.9 30.9 -4.4 -4.4 -9.81 -9.81

4.004.00 21.6 21.6 -14.2 -14.2 -9.81 -9.81

5.005.00 2.50 2.50 -24.0 -24.0 -9.81 -9.81

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