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AP Physics Chapter 2

Motion Along a Straight Line

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AP Physics

Turn in Chapter 1 Homework, Worksheet, & Lab

Take quiz Lecture Q&A

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Motion

Kinematics: Study of motion, emphasizing on describing motion

Simplest motion: 1-Dimensional Point Particle: no rotation

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Position

Position: Location, where an object is Symbol: x Unit: meter How to describe a position in 1-D?

Frame of Reference (numbered line) Reference point: origin Direction:

o Positive direction: We can define the positive direction to be any direction we want, normally direction of motion

o Negative direction: Opposite to positive direction

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Position is a ______

Direction: negative sign indicates direction only (more later)

Magnitude: how far from the origin

vector

More on Chapter 3

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Position and Frame of Reference

EastWestDowntown, LAHami

10 miles

-10

+10

0

(- indicates only direction, west.)

Frame 2: Define West positive, and x = 0 at Downtown LA, then position of Hami is _____ miles.

Frame 1: Define East positive, and x = 0 at Downtown LA, then position of Hami is _____ miles.

Frame 3: Define East positive, and x = 0 at Hami, then position of Hami is _____ miles.

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So, Representing Position …

The representation of a position depends on the choice of the frame of reference.

The same position can be expressed in different ways in different frame of reference.

But the physical meaning of position does not change, regardless the choice of frame of reference. Hami still is 10 miles west of Downtown LA, no more or less. If you want to go from Downtown LA to Hami, you still have to

go 10 miles west.

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Displacement

Displacement: Change of position Symbol: x, Δ = change x = x2 - x1 = xf – xi

Unit: same as position, meter Displacement is also a vector:

– Magnitude: how far– Direction: Negative sign indicates direction only, it has nothing

to do with magnitude.

Displacement has nothing to do with the actual path. It depends only on the initial and final positions.

A –3m displacement is ___________ a 2m displacement.larger than

++2m-3m

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Distance (d)

Distance Displacement Distance is a scalar. (magnitude only) Distance is equal to Magnitude of Displacement when

there is no change in direction. d = |x|

Distance is not necessarily equal to the magnitude of total displacement.

Displacement cares only end points; distance cares both end points and the actual path.

Representations of position and displacement depend on frame of reference, but distance does not depend on frame of reference.

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Total Distance

Whenever there is a change in direction, total distance will not be the same as the magnitude of total displacement.

2 0 2f ix x x m m When you go from x = 0 to 3m then back to 2m,

Your total displacement is __ m.

Your total distance traveled is __ m.

2

4 1 2 3 1 4d d d m m m

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Total Distance, dtot

When there is no change in direction:

xd

1 2 1 2... ...totald d d x x

When there is change in direction:

where d1 and d2 are distances of segments in which there is no change in direction.

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Average Velocity

Average velocity: ratio of displacement x that occurs during a particular time interval t to that time interval

f i

f i

x xxv

t t t

Standard Unit: m/s

Constant velocity: v v

x v t

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Average speed,

Average speed is distance traveled during a time interval divided by the time interval.

totalds

t

s or v

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Example: During a hard sneeze, your eyes might shut for 0.50 s. If you are driving a car at 90 km/h, how far does it move during that time?

Given: t = 0.50s, v = 90km/h x = ?

90kmv

h

x v t

1000

1

m

km

1

3600

hr

s

25m

s

25 0.50 12.5 13.m

s m ms

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Practice: Boston Red Sox pitcher Roger Clemens could routinely throw a fastball at a horizontal speed of 160 km/h. How long did the ball take to reach home plate 18.4 m away?

Given: s

m

s

h

km

m

h

kmv 4.44

3600

1

1

1000160

mx 4.18,

t = ?

x v t

18.40.414

44.4

x mt s

mvs

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Practice: 30-11

You drive on Interstate 10 from San Antonio to Houston, half the time at 55 km/h and the other half at 90 km/h. On the way back you travel half the distance at 55 km/h and the other half at 90 km/h. What is your average speed

a) from San Antonio to Houston,b) from Houston back to San Antonio, andc) for the entire trip?d) What is your average velocity for the entire trip?e) Graph x versus t for (a), assuming the motion is all in the positive

x direction. Indicate how the average velocity can be found on the graph.

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Practice:

a)Let T be the total number of hours from San Antonio to Houston, then

the distance of the first T/2 hours is (55 T/2) km,

and that of the second T/2 hours is (90 T/2) km.

Total distance is (55 T/2) + (90 T/2) = 72.5T km.

Then average speed is

72.5totd Ts

t

km

T73

km

hhours

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Practice: (2)

b) Let D km be the total distance. Then the time for the first D/2 km is

d st

Similarly the time for the second D/2 km is D/180 h, then total time is

180 110

110 180 180 110tot

D Dt h D h

Then the average speed is

tot

tot

ds

t

dts

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Dkm

t 55km 110

Dh

h

D

180 110

180 110

km

D

180 11068

180 110

km km

h hhours

1 2

1 2

1 2

1 2

2 2

2 2

2

v vv

v v

v v

v v

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Practice: (3)

c)

Similar to Part b), the average speed for the entire trip is

2 73 2 6870

2 73 2 68

km

h

d)

Total displacement is 0 for round trip. Then average velocity is

00totxv

t t

2 90 2 55180 110

180 110 2 90 2 55

km

h

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Practice: (4)

t

x

Average velocity can be found by finding the slope of the line (in red) connecting the end points.

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Instantaneous velocity

Instantaneous velocity is the average velocity when the time interval becomes very, very small, essentially zero.

0limt

x dxv

t dt

Instantaneous velocity is the time-derivative of position function.

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Derivatives

C is constant, and x and y are functions of t.

1

0

sin cos

cos sin

n ndct cnt

dtdc

dtd dx dyx y

dt dt dtd

ct c ctdtd

ct c ctdt

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Try

36d

tdt

23 2 6, ?dx

x t tdt

6 2dx

tdt

218t

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Instantaneous velocity

Tangent linex

tAs the time interval becomes smaller and smaller, average velocity becomes instantaneous velocity, which is the slope of the tangent line.

ti t2 t1t3t4

dslope

t

Slope = average velocity from time ti to t1

What does the slope of this line mean?

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Derivative and slope

The derivative of a curve at any point is the slope of its tangent line at that point.

Instantaneous velocity v is the time derivative of position x. On a position-time graph, the instantaneous velocity at any time is the slope of the line tangent to the curve at that time.

If position graph is a straight line:

slope is _________, andconstant

_________ is constant.velocity

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How far does the runner whose velocity-time graph is shown below travel in 16s?

Displacement = Area under curve of v-t graph

1

8 108 72

2

s s mA m

s

0 161210

t (s)

4

8

v (m/s)

trapezoid

upper length + lower lengthA × height

2

2

4 82 12

2

m m

s sA s m

3 4 4 16m

A s ms

1 2 3 72 12 16 100totA A A A m m m m

100d m

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A1

A1

1 2

1 2

x A A

d A A

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Acceleration

Average acceleration f i

f i

v

t t

va

v

t

0limt

v dva

t dt

Instantaneous acceleration

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2

dd x dx

adt dt

d

dt

xdt

2

2dxv

dt

Think of acceleration as a push as of now, though not exactly correct.

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Velocity-time graph

On a velocity-time graph, the slope of the line tangent to the curve at any time is the instantaneous acceleration at that time.

If the graph is a straight line, acceleration is constant.

Displacement is area under curve

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Position, Velocity, and acceleration Graphs

Position (vs. Time) graph

v = slope

a = slope

x = Area under curve

v = Area under curve

Velocity (vs. Time) graph

Acceleration (vs. Time) graph

Derivative

Integrate

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Speeding Up or Slowing down

v and a are in the same direction (or have the same sign) ___________

v: speeding up

v: speeding up

v:

v: slowing down

slowing down

a: a:

a: a:

v and a are in the opposite direction (or have the opposite signs) ____________

speeding up

slowing down

+

+

-

-

+

-

-

+

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Directions of acceleration and velocity

Sign of a vector indicates direction only. If the signs of the velocity and acceleration of a particle

are the same (same direction), the speed of the particle increases (speeding up).

If the signs are opposite (opposite directions), the speed decrease (slowing down).– A negative acceleration does not necessarily mean

slowing down.– Deceleration = slow down

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Speeding up or slowing down

Speeding up:

Slowing down:

in the same

opposite to

Acceleration is ___________ direction of velocity.

Acceleration is __________ the direction of velocity.

Then we can determine the sign of acceleration depending on what direction has been already defined as the positive direction

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Practice 35-79

If the position of an object is given by x = 2.0t3, where x is measured in meters and t in seconds, find (a) the average velocity and (b) the average acceleration between t = 1.0s and t = 2.0 s. Then find (c) the instantaneous velocities and (d) the instantaneous accelerations at t = 1.0s and t = 2.0 s. (e) Compare the average and instantaneous quantities and in each case explain why the larger one is larger. (f) Graph x versus t and v versus t, and indicate on the graphs your answers to (a) through (d).

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Practice (2)

a) 1.0 ,it s

xv

t

2.0 ,ft s

332.0 2.0 1.0 2.0i ix t m

332.0 2.0 2.0 16f fx t m

?v

32.0x t

f i

f i

x x

t t

16 2.014.

2.0 1.0

m m m

s s s

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Practice: (3)

b)

?a

1.0 , 2.0 ,i ft s t s

2

24 6.018

2.0 1.0f i

f i

m mv v ms sat t s s s

32.0x t

26.0 1.0 6.0 ,i

mv

s

26.0 ,dx

v tdt

26.0 2.0 24.f

mv

s

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Practice: (4)

c)26.0 ,

dxv t

dt

d)26.0

dxv t

dt

26.0 1.0 6.0 ,i

mv

s 2

6.0 2.0 24.f

mv

s

12.0 ,dv

a tdt

212.0 1.0 12. ,i

ma

s 212.0 2.0 24.f

ma

s

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Practice: (5)

d)

i fv v v

Because both the velocity and acceleration are increasing, so the final values are the largest ones.

i fa a a

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Practice: (6)

f) x vs. t

0

5

10

15

20

25

30

35

0 0.5 1 1.5 2 2.5 3

t (s)

x (m

)

a) Slope of red line is the average velocity from 1s to 2s

c) Slope of blue line is the instantaneous velocity at 1 s

c) Slope of green line is the instantaneous velocity at 2 s

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Practice: (7)

v vs. t

05

10152025303540

0 0.5 1 1.5 2 2.5

t (s)

v (m

/s)

b) Slope of red line is the average acceleration from 1s to 2s

d) Slope of blue line is the instantaneous acceleration at t =1s

d) Slope of green line is the instantaneous acceleration at t = 2s

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Constant Acceleration Motion

a = constant Let initial time

t = 0, then at any time t, velocity and position are given by

00

0

20 0

2 20 0

0

20

1

2

1

2

2

21

2

v vx x v v t v

x

v v at

x x v t at

v v a x

x at

x

vt

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Free-Fall Motion

• Assume no air resistance. (Valid when speed is not too fast.)

• a = g, downward (g = 9.81 m/s2) Acceleration can be positive or negative, depending on what we

define as the positive direction. g is always a positive number, equivalent to 9.81 m/s2. Does not matter if the object is on its way up, on its way down,

or at the very top.

g is acceleration due to gravity (It is not gravity.) g does not depend on mass of object.

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Terms

Released, drop– If hand not moving, then vi =0.

– If hand moving, then vi 0

a = g, downward, regardless of direction of velocity– on the way up– on the way down, or– at the very top

vi = 0 relative to hand

V = 0

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Signs of v and a

v: a:

v

v: a:

a:

Define: up = +

v:

Define: down = +

a:

v a:

v: a:

+ -

= 0 -

- -

- +

= 0 +

+ +

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Free-Fall motion equations

These equation are valid only when downward is defined as the positive direction.

Not valid when upward is defined as the positive direction. (Must replace every g with –g.)

No need to remember these equations.

2 2

2

2

1

2

f i f

f i

i f f

v v gt

v v g x

x v t gt

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Practice: A jumbo jet must reach a speed of 360 km/h (=225 mi/h) on the runway for takeoff. What is the least constant acceleration needed for takeoff from a 1.80 km runway?

?,1800

,1001

1000

3600

1360,0

amx

s

m

km

m

s

h

h

kmvv fi

2 2 2f iv v a x

22

2 2

2

100 02.78

2 2 1800f i

mv v ms

ax m s

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Practice: A stone is thrown down from a bridge 43.9 m above the water and splashes the water 2.0 s later. What is the initial speed of the stone? With what speed the stone hit the water?

2

2

143.9 9.8 2.0

212

2.0

mm s

mss s

iv at

Let down = +. Then

?, , 43.9 , 2.0 , ?i fv a g x m t s v

21

2ix v t at

iv 21

2x at

t

fv 212 9.8 2.0 32m m m

ss s s

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Practice: A rock is thrown straight upward with an initial speed of 10 m/s from a window that is 20 m high. A) How much higher can it go? B) How fast is it moving when it hits the ground?

10 , 0,i i

mv x a g

s

Let up = +, x = 0 at window.

) ?gb v ) ?ta x

2 2 2t i t iv v a x x

2 2

22

2

2

0 100 5.1

2 9.8

t it i

v vx x

a

m

sm

m

s

, 0tv 2 2 2g i g iv v a x x

2

2

2

2

10 2 9.8 20 0

22

g i g iv v a x x

m mm

s s

m

s

, 20gx m

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Practice: A rocket-driven sled running on a straight, level track is used to investigate the physiological effects of large accelerations on humans. One such sled can attain a speed of 1600 km/h in 1.8 s starting from rest. Find (a) the acceleration (assumed constant) in g units and (b) the distance traveled.

sts

m

h

kmvv fi 8.1,4441600,0

) ?a a

) ?b x

or

x v t

f iv v at

2

444 0247

1.8f i

mv v msat s s

2247

m

s

2

25.29.8

gg

ms

2 2 2f iv v a x 2

2 2

2

444 0

2 2 247

399

f i

mv v s

xmas

m

2i fv v

t

0 4441.8 400

2

m

s s m