physics 201: lecture 2, pg 1 lecture 2 chapter chapter 2.1-2.4 define position, displacement &...
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Physics 201: Lecture 2, Pg 1
Lecture 2ChapterChapter 2.1-2.4
Define Position, Displacement & Distance
Distinguish Time and Time Interval
Define Velocity (Average and Instantaneous), Speed
Define Acceleration
Understand algebraically, through vectors, and graphically the relationships between position, velocity and acceleration
Comment on notation
Physics 201: Lecture 2, Pg 2
Informal Reading Quiz
Displacement, position, velocity & acceleration are the main quantities that we will discuss today.
Which of these 4 quantities have the same units
A. Velocity & position
B. Velocity & acceleration
C. Acceleration & displacement
D. Position & displacement
E. Position & acceleration
Physics 201: Lecture 2, Pg 3
Range of LengthsDistance Length (m)Radius of Visible Universe 1 x 1026
To Andromeda Galaxy 2 x 1022
To nearest star 4 x 1016
Earth to Sun 1.5 x 1011
Radius of Earth 6.4 x 106
Willis Tower 4.5 x 102
Football Field 1 x 102
Tall person 2 x 100
Thickness of paper 1 x 10-4
Wavelength of blue light 4 x 10-7
Diameter of hydrogen atom 1 x 10-10
Diameter of proton 1 x 10-15
Physics 201: Lecture 2, Pg 4
Range of Times
Interval Time (s)
Age of Universe 5 x 1017
Age of Grand Canyon 3 x 1014
Avg age of college student 6.3 x 108
One year 3.2 x 107
One hour 3.6 x 103
Light travel from Earth to Moon 1.3 x 100
One cycle of guitar A string 2 x 10-3
One cycle of FM radio wave 6 x 10-8
One cycle of visible light 1 x 10-15
Time for light to cross a proton 1 x 10-24
World’s most accurate timepiece: Cesium fountain Atomic Clock
Lose or gain one second in some 138 million years
Physics 201: Lecture 2, Pg 5
One-Dimension Motion (Kinematics) Position, Displacement, Distance
Position: Reflects where you are. KEY POINT 1: Magnitude, Direction, Units KEY POINT 2: Requires a reference point (Origin)
Origins are arbitrary
Physics 201: Lecture 2, Pg 6
One-Dimension Motion (Kinematics) Position, Displacement, Distance
Position: Reflects where you are. KEY POINT 1: Magnitude, Direction, Units KEY POINT 2: Requires a reference point (Origin)
Origins are arbitrary
Example: Where is Boston ?
Choose origin at New YorkNew York
Boston is 212 miles northeast of New York
OR
Boston is 150 miles east and 150 miles north of New York
Boston
New York (Origin)
Physics 201: Lecture 2, Pg 7
One-Dimension Motion (Kinematics) Position, Displacement, Distance
Getting from New York to Boston requires a PATH Path defines what places we pass though
Displacement: Change in position Requires a time interval Any point on the path must be
associated with a specific time
( t1, t2, t3, ….) Path 1 and Path 2 have the
same change in position so theythe same displacement.
However the distance travelled isdifferent.
Boston
New York
Path 1
Path 2
Physics 201: Lecture 2, Pg 8
Motion in One-Dimension (Kinematics) Position
Position along a line; references xi and ti :
At time = 0 seconds Pat is 10 meters to the right of the lamp Origin lamp Positive direction to the right of the lamp Position vector ( xi , ti) or (10 m, 0.0 s) Particle representation
10 meters
PatO
-x +x10 meters
Physics 201: Lecture 2, Pg 9
Displacement
One second later Pat is 15 meters to the right of the lamp
At t = 1.0 s the position vector is ( xf , tf ) or (15 m, 1.0 s) Displacement is just change in position
x ≡ xf – xi
There is also a change in time
t ≡ tf – ti
10 meters 15 meters
Patxi
O
xfΔx
Physics 201: Lecture 2, Pg 10
Displacement Putting it all together
x = xf - xi = 5 meters to the right !t = tf - ti = 1 second
Relating x to t yields average velocity
O Patxi
10 meters 15 meters
xfΔx
Physics 201: Lecture 2, Pg 11
Average Velocity
Changes in position vs Changes in time
)increment time(
)ntdisplaceme( velocityaverageav, t
xv gx
• Average velocity = displacement per time increment , includes BOTH magnitude and direction
• Pat’s average velocity was 5 m / s to the right
)sec 1()right the tom 5(
avg, tx
vx
Physics 201: Lecture 2, Pg 12
Average Speed
Average speed, vavgavg, reflects a magnitude “How fast” without the direction.
References the total distance travelled
t
d
tv
) timetotal(
path along taken distancespeed averageavg
• Pat’s average speed was 5 m / s
NOTE: Serway’s notation varies from other texts(There really is no standard)
Physics 201: Lecture 2, Pg 13
Pat on tour (graphical representation) Pat is walking from and to the lamp (at the origin).
(x1 , t1) = (10 m, 0.0 sec)
(x2 , t2) = (15 m, 1.0 sec)
(x3 , t4) = (30 m, 2.0 sec)
(x4 , t4) = (10 m, 3.0 sec)
(x5 , t5) = ( 0 m, 4.0 sec)
Compare displacement distance avg. vel. avg. speed
t = 1 s x1,2 = x2 – x1 = 5 m d = 5 m vx,avg= 5 m/s vx,avg= 5 m/s
t = 2 s x1,3 = x3 – x1 = 20 m d = 20 m vx,avg= 10 m/s vx,avg= 10 m/s
t = 3 s x1,4 = x4 – x1 = 0 m d = 40 m vx,avg= 0 m/s vx,avg= 13 m/s
Here d = |x1,2 | + |x2,3 |+ |x3,4 | = 5 m + 15 m + 20 m = 40 mSpeed and velocity measure different things!
t (seconds)x
(met
ers
)10
30
20
1 2 430
Physics 201: Lecture 2, Pg 14
Calculating path distance in general
dtdxdf
i
t
tdtdx || ||
|| ixd
Physics 201: Lecture 2, Pg 15
Exercise 2 Average Velocity
x (meters)
t (seconds)
2
6
-2
4
What is the magnitude of the average velocity over the first 4 seconds ?
(A) -1 m/s (D) not enough information to decide.
(C) 1 m/s(B) 4 m/s
1 2 430
Physics 201: Lecture 2, Pg 16
Average Velocity Exercise 3 What is the average velocity in the last second (t = 3 to 4) ?
A. 2 m/s
B. 4 m/s
C. 1 m/s
D. 0 m/s
x (meters)
t (seconds)
2
6
-2
4
1 2 43
Physics 201: Lecture 2, Pg 17
Average Speed Exercise 4
What is the average speed over the first 4 seconds ?0 m to -2 m to 0 m to 4 m 8 meters total
A. 2 m/s
B. 4 m/s
C. 1 m/s
D. 0 m/s
x (meters)
t (seconds)
2
6
-2
4
1 2 43
turning point
Physics 201: Lecture 2, Pg 18
Instantaneous velocity
• Limiting case as the change in time 0
dtdxv t
x
tx
)time()ntdisplaceme(
0lim
x
t0
• As t 0 velocity is the tangent to the curve (& path)
• Dashed green line is vx
• Yellow lines areaverage velocities instantaneous velocity at
t = 0 s
Physics 201: Lecture 2, Pg 19
Instantaneous speed
• Just the magnitude of the instantaneous velocity
sv dtdx
tx
tx
|||lim||| )time(
)ntdisplaceme(
0
Physics 201: Lecture 2, Pg 20
What is the instantaneous velocity at the fourth second?
(A) 4 m/s (D) not enough information to decide.
(C) 1 m/s(B) 0 m/s
x (meters)
t (seconds)
2
6
-2
4
1 2 43
Exercise 5Instantaneous Velocity
Physics 201: Lecture 2, Pg 21
Special case: Instantaneous velocity is constant
• Slope is constant over a time t.
tx
dtdxvx
constant
x
t0
t
x (xi , ti)
(xf , tf)
Physics 201: Lecture 2, Pg 22
Special case: Instantaneous velocity is constant
• Slope is constant over a time t.
x
t0
t
x (xi , ti)
(xf , tf)
ixf
xif
xtvx
tvxx
• Given t, xi and vx we can deduce xf
and this reflects the area under the velocity curve
txx
txv if
x
Physics 201: Lecture 2, Pg 23
Now multiple vx ; Pat’s velocity plot
t (seconds)x
(met
ers
)10
30
20
1 2 430
v x (m
/s) 10
20
1 2 430
-10
-20t (seconds)
(x1 , t1) = (10 m, 0.0 s)
(x2 , t2) = (15 m, 1.0 s)
(x3 , t3) = (30 m, 2.0 s)
(x4 , t4) = (10 m, 3.0 s)
(x5 , t5) = ( 0 m, 4.0 s)
Physics 201: Lecture 2, Pg 24
Home exercise 6 (and some things are easier than they appear)
A marathon runner runs at a steady 15 km/hr. When the runner is 7.5 km from the finish, a bird begins flying from the runner to the finish at 30 km/hr. When the bird reaches the finish line, it turns around and flies back to the runner, and then turns around again, repeating the back-and-forth trips until the runner reaches the finish line.
How many kilometers does the bird travel?
A. 10 km
B. 15 km
C. 20 km
D. 30 km
Physics 201: Lecture 2, Pg 25
Objects with slowly varying velocities
dt
dxvx
] offunction a is [ )( txtxx x
vx
t
t
Change of the change….
changes in velocity with time
give average acceleration
tv
a xx
avg,
t
vx
Physics 201: Lecture 2, Pg 26
And finally instantaneous acceleration
2
2
adt
xddtdvx
x
t
ax
t
x
vx
t
t
tv
a xx
avg,
dtdv
a xx
tx tv
0lim
Physics 201: Lecture 2, Pg 27
Example problem
A car moves to the right first for 2.0 sec at 1.0 m/s and then 4.0 seconds at 2.0 m/s.
What was the average velocity?
Two legs with constant velocity but ….
vx
t
221
avgx,
vvv
Physics 201: Lecture 2, Pg 28
Example problem
A particle moves to the right first for 2.0 seconds at 1.0 m/s and then 4.0 seconds at 2.0 m/s.
What was the average velocity?
Two legs with constant velocity but ….
We must find the total displacement (x2 –x0) And x1 = x0 + v0 (t1-t0) x2 = x1 + v1 (t2-t1) Displacement is (x2 - x1) + (x1 – x0) = v1 (t2-t1) + v0 (t1-t0) x2 –x0 = 1 m/s (2 s) + 2 m/s (4 s) = 10 m in 6.0 s or 1.7 m/s
vx
t2vv
v 21Avg
Physics 201: Lecture 2, Pg 29
Position, velocity & acceleration
All are vectors! Cannot be used interchangeably (different units!)
(e.g., position vectors cannot be added directly to velocity vectors)
But the directions can be determined “Change in the position” vector vs. time gives the
direction of the velocity vector “Change in the velocity” vector vs. time gives the
direction of the acceleration vector Given x(t) vx(t) ax (t) Given ax (t) vx (t) x(t)
v
a
Physics 201: Lecture 2, Pg 30
Assignment
Reading for Tuesday’s class
» All of Chapter 2
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