Transcript Physics 207: Lecture 2 Notes

Lecture 2
Chapter 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 1
Displacement, position, velocity & acceleration are the
main quantities that we will discuss today.
Informal Reading Quiz
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 2
Range of Lengths
Distance
Radius of Visible Universe
To Andromeda Galaxy
To nearest star
Earth to Sun
Radius of Earth
Willis Tower
Football Field
Tall person
Thickness of paper
Wavelength of blue light
Diameter of hydrogen atom
Diameter of proton
Length (m)
1 x 1026
2 x 1022
4 x 1016
1.5 x 1011
6.4 x 106
4.5 x 102
1 x 102
2 x 100
1 x 10-4
4 x 10-7
1 x 10-10
1 x 10-15
Physics 201: Lecture 2, Pg 3
Range of Times
Interval
Age of Universe
Age of Grand Canyon
Avg age of college student
One year
One hour
Light travel from Earth to Moon
One cycle of guitar A string
One cycle of FM radio wave
One cycle of visible light
Time for light to cross a proton
Time (s)
5 x 1017
3 x 1014
6.3 x 108
3.2 x 107
3.6 x 103
1.3 x 100
2 x 10-3
6 x 10-8
1 x 10-15
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 4
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 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
Example: Where is Boston ?
Choose origin at New 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 6
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 they
the same displacement.
However the distance travelled is
different.
Boston
Path 1
New York
Path 2
Physics 201: Lecture 2, Pg 7
Motion in One-Dimension (Kinematics)
Position

Position along a line; references xi and ti :
10 meters
 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
-x
+x
10 meters
O
Pat
Physics 201: Lecture 2, Pg 8
Displacement

One second later Pat is 15 meters to the right of the lamp
10 meters


At t = 1.0 s the position vector is ( xf , tf ) or (15 m, 1.0 s)
Displacement is just change in position


15 meters
x ≡ xf – xi
There is also a change in time

t ≡ tf – ti
O
xi
Δx
xf
Pat
Physics 201: Lecture 2, Pg 9
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
15 meters
10 meters
O
xi
Δx
xf
Pat
Physics 201: Lecture 2, Pg 10
Average Velocity
Changes in position vs Changes in time
•
Average velocity = displacement per time increment ,
includes BOTH magnitude and direction
x(displaceme nt )
vx ,avg average velocity  
t ( time increment )
vx ,avg
•
x(5 m to the right )

t (1 sec)
Pat’s average velocity was 5 m / s to the right
Physics 201: Lecture 2, Pg 11
Average Speed
Average speed, vavg, reflects a magnitude
 “How fast” without the direction.
 References the total distance travelled

distance taken along path d
v avg average speed  

t ( total time )
t
•
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 12
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)
x (meters)

30
20
10
0
1
2
3
4
t (seconds)
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 m
Speed and velocity measure different things!
Physics 201: Lecture 2, Pg 13
Calculating path distance in general
d 
 |x |
i
tf
d   | dx |  |
ti
dx
dt
| dt
Physics 201: Lecture 2, Pg 14
Exercise 2 Average Velocity
x (meters)
6
4
2
0
-2
1
2
3
4 t (seconds)
What is the magnitude of the average velocity over the first 4 seconds ?
(A) -1 m/s
(B) 4 m/s
(C) 1 m/s
(D) not enough
information to
decide.
Physics 201: Lecture 2, Pg 15
Average Velocity Exercise 3
What is the average velocity in the last second (t = 3 to 4) ?
x (meters)
6
4
2
-2
A.
B.
C.
D.
1
2
3
4 t (seconds)
2 m/s
4 m/s
1 m/s
0 m/s
Physics 201: Lecture 2, Pg 16
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
x (meters)
6
4
2
A.
B.
C.
D.
2 m/s
4 m/s
1 m/s
0 m/s
-2
1
2
3
4 t (seconds)
turning point
Physics 201: Lecture 2, Pg 17
Instantaneous velocity
•
Limiting case as the change in time  0
vx  lim
t 0
•
x ( displaceme nt )
t ( time )
x
Yellow lines are
average velocities
•
dx

dt
instantaneous velocity at
t=0s
0
t
As t  0 velocity is the tangent to the curve (& path)
Dashed green line is vx
Physics 201: Lecture 2, Pg 18
•
Instantaneous speed
•
Just the magnitude of the instantaneous velocity
| vx || lim
t 0
x ( displaceme nt )
t ( time )
||
dx
dt
| s
Physics 201: Lecture 2, Pg 19
Exercise 5
Instantaneous Velocity
x (meters)
6
4
2
-2
1
2
3
4 t (seconds)
What is the instantaneous velocity at the fourth second?
(A) 4 m/s
(B) 0 m/s
(C) 1 m/s
(D) not enough
information to
decide.
Physics 201: Lecture 2, Pg 20
Special case: Instantaneous velocity is constant
•
Slope is constant over a time t.
vx  constant  dx  x
dt
t
x
(xf , tf)
x
(xi , ti)
t
0
Physics 201: Lecture 2, Pg 21
t
Special case: Instantaneous velocity is constant
•
Slope is constant over a time t.
x

x
f
i

x
vx 

t
t
x
(xf , tf)
x f  xi  vx t
x f  vx t  xi
•
x
(xi , ti)
t
0
Given t, xi and vx we can deduce xf
and this reflects the area under the velocity curve
Physics 201: Lecture 2, Pg 22
t
30
20
10
0
1
2
3
4
3
4
t (seconds)
20
vx (m/s)
(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)
x (meters)
Now multiple vx ; Pat’s velocity plot
10
0
-10
1
2
-20
t (seconds)
Physics 201: Lecture 2, Pg 23
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 24
Objects with slowly varying velocities
x  x(t ) [ x is a function of t ]
x
vx
dx

dt
Change of the change…. vx
changes in velocity with time
vx
give average acceleration

a x ,avg
vx

t
t
t
t
Physics 201: Lecture 2, Pg 25
And finally instantaneous acceleration
a x ,avg
ax 
vx

t
vx
lim
t 0 t
x
t
vx
dvx

dt
2
dvx d x
ax 
 2
dt
dt
t
ax
Physics 201: Lecture 2, Pg 26
t
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?
vx
t

Two legs with constant velocity but ….
vx, avg
v1  v2

2
Physics 201: Lecture 2, Pg 27
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?
v Avg
v1  v 2

2
vx
t

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



Physics 201: Lecture 2, Pg 28
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

v vs. time gives the
 “Change in the position” vector
direction of the velocity vector 
 “Change in the velocity” vectoravs. time gives the
direction of the acceleration vector
 Given x(t)  vx(t)  ax (t)
 Given ax (t)  vx (t)  x(t)

Physics 201: Lecture 2, Pg 29
Assignment

Reading for Tuesday’s class
» All of Chapter 2
Physics 201: Lecture 2, Pg 30