Transcript Physics 207, Lecture 3, Sept. 10

Physics 207, Lecture 3, Sept. 10

Goals (finish Chap. 2 & 3)
 Understand the relationships between position,
velocity & acceleration in systems with 1-dimensional
motion and non-zero acceleration (usually constant)
 Solve problems with zero and constant acceleration
(including free-fall and motion on an incline)
 Use Cartesian and polar coordinate systems
 Perform vector algebra
Assignment:
1. For Monday: Read Chapter 4
2. Homework Set 2 (due Wednesday 9/17)
Physics 207: Lecture 3, Pg 1
Position, velocity & acceleration for motion
along a line

If the position x is known as a function of time, then
we can find both the instantaneous velocity vx and
instantaneous acceleration ax as a function of time!
x
x  x(t ) [ x is a function of t ]
dx
vx 
dt
dv x d 2 x
ax 
 2
dt
dt
t
vx
t
ax
t
Physics 207: Lecture 3, Pg 2
Position, displacement, velocity & acceleration

All are vectors and so vector algebra is a must !
These cannot be used interchangeably (different units!)
(e.g., position vectors cannot be added directly to velocity
vectors)
 But we can determined directions
 “Change in the position” vector gives the direction of

the velocity vector
v
 “Change in the velocity” vector gives the direction of
the acceleration vector 

a


Given x(t)  vx(t)  ax (t)
Given ax (t)  vx (t)  x(t)
Physics 207: Lecture 3, Pg 3
And given a constant acceleration we
can integrate to get explicit vx and ax
x  x(t ) [ x is a function of t ]
dx
vx 
dt
dv x d 2 x
ax 
 2
dt
dt
x0
t
vx
x  x0  v x t  a x t
0
x
1
2
0
2
t
ax
v x  v x  a x t
0
a x  const
t
Physics 207: Lecture 3, Pg 4
Exploiting
( x  x0 )  t
2
A “biology” experiment
 Hypothesis: Older people have slower reaction
times
 Distance accentuates the impact of time differences
 Equipment: Ruler and four volunteers
 Older student
 Younger student
 Record keeper
 Statistician
 Expt. require multiple trials to reduce statistical
errors.

Physics 207: Lecture 3, Pg 5
Rearranging terms gives two other relationships

For constant acceleration:
x  x0  v x t  12 a x t 2
0
v x  v x  a x t
0
a x  const

From which we can show (caveat: a constant acceleration)
v 2x  v x2  2a x (x  x 0 )
0
v x (avg)
1
 (v x  v x )
2
ax
0
t
Physics 207: Lecture 3, Pg 6
Acceleration

Changes in a particle’s motion often involve acceleration
 The magnitude of the velocity vector may change
 The direction of the velocity vector may change
(true even if the magnitude remains constant)
 Both may change simultaneously
a
v0
v1
v2
v3
v4
v5
a
a
a
a
a
a
t
t
a t = area under curve = v
v
0
v(t)=v0 + a t
t
Physics 207: Lecture 3, Pg 7
Acceleration has its limits
“High speed motion picture camera frame: John Stapp is caught in the teeth of
a massive deceleration. One might have expected that a test pilot or an
astronaut candidate would be riding the sled; instead there was Stapp, a mild
mannered physician and diligent physicist with a notable sense of humor.
Source: US Air Force photo
Physics 207: Lecture 3, Pg 8
Free Fall

When any object is let go it falls toward the ground !!
The force that causes the objects to fall is called
gravity.

This acceleration on the Earth’s surface, caused by
gravity, is typically written as “little” g

Any object, be it a baseball or an elephant,
experiences the same acceleration (g) when it is
dropped, thrown, spit, or hurled, i.e. g is a constant.
Physics 207: Lecture 3, Pg 9
Exercise 1
Motion in One Dimension
When throwing a ball straight up, which of the
following is true about its velocity v and its
acceleration a at the highest point in its path?
A.
B.
C.
D.
Both v = 0 and a = 0
v  0, but a = 0
v = 0, but a  0
None of the above
y
Physics 207: Lecture 3, Pg 10
Exercise 1
Motion in One Dimension
When throwing a ball straight up, which of the following is
true about its velocity v and its acceleration a at the highest
point in its path?
A.
B.
C.
D.
Both v = 0 and a = 0
v  0, but a = 0
v = 0, but a  0
None of the above
y
Physics 207: Lecture 3, Pg 11
Exercise 2
More complex Position vs. Time Graphs
In driving from Madison to Chicago, initially my speed is at a
constant 65 mph. After some time, I see an accident ahead of me on
I-90 and must stop quickly so I decelerate increasingly “fast” until I
stop. The magnitude of my acceleration vs time is given by,
•
t
Question: My velocity vs time graph looks
like which of the following ?
a
A.

v
t
B.
C.



v

v
Physics 207: Lecture 3, Pg 12
Exercise 2
More complex Position vs. Time Graphs
In driving from Madison to Chicago, initially my speed is at a
constant 65 mph. After some time, I see an accident ahead of me on
I-90 and must stop quickly so I decelerate increasingly fast until I
stop. The magnitude of my acceleration vs time is given by,
•
t
Question: My velocity vs time graph looks
like which of the following ?
a
A.

v
t
B.
C.



v

v
Physics 207: Lecture 3, Pg 13
Gravity facts:

g does not depend on the nature of
the material !
 Galileo (1564-1642) figured this
out without fancy clocks & rulers!

Feather & penny behave just the
same in vacuum

Nominally,
g = 9.81 m/s2
At the equator
g = 9.78 m/s2
At the North pole g = 9.83 m/s2
Physics 207: Lecture 3, Pg 14
Gravity Map of the Earth
(relief exaggerated)
A person off the
coast of India would
weigh 1% less
than at most other
places on earth.
Physics 207: Lecture 3, Pg 15
Gravity map of the US
Red: Areas of stronger local g
Blue: Areas of weaker local g
Due to density variations of the Earth’s crust and mantle
Physics 207: Lecture 3, Pg 16
Exercise 3 1D Freefall

Alice and Bill are standing at the top of a cliff of
height H. Both throw a ball with initial speed v0,
Alice straight down and Bill straight up. The speed
of the balls when they hit the ground are vA and vB
respectively.
A.
vA < vB
Alice
B.
vA = v B
v0
Bill
v0
C.
vA > vB
H
vA
vB
Physics 207: Lecture 3, Pg 17
Exercise 3 1D Freefall

Alice and Bill are standing at the top of a cliff of
height H. Both throw a ball with initial speed v0,
Alice straight down and Bill straight up. The speed
of the balls when they hit the ground are vA and vB
respectively.
A.
vA < vB
Alice
B.
vA = v B
v0
Bill
v0
C.
vA > vB
H
vA
vB
Physics 207: Lecture 3, Pg 18
Problem Solution Method:
Five Steps:
1)
Focus the Problem
-
2)
Describe the physics
-
3)
what are the relevant physics equations
Execute the plan
-
5)
what physics ideas are applicable
what are the relevant variables known and unknown
Plan the solution
-
4)
draw a picture – what are we asking for?
solve in terms of variables
solve in terms of numbers
Evaluate the answer
-
are the dimensions and units correct?
do the numbers make sense?
Physics 207: Lecture 3, Pg 20
Example of a 1D motion problem



A cart is initially traveling East at a constant speed of
20 m/s. When it is halfway (in distance) to its destination
its speed suddenly increases and thereafter remains
constant. All told the cart spends a total of 10 s in transit
with an average speed of 25 m/s.
What is the speed of the cart during the 2nd half of the trip?
Dynamical relationships:
x  x0  v x t  a x t
0
v x  v x  a x t
0
a x  const
And
1
2
2
v 2x  v x2  2a x (x  x 0 )
0
v x (avg)
1
 (v x  v x )
2
0
x(displaceme nt )
vaverage velocity  
t ( total time )
Physics 207: Lecture 3, Pg 21
The picture
x0


t0
v0
v1 ( > v0 )
a0=0 m/s2
a1=0 m/s2
x1 t1
x
x x2  x0
v

t t2  t0
t2
2
Plus the average velocity
Knowns:
 x0 = 0 m
 t0 = 0 s
0
x
 v0 = 20 m/s
vx  vx
 t2 = 10 s
ax  0
 vavg = 25 m/s
 relationship between x1 and x2
Four unknowns x1 v1 t1 & x2 and must find v1 in terms of knowns
x  x  v t
0
0

Physics 207: Lecture 3, Pg 22
x  x0  v x t
Using
0
x0
v0
v1 ( > v0 )
a0=0 m/s2
a1=0 m/s2
t0
x1  x0  v0 (t1  t0 )
Four
unknowns
 Four
relationships

x1 t1
x
x2  x1  v1 (t2  t1 )
t2
2
x x2  x0
v

t t2  t0
x1  ( x2  x0 )
1
2
Physics 207: Lecture 3, Pg 23
x0  0
Using
x0
v0
v1 ( > v0 )
a0=0 m/s2
a1=0 m/s2
t0
x1 t1
1
x1  v0 t1
Eliminate
unknowns
first t1
3

next x1
t0  0
1&2
3
2
x
x2  x1  v1 (t2  t1 )
x1  x2
1
2
t2
2
x2
v
t2
4
x2  x1  v1 (t2  )
x1
v0
x2 
x  v1 (t2  )
1
2 2
x2
v0
1
2
Physics 207: Lecture 3, Pg 24
Now Algebra and Relationship 4
x0

4
v1 ( > v0 )
a0=0 m/s2
a1=0 m/s2
t0
x1 t1
Algebra to simplify
x2 
1
2
v0
1
2
x2  v1 (t2  )
x2
v0
1
2
x2  v1 (t2  )
x2
v0
x2
v
t2
x
t2
2
1
2
vt2  v1 (2t2  )
v t2
v0
v1  (
v v0
2 v0  v
)
v  v1 (2  vv )
0
Physics 207: Lecture 3, Pg 25
Fini
x0


v0
v1 ( > v0 )
a0=0 m/s2
a1=0 m/s2
t0
Plus the average velocity
Given:
 v0 = 20 m/s
 t2 = 10 s
 vavg = 25 m/s
x1 t1
x
2
v1 
v v0
2 v0  v
v1 
25 m/s  20 m/s
2 20 m/s  25 m/s

t2
500 m/s
15
 33.3 m/s
Physics 207: Lecture 3, Pg 26
Tips:

Read !
Before you start work on a problem, read the
problem statement thoroughly. Make sure you
understand what information is given, what is
asked for, and the meaning of all the terms
used in stating the problem.

Watch your units (dimensional analysis) !
Always check the units of your answer, and
carry the units along with your numbers during
the calculation.

Ask questions !
Physics 207: Lecture 3, Pg 27
See you Monday
(Chapter 3 on Monday….)
Assignment:

For Monday, Read Chapter 4

Mastering Physics Problem Set 2
Physics 207: Lecture 3, Pg 32