Physics 207, Lecture 3

Today (Finish Ch. 2 & start Ch. 3)
 Understand 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
Reading Assignment:
For Wednesday: Read Chapter 3 (carefully) through 4.4
Physics 207: Lecture 3, Pg 1
“2D” Position, Displacement
time (sec) 1
position -2,2
(x,y meters)
2
-1,2
3
0,2
4
1,2
5
y
6
2,2
3,2
x
position vectors
origin
displacement vectors
Physics 207: Lecture 3, Pg 2
Position, Displacement, Velocity
time (sec) 1
2
3
4
5
y
6
x
vx (avg)
x f  xi x


t f  ti
t
12
23
34
displacement vectors
velocity vectors
45
56
Velocity always has same magnitude & length  CONSTANT
x(t )  xi  vx t
vx  vx final  vx initial  0
a x  vx / t  0 m/s
2
Physics 207: Lecture 3, Pg 3
Acceleration


Particle motion often involves non-zero 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
E.g., a “particle” with smoothly decreasing speed
v1
v0

v
v1

v0  v  v1
v2

v

v
 
v  v1  v0
v5
v4
v3

v

v

aavg

v

t
Physics 207: Lecture 3, Pg 4
Average & Instantaneous Acceleration

Average acceleration

The instantaneous acceleration is the limit of the
average acceleration as ∆v/∆t approaches zero

Note: Position, velocity & acceleration are all vectors,
they cannot be added directly to one another
(different dimensional units)


 
v (t f )final  v (ti )initial v f  vi

t f  ti
t
Physics 207: Lecture 3, Pg 5
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 6
Going the other way….

Particle motion with constant acceleration
 The magnitude of the velocity vector changes

A particle with smoothly decreasing speed:
v x  v x  a x t
vx
ax 
t
0
v0
v1
v2
v3
v4
v5
a
a
a
a
a
a
v
a
t
0
ti
vf = vi + a t
= vi + a (tf - ti )
t
tf
0
a t = area under curve = v (an integral)
t
Physics 207: Lecture 3, Pg 7
So if constant acceleration we can integrate
to get explicit v and a
x
x0
x  x0  v x t  a x t
0
v x  v x  a x t
0
a x  const
1
2
2
t
vx
v0
t
ax
t
Physics 207: Lecture 3, Pg 8
Rearranging terms gives two other relationships

If constant acceleration then we also get:
v  v  2a x (x  x 0 )
2
x
v x (avg)
2
x0
1
 (v x  v x )
2
0
Physics 207: Lecture 3, Pg 9
An example problem

A particle moves to the right first for 2 seconds at 1 m/s
and then 4 seconds at 2 m/s.

What was the average velocity?
vx
vAvg
v1  v2

2
t

Two legs with constant velocity but ….

We must find the 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 seconds or 5/3 m/s



Physics 207: Lecture 3, Pg 10
x  x0  v x t  a x t
1
2
0
2
A particle starting at rest & moving along a line
with constant acceleration has a displacement
whose magnitude is proportional to t2
( x  x0 )   a x t
1
2
2
1. This can be tested
2. This is a potentially useful result
Physics 207: Lecture 3, Pg 11
Speed can’t really kill but acceleration may…
“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 12
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.
a y  -g
y(t )  y0  v y t  g t
0
1
2
Physics 207: Lecture 3, Pg 13
2
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
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 15
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
most like which of the following ?
a
A.

v
t
B.
C.



v

v
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.
v A < vB
Alice
B.
v A = vB
v0
Bill
v0
C.
v A > vB
H
vA
vB
Physics 207: Lecture 3, Pg 17
Exercise 3 1D Freefall : Graphical solution

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.
cliff
v0
vx
turnaround
point
back
at
cliff
v= -g t
t
identical displacements
(one + and one -)
-v0
vground
ground
ground
Physics 207: Lecture 3, Pg 18
Home Exercise,1D Freefall
The graph at right shows the y
velocity versus time graph for a
ball. Gravity is acting downward
in the -y direction and the x-axis
is along the horizontal.
Which explanation best fits the
motion of the ball as shown by
the velocity-time graph below?
A.
B.
C.
D.
E.
The ball is falling straight down, is caught, and is then thrown straight
down with greater velocity.
The ball is rolling horizontally, stops, and then continues rolling.
The ball is rising straight up, hits the ceiling, bounces, and then falls
straight down.
The ball is falling straight down, hits the floor, and then bounces
straight up.
The ball is rising straight up, is caught and held for awhile, and then is
thrown straight down.
Physics 207: Lecture 3, Pg 19
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
See you Wednesday
Assignment:

For Wednesday, Read through Chapter 4.4
Physics 207: Lecture 3, Pg 32