Transcript Lecture 2.OneDimensi.. - Faculty Web Sites at the University of Virginia

One Dimensional Motion –
Velocity & Acceleration
Lecturer:
Professor Stephen T. Thornton
Reading Quiz: The red and black lines
represent the motion of two cars. Which of
the following statements is most true?
A)
B)
C)
D)
E)
Red and black car have the same acceleration.
Red car always has a higher speed.
Black car always has a higher speed.
Black car always has a lower acceleration.
Red car always has a lower acceleration.
Answer: E
The red car has a constant velocity and
zero acceleration. Only the black car has
acceleration, and it is positive.
Last time
Went through class procedures,
WebAssign, etc.
Discussed length, mass, and time.
Significant figures, scientific notation.
Problem solving.
Today
Distance, displacement.
Average speed and average velocity.
Instantaneous velocity
Acceleration: average, instantaneous
Motion with constant acceleration
Reference Frames and Displacement
Any measurement of position, distance, or speed
must be made with respect to a reference frame.
For example, if you are sitting on a train and someone
walks down the aisle, the person’s speed with respect to
the train is a few miles per hour, at most. The person’s
speed with respect to the ground is much higher.
Copyright © 2009 Pearson Education, Inc.
One-Dimensional Coordinate System
Distance = total length traveled
Displacement
Definition:
displacement = change in position
= final position – initial position
Δx = xf - xi
One-Dimensional Motion Along the x-axis
finish
start
What is the total distance traveled (0-4s)?
What is the displacement? Do quiz next.
Conceptual Quiz: What is the total distance
traveled?
A)
B)
C)
D)
E)
-8m
-4 m
-2 m
+4 m
+8 m
finish
start
Conceptual Quiz: What is the total distance
traveled?
A)
B)
C)
D)
E)
-8m
-4 m
-2 m
+4 m
+8 m
finish
start
Conceptual Quiz: What is the total
displacement?
A)
B)
C)
D)
E)
-8m
-4 m
-2 m
+4 m
+8 m
finish
start
Conceptual Quiz: What is the total
displacement?
A)
B)
C)
D)
E)
-8m
-4 m
-2 m
+4 m
+8 m
Δx = xf – xi = (-1 m) – (1 m) = -2 m
finish
start
Average Speed
distance
average speed =
elapsed time
Note that this is always a positive
number.
Average Velocity
Velocity is different than speed,
because velocity is a vector.
displacement
average velocity =
elapsed time
x x f  xi
vav 

t t f  ti
One-Dimensional Motion Along the x-axis
finish
start
What is average velocity from 0 to 4 s?
x x f  xi (1m) - (1m) 2m
vav 



 0.5 m/s
t t f  ti
4s-0s
4s
direction?
Motion along the x axis
represented with an x-versus-t
graph (same motion as before)
Average Velocity on an x-Versus-t Graph
Look at average
from point A to B
vav  x  0
t
Average Velocity on an x-versus-t Graph
Now look
from 2 s to 3 s
vav  x  0
t
Instantaneous Velocity
x dx
v  lim

t 0 t
dt
In this way we can find the
velocity at any particular
instant of time.
What is the instantaneous
velocity at t = 1 s?
motion
Constant Velocity Graph
straight line
Instantaneous velocity is equal to average
velocity when velocity is constant.
Graphical Interpretation of Average
and Instantaneous Velocity
Observable Universe. Recent findings in
astrophysics suggest that the observable
Universe can be modeled as a sphere of radius
R = 13.7 x 109 light-years with an average
mass density of about 1 x 10-26 kg/m3 where
only about 4% of the Universe’s total mass is
due to “ordinary” matter (such as protons,
neutrons, and electrons). Use this information
to estimate the total mass of ordinary matter in
the observable Universe.
1lightyear  9.46  10 m.
15
Airplane Motion. An airplane
travels 3100 km at a speed of 720
km/h, and then encounters a tailwind
that boosts its speed to 990 km/h for
the next 2800 km. What was the total
time for the trip? What was the
average speed of the plane for this
trip?
Acceleration
Just like velocity is given by the rate
of change of position with respect to
time, the acceleration is given by the
rate of change of velocity with
respect to time.
We are still dealing with onedimensional motion, so vector
direction is simple.
We use signs to denote the directions of both
velocity and acceleration along a particular axis.
x
 When v is +, motion is to right.
 When v is -, motion is to left.
 When motion is to the right, and a is +, then object
speeds up (accelerates) to the right (case II).
 When motion is to the left,
and a is +, then object is
slowing down and will
eventually turn to the right.
(not shown here)
Graphical Interpretation of Average
and Instantaneous Acceleration
If you don’t
remember
about tangents,
please review!
Average Acceleration
v v f  vi
aav 

t t f  ti
We must be very careful with
units. What are they?
m/s2
Run physlet
http://physics.bu.edu/~duffy/semester1/semest
er1.html
Constant velocity, constant acceleration
Play with these physlets at home and try the
different cases. Time is too short here.
Instantaneous Acceleration
x dx
v  lim

t 0 t
dt
v dv d  dx  d x
a  lim

   2
t 0 t
dt dt  dt  dt
2
Similarity between velocity and
acceleration is clear.
Conceptual Quiz: The graph shows position as
a function of time for two trains running on
parallel tracks. Which of the following is true?
A) At time tB, both trains have the same velocity.
B) Both trains speed up all the time.
C) Both trains have the
same velocity at some
time before tB.
D) Somewhere on the
graph, both trains
have the same
acceleration.
time
Conceptual Quiz: The graph shows position as
a function of time for two trains running on
parallel tracks. Which of the following is true?
A) At time tB, both trains have the same velocity.
B) Both trains speed up all the time.
C) Both trains have the
same velocity at some
time before tB.
D) Somewhere on the
graph, both trains
have the same
acceleration.
time
Motion with Constant Acceleration
If the acceleration is constant, then we have
v  v0  at
where v = v0 at t = 0. This result is easy to show
from our definition of a.
The Average Velocity
vf
vi
constant acceleration
Constant acceleration
Another important result:
1
1
vav   v0  v   (vi  v f )
2
2
Note that I have used vi and vf , which is
more general than using v0 and v, because
we may want to find the average between
some initial and final position other than v0
and a general v.
Let’s determine some important equations.
x x f  xi x  x0
vav 


t
t f  ti
t 0
x  x0
vav 
t
or vavt  x  x0
solve for x :
x  x0  vavt
Insert our previous result for vav
1
x  x0  (v0  v)t
2
Only for constant acceleration!!!!!
1
x  x0  (v0  v )t
2
This is a very important equation.
It relates the position x to the
velocity v as a function of time t.
But we also know the relationship
between velocity v and
acceleration a. It was
v  v0  aconstant t
1
x  x0  (v0  v)t
2
1
x  x0  v0  (v0  aconstant t ) t
2
1
2
x  x0  v0t  aconstant t
2
v  v0  at
solve for t ; a is constant
v  v0
t
a
1
We had x  x0  (v0  v)t
2
Substitute in for t from above
1
 v  v0 
x  x0  (v  v0 ) 

2
 a 
v v
x  x0 
or
2a
2
2
0
v  v  2a  x  x0 
2
2
0
We now have all the equations we need to
solve constant-acceleration problems.
v  v0  at
1 2
x  x0  v0t  at
2
v  v  2a  x  x0 
2
2
0
1
1
vav   v0  v   (vi  v f )
2
2
Make sure you look carefully at
the examples in the text. I will
not go through those.
Conceptual QuizI
You drive for 30 minutes at 30
A) more than 40 mi/hr
mi/hr and then for another 30
B) equal to 40 mi/hr
minutes at 50 mi/hr. What is your
average speed for the whole trip?
C) less than 40 mi/hr
Conceptual Quiz
You drive for 30 minutes at 30
A) more than 40 mi/hr
mi/hr and then for another 30
B) equal to 40 mi/hr
minutes at 50 mi/hr. What is your
C) less than 40 mi/hr
average speed for the whole trip?
It is 40 mi/hr in this case. Because the average speed is distance/time
and you spend the same amount of time at each speed, then your
average speed would indeed be 40 mi/hr.
Conceptual Quiz
You drive 4 miles at 30 mi/hr and
A) more than 40 mi/hr
then another 4 miles at 50 mi/hr.
B) equal to 40 mi/hr
What is your average speed for
C) less than 40 mi/hr
the whole 8-mile trip?
Conceptual Quiz
You drive 4 miles at 30 mi/hr and
A) more than 40 mi/hr
then another 4 miles at 50 mi/hr.
B) equal to 40 mi/hr
What is your average speed for
C) less than 40 mi/hr
the whole 8-mile trip?
It is not 40 mi/hr! Remember that the average speed is distance/time.
Because it takes longer to cover 4 miles at the slower speed, you are
actually moving at 30 mi/hr for a longer period of time! Therefore, your
average speed is closer to 30 mi/hr than it is to 50 mi/hr.