Lecture 2:
One Dimensional Motion
Position, Distance, and Displacement
Before describing motion, you must set up a coordinate
system – define an origin and a positive direction.
The distance is the total length of
travel; if you drive from your house to
the grocery store and back, you have
covered a distance of 8.6 mi. Your net
displacement is zero.
Position, Distance, and Displacement
If you drive from your house to the
grocery store and then to your friend’s
house, your displacement is -2.1 mi and
the distance you have traveled is 10.7 mi.
Walking the Dog
You and your dog go for a walk to the park.
On the way, your dog takes many side
trips to chase squirrels or examine fire
hydrants. When you arrive at the park, do
you and your dog have the same
displacement?
a) yes
b) no
Walking the Dog
You and your dog go for a walk to the park.
On the way, your dog takes many side
trips to chase squirrels or examine fire
hydrants. When you arrive at the park, do
you and your dog have the same
a) yes
b) no
displacement?
Yes, you have the same displacement. Because you and your dog had
the same initial position and the same final position, then you have (by
definition) the same displacement.
Follow-up: have you and your dog traveled the same distance?
Average Speed
The average speed is defined as the distance
traveled divided by the time the trip took:
Average speed = distance / elapsed time
If you drive from your house to
the grocery store and then to your
friend’s house, your displacement
10.7 miles
sav =
is -2.1 mi and the distance you
0.5 hours
have traveled is 10.7 mi. The trip
= 21.4 miles/hr
takes 30 minutes (with traffic and
lights). What is your average speed?
Average Velocity
Average velocity = displacement / elapsed time
If you return to your starting point, your average
velocity is zero.
Walking the Dog II
You and your dog go for a walk to the
park. On the way, your dog takes
many side trips to chase squirrels or
examine fire hydrants. When you
arrive at the park, you affectionately
pat your dog’s head.
Which statement correctly describes
your average speed and velocity
relative to that of your dog, on the
trip from home to the park?
a) Average speed and average
velocity were both different
b) Average speed was the
same, but average velocity
was different
c) Average speed was different,
but average velocity was the
same
d) Average speed and average
velocity were the same
Walking the Dog II
You and your dog go for a walk to the
park. On the way, your dog takes
many side trips to chase squirrels or
examine fire hydrants. When you
arrive at the park, you affectionately
pat your dog’s head.
Which statement correctly describes
your average speed and velocity
relative to that of your dog, on the
a) Average speed and average
velocity were both different
b) Average speed was the
same, but average velocity
was different
c) Average speed was different,
but average velocity was the
same
d) Average speed and average
velocity were the same
trip from home to the park?
Your dog’s many side trips mean that he travelled more distance than you
in the same amount of time, so his average speed must have been
greater. However, his net displacement was the same (starting and
ending at the same place), so his average velocity must have been the
same.
Averaging Speed
Is the average speed of the red car
a) 40.0 mi/h,
b) more than 40.0 mi/h, or
c) less than 40.0 mi/h?
Hint: is t1 = t2?
Position vs. Time
Consider this motion sequence,
plotted here in one-dimension...
...and here as
an x-t graph
average velocity
Instantaneous Velocity
Evaluating the average velocity over a shorter and
shorter period of time, one approaches the
“instantaneous velocity”.
The instantaneous velocity is tangent to the curve.
Graphical Interpretation of Average and
Instantaneous Velocity
Position v. Time I
The graph of position versus
time for a car is given below.
What can you say about the
velocity of the car over time?
a) it speeds up all the time
b) it slows down all the time
c) it moves at constant velocity
d) sometimes it speeds up and
sometimes it slows down
e) not really sure
x
t
Position v. Time I
The graph of position versus
time for a car is given below.
What can you say about the
velocity of the car over time?
a) it speeds up all the time
b) it slows down all the time
c) it moves at constant velocity
d) sometimes it speeds up and
sometimes it slows down
e) not really sure
The car moves at a constant velocity
x
because the x vs. t plot shows a straight
line. The slope of a straight line is
constant. Remember that the slope of x
vs. t is the velocity!
t
Position v. Time II
a) it speeds up all the time
The graph of position vs. time
b) it slows down all the time
for a car is given below. What
c) it moves at constant velocity
can you say about the velocity
of the car over time?
d) sometimes it speeds up and
sometimes it slows down
e) not really sure
x
t
Position v. Time II
a) it speeds up all the time
The graph of position vs. time
b) it slows down all the time
for a car is given below. What
c) it moves at constant velocity
can you say about the velocity
of the car over time?
d) sometimes it speeds up and
sometimes it slows down
e) not really sure
The car slows down all the time because
the slope of the x vs. t graph is
diminishing as time goes on. Remember
that the slope of x vs. t is the velocity! At
large t, the value of the position x does
not change, indicating that the car must
be at rest.
x
t
Acceleration
Average acceleration:
Instantaneous acceleration:
Graphical Interpretation of Average and
Instantaneous Acceleration
Sign of Acceleration
Does deceleration mean “negative acceleration”?
No. It means “decreasing speed”.
Speed
Speed increasing
Decreasing
Acceleration
Acceleration (increasing speed) and deceleration
(decreasing speed) should not be confused with
the direction (or sign) of velocity and acceleration:
Graphical Interpretation of Average and
Instantaneous Acceleration
Constant Acceleration
If the acceleration is constant, the velocity
changes linearly with time:
Average velocity:
v
Propeller car
a:
v
c:
t
t
v
v
b:
d:
t
e: Not Sure
Which of the above plots
represents the v vs. t
graph for the motion of
the propeller car after it
was pushed?
t
v
Propeller car
a:
v
c:
t
t
v
v
b:
d:
t
e: Not Sure
Which of the above plots
represents the v vs. t
graph for the motion of
the propeller car after it
was pushed?
The car has a negative initial
velocity, but a constant
positive acceleration. So the
plot should start at a negative
value, and increase linearly
with increasing time.
t
Motion with Constant Acceleration
Position as a function of time
Motion with Constant Acceleration
The relationship between position and time
follows a characteristic curve.
My guess: a ~ 7 m/s2
Propeller car
displacement
Which of the displayed
plots represents the x
vs. t graph for the
motion of the propeller
car after it was
pushed?
x
t
Propeller car
displacement
Which of the displayed
plots represents the x
vs. t graph for the
motion of the propeller
car after it was
pushed?
x
t
The car starts at zero position, with a negative initial
velocity, so the displacement is growing in the negative
direction. The slope is changing linearly with time to
become more positive. It should be a smooth change, go
through a minimum, and then change more and more
quickly with increasing timex = v0t + 1/2 at2
Motion with Constant Acceleration
Velocity as a function of position
Stopping Distance
deaccelerating with constant a, take final velocity = 0,
find (x-x0) = “stopping distance”
1/2 Stopping Distance
v = 0.7 v0
3/4 Stopping Distance
Tip for save driving: if you double your speed, what
happens to your stopping distance?
Hit the Brakes!
Freely Falling Objects
Free fall from rest:
Free fall is the motion of
an object subject only to
the influence of gravity.
The acceleration due to
gravity is a constant, g.
g = 9.8 m/s2
For free falling objects,
assuming your y axis is
pointing up, a = -g = -9.8 m/s2
Reaction Time
In the peculiar units used
by Professor Norum, his
reaction time is measured
to be 20 cm. What is the
corresponding reaction
time, in seconds? [Use
g=10 m/s2]
a) 0.1 seconds
b) 0.2 seconds
c) 2 seconds
d) not enough information to work the
problem
e) not really sure
With v0=0 and x0 = 0, position as a function of time is:
20 cm = 0.2 m, so:
t = 0.2 seconds
1-D motion of a vertical projectile
1-D motion of a vertical projectile
v
a:
t
v
b:
t
v
c:
v
t
d:
t
- Clickers, everyday, in class.
- Assignment 1 on MasteringPhysics. Due
Monday, August 29, at midnight!
- Reading, for next class (Chapter 2, 3.1-3.6)
- When you exit, please use the REAR doors!
Air Track
The car undergoes constant
acceleration on the tilted
airtrack. Where will the velocity
be twice that measured at
0.5m?
a) 1 m
b) 1.4 m
c) 2 m
d) not enough information to
solve the problem
e) not really sure
With v0=0 and x0 = 0, velocity as a function of position is:
We want v2 / v1 = 2:
x2 = 4x1 = 2 m