Physics 1110: Motion in 1D
Announcements:
•  Wasn’t around yesterday;
Had a 65 ft spruce tree crash
thru my roof in the wind
storm
Good talk on Sat at 2pm in
G1B30 by Professor Emeritus
Albert Bartlett
•  CAPA homework due Tuesday at 10pm
Web page: http://www.colorado.edu/physics/phys1110/phys1110_sp12/
1
Vectors
•  Vectors have magnitude and direction
–  velocity, acceleration, displacement
•  Scalars only have magnitude
–  speed, temperature, mass, volume
•  Vector quantities are indicated by bold and/or an
arrow above:
•  Can use arrow to represent direction & magnitude
2
Reference frame
•  To solve any problem, need to pick a reference
frame and conduct all calculations in that frame
Standard 2D reference frame is
motionless with respect to the Earth
-x
and has x and y axes as shown
+y
+x
-y
•  For 1D motion, we choose the x-axis
to lie along the direction of motion
so the y-axis is unnecessary
3
1D kinematics (motion)
•  Define two quantities related to change in position:
–  Distance is a scalar which can be thought of as the
ground covered by the chosen path
–  Displacement is a vector between the start and
end points whose magnitude is the distance
between the start and end points:
•  In 1D, can simplify to
. The vector
direction can be along positive or negative axis so
the signs of x2 and x1 are still relevant.
4
Displacement and Distance
•  A person swims one lap (roundtrip) in an Olympic-sized pool (50m
long). What is the distance traveled
and the displacement?
•  Place origin at the starting point
–  Outbound trip:
–  Return trip:
Distance = 100 m
-100
-100
50
50
0
0
50
50
x
100 (m)
x
100 (m)
Displacement = Δx = x2 - x1 = 0 - 0 = 0 m
5
Clicker question 1
Set frequency to BA
An ant crawling along the floor follows a semi-circular path, going
half way around the circumference of a circle of radius R.
The
distance
traveled,
and
the
displacement
of
the
ant,
are
respectively:
A: π R and π R
B: 2 R and π R
C: π R and 2 R
D: π R and zero
E: none of these
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Clicker question 2
Set frequency to BA
An object goes from one point in space to another.
After it arrives at its destination, the size of its
displacement is:
A:
either greater than or equal to
B: always equal to
C:
either smaller than or equal to
D: could be smaller or larger
than the distance it traveled.
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Speed and velocity
•  Speed (scalar) is distance divided by time
•  Velocity (vector) is displacement
divided by time:
•  In 1D we drop the vector
notation but remember to
keep the sign of x1 and x2.
8
Clicker question 3
Set frequency to BA
A person starts in Boulder, drives to Denver (50 km away) in 1 hour,
stays in Denver 1 hour, then speeds back to Boulder in 30 minutes.
50
km
in
1
hour
Denver
start
5inish
50
km
in
1/2
hour
What is the average speed of the round
trip?
A:
B:
C:
D:
E:
25 km/hr
67 km/hr
40 km/hr
75 km/hr
none of these
(wait
1
hour)
Ave speed = total distance/total
time = 100 km / 2.5 hours
= 40 km/hr
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Clicker question 4
Set frequency to BA
Q. In 2002, Steve Fosset traveled 32000 km around
the world in 13 days (1.0×106 s) in a balloon, landing
at his started point. What was his average velocity?
A. 0.032 m/s
B. 32 m/s
C. 0.0 m/s
D. 0.03 m/s
Since he started and ended at the same
place his displacement was 0 so
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Clicker question 5
Set frequency to BA
Rank order the speeds at times 1, 2, and 3 from the slowest to the fastest
A:
B:
C:
D:
E:
V1 < V2 < V3
V2 < V1 < V3
V3 < V1 < V2
V3 < V2 < V1
V2 < V3 < V1
Speed = |velocity|, the magnitude of velocity. Velocity is the slope of the "x vs t"
curve. At point 3, the slope is zero (it's flat). At point 2, it has a small (negative)
slope. The speed there is small and positive. At point 1, it has a large positive
slope. The speed there is large.
So, V3 < V2 < V1
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Displacement as a function of time
•  Slight label change for average velocity gives
where x0 and t0 are the starting
position and time. Can set t0 = 0.
•  Solving for x gives
•  If the velocity is constant:
is the initial velocity
•  If the velocity is not constant, x = x 0 +
where x 0 is the initial position at t=0
€
€
where
t
∫ v(t)dt
0
12
Clicker question 6
Set frequency to BA
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: 20 km
C: 15 km E: Not sure/impossible
to decide
The runner goes at 15 km/hr, and has to run 7.5 km to
finish. How long does that take her? Well, 1/2 an hour
(v*t = d, so t = d/v = 7.5 km / 15 km/hr = .5 hr)
And during that half hour, the little bird is flying back and
forth at a steady 30 km/hr. If it goes 30 km/hr * 1/2 an
hour = 15 km, that's how far the bird went!
13
x vs t graphs
•  A bicyclist goes 100 m in a straight line at constant
speed in 10 s. How can we represent this motion?
x(m)
•  Can plot the position x versus
100
time t over the 10 seconds
•  What is the average velocity
and speed of the bicyclist?
0
10 t(s)
Speed is
the same
•  Note this is the slope of the above graph
14
Non-constant velocity
Constant velocity v(m/s)
x(m)
100
10
0
10 t(s)
10 t(s)
Non-constant velocity v(m/s)
x(m)
100
0
0
20
10 t(s)
Slope of tangent line at each time
0
10 t(s)
gives instantaneous velocity
15
Clicker question 7
Set frequency to BA
Q. A train moves along a straight track. The
graph shows the train position as a function
of time. The graph shows that the train:
A. speeds up the whole time
B. slows down the whole time
C. moves at constant velocity
D. speeds up part of the time and
slows down part of the time
E. none of the above
The slope is largest at t = 0 and continually decreases as
time increases so the velocity is decreasing (but positive)
16
Acceleration
Instantaneous
velocity:
For constant velocity case:
so
We define acceleration as change of velocity over time
so
in 1D
Instantaneous acceleration is
Acceleration is slope of velocity vs time graph
For constant velocity,
17
Position, velocity, acceleration vs t
Constant velocity
x(m)
100
v(m/s)
a(m/s2)
10
0
2
0
10 t(s)
10 t(s)
Constant acceleration v(m/s)
x(m)
100
0
10 t(s)
a(m/s2)
20
2
0
10 t(s)
0
10 t(s)
0
10 t(s)
18
Set frequency to BA
Clicker question 8
Q. An object’s velocity vs time graph
is shown on the right. What best
describes the car’s acceleration vs
time?
a
a
A
B
t
t
a
a
C
t
D
t E. None of these
19
Clicker question 9
Set frequency to BA
A particle starts at the origin. Below is a graph of
velocity vs. time.
x is the AREA under the "v vs t"
curve. The area under the triangle
is (1/2) base*height = (1/2)*3 sec
*6 m/s=9 m.
Another approach is to observe that
"straight line v vs t" means
CONSTANT acceleration, so you
What is the approximate position at t=3 can use delta x = (1/2)* a * t^2.
Here, a is the slope of the line,
sec?
which is 6 m/s / (3 sec) = 2 m/s^2.
A: 3 m
B: 6 m
So Delta x = (1/2)*(2 m/s^2)*(3
C: 9 m
D: 18 m
sec)^2 = 9 m. Same answer, Will
E: None of these/not enough information.
derive next lecture!
20
Summary
•  Read Chapter 2 of H+R – “Motion along a
Straight Line”
•  Register your Clickers
•  Think about attending the talk on Saturday by
Al Bartlett
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