Issues/Bugs of the week:
●  Two issues this week, both due to relevant read/write
access settings not being correct:
●  Sakai Bonus Problem W1/Q5 did not display the
formula to be evaluated:
(hidden resource files are inaccessible, not just hidden)
●  Lecture 3 is now posted on the web
(write access to the directory has been corrected)
UF PHY2053, Lecture 3: Motion in One Dimension
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Sakai problem W1/Q5
UF PHY2053, Lecture 3: Motion in One Dimension
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Sakai problem W1/Q5
UF PHY2053, Lecture 3: Motion in One Dimension
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Sakai problem W1/Q5
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UF PHY2053, Lecture 3: Motion in One Dimension
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PHY2053, Fall 2013
Lecture 3:
Motion in One Dimension
UF PHY2053, Lecture 3: Motion in One Dimension
5
Clicker Question #1 (5 min)
“Half the distance”
UF PHY2053, Lecture 3: Motion in One Dimension
6
Clicker Question #1a (5 min)
A car is driving from city A to city B. It drives the first
half of the distance at a constant velocity of 15 m/s, and
then drives the second half of the distance at a constant
velocity of 30 m/s. What is the average velocity of the
car for this trip?
A.  17.5 m/s
B.  20.0 m/s ✔
C.  22.5 m/s
D.  This is not the correct answer
E.  27.5 m/s
UF PHY2053, Lecture 3: Motion in One Dimension
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The distance between cities is not specified. We can define it to be a value
“d”, and the final answer should not depend on “d”. The total driving time
(Δt) consists of two segments. We can denote them as Δt1 and Δt2, such
that Δt = Δt1 + Δt2. The average velocity is:
vavg
d
=
=
t
d
t1 +
t2
We need to compute Δt1 and Δt2. Δt1 is the time it took to drive half the
distance (d/2) at 15 m/s: Δt1
d/2
15 m/s =
→
t1
d
. Similarly,
t1 =
2 ⇥ 15 m/s
Then, the total driving time Δt = Δt1 + Δt2 is:
t=
t1 +
d
t2 =
2 ⇥ 30 m/s
d
d
2d + d
3d
d
t2 =
+
=
=
=
30 m/s 60 m/s
60 m/s
60 m/s
20 m/s
and the avg. velocity vavg
d
=
=
t
UF PHY2053, Lecture 3: Motion in One Dimension
d
1
d
20 m/s
d ⇥ 20 m/s
=
= 20 m/s
d⇥1
8
The Inverse Problem:
From Velocity to Position
UF PHY2053, Lecture 3: Motion in One Dimension
9
Computing positions from
velocity distributions
•  vx = 5 m/s
•  for 2 seconds (6s – 4s)
•  Change in position:
•  Δx = 2s × 5m/s = 10 m
•  Notice: if vx were
negative, this would
indicate movement in the
negative direction
UF PHY2053, Lecture 3: Motion in One Dimension
10
A more complex distribution..
•  Nontrivial shape
•  How would we compute
the change in position?
•  Start from a rough
approximation..
UF PHY2053, Lecture 3: Motion in One Dimension
11
A more complex distribution..
•  Rough approximation:
assume that the velocity
is constant in increments
of one second.
•  From t = 2s to t = 3s,
velocity ~ 4 m/s, distance
traveled ~ 4 m.
UF PHY2053, Lecture 3: Motion in One Dimension
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A more complex distribution..
•  Rough approximation:
assume that the velocity
is constant in increments
of one second.
•  From t=2s to t=3s,
velocity ~ 4 m/s, distance
traveled ~ 4 m.
•  Do this for the entire
time range of motion;
sometimes undershoot,
sometimes overshoot
UF PHY2053, Lecture 3: Motion in One Dimension
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A more complex distribution..
•  Improve approximation:
assume that the velocity
is constant in increments
of one half second.
•  Better description of
change of position, more
precise calculation
UF PHY2053, Lecture 3: Motion in One Dimension
14
A more complex distribution..
•  Improve approximation:
assume that the velocity
is constant in increments
of one half second.
•  Better description of
change of position, more
precise calculation
•  Computation with
increments of a quarter
second would be better;
•  Can continue to infinity..
UF PHY2053, Lecture 3: Motion in One Dimension
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A more complex distribution..
•  Can continue to infinity..
•  At which point we find
that the distance traveled
equals the area under the
graph of velocity vs. time
Δx
UF PHY2053, Lecture 3: Motion in One Dimension
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Clicker Problem #2 [3 min]
Average Velocity from
Velocity vs. Time Graph
UF PHY2053, Lecture 3: Motion in One Dimension
17
vx [m/s]
Clicker Problem #2a [2 min]
•  For the velocity
distribution on the left,
the average velocity
from t = 2s to t = 8 s is:
10
9
8
7
6
5
4
3
2
1
0
0
1 2
3
4
5 6
7
8 9 10
A.  Not the answer
B.  6.0 m/s
C.  5.5 m/s
D.  5.0 m/s
E.  4.0 m/s ✔
t [s]
UF PHY2053, Lecture 3: Motion in One Dimension
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UF PHY2053, Lecture 3: Motion in One Dimension
19
Acceleration
UF PHY2053, Lecture 3: Motion in One Dimension
20
Average and Instantaneous
Acceleration
•  the rate of change of velocity has physical relevance
•  By analogy with velocity, define average acceleration:
Δvx
aav,x ≡ Δt =
vxf − vxi
tf − ti
•  and instantaneous acceleration:
Δvx
vxf - vxi
= lim
ax ≡ lim
t →t
Δt
tf - ti
Δt→0
f
UF PHY2053, Lecture 3: Motion in One Dimension
i
21
Graphical Representation
B
A
UF PHY2053, Lecture 3: Motion in One Dimension
•  By analogy with
velocity in the graph of
position vs. time:
•  Average acceleration
between A and B is the
slope of the blue line
connecting them
•  Instantaneous
acceleration is the slope
of the tangent at the
point of interest (A)
22
Graphical Representation
•  By analogy with
velocity in the graph of
position vs. time:
B
Graphs of Velocity vs. Time
be particularly
•  will
Average
acceleration
useful and interesting:
between A and B is the
From the area under the graph
slopeobtain
of theΔx
blue line
From the slope of the chord obtain
avg. accel.
connecting
them
A
From the slope of the tangent,
inst. accel.
•  Instantaneous
acceleration is the slope
of the tangent at the
point of interest (A)
UF PHY2053, Lecture 3: Motion in One Dimension
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Motion along a line
with constant acceleration
UF PHY2053, Lecture 3: Motion in One Dimension
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Graphical representation
•  Special case: slope of the velocity
vs. time graph is constant!
•  Take advantage of this and derive
some practical formulas
•  First, the obvious one:
vx = vf x
UF PHY2053, Lecture 3: Motion in One Dimension
vix = ax
t
25
Position vs. Time for Const. Accel.
•  Reminder: Change in position is
the area under the velocity graph
•  Triangular area:
1
vx ⇥
t=
2
•  Rectangular area:
vix
1
2
ax ( t)
2
t
•  Total area = Rectangle + Triangle
for constant ax:
x = vix
UF PHY2053, Lecture 3: Motion in One Dimension
t+
1
2
ax ( t)
2
26
Velocity Changes ↔ Displacement
From:
vx = vf x
Follows:
Use in:
vix = ax
t=
x = vix
t
vx
ax
t+
1
2
ax ( t)
to obtain:
for constant ax:
2
vf x
UF PHY2053, Lecture 3: Motion in One Dimension
2
vix
= 2 ax ( x)
2
27
2
The full math:
in
x = vix
x = vix
x=
x=
(vf x
t+
vix )
ax
vix (vf x
x=
x=
Use
+
vix )
ax
2 vix vf x
2
2 vix
2ax
+ 2 vix vf x
vf2 x
vf2 x
2ax
2
vix
UF PHY2053, Lecture 3: Motion in One Dimension
+
vx
t=
1
2
1
2
ax
ax ( t)
ax
+
✓
vf x
◆2
vix )2
2 ax
2
2 vf x vix + vix
2 vix vf x
vf2 x
vix
ax
(vf x
vf2 x
2ax
2
2 ax
2
+ vix
2
2 vix
+
2
vix
= 2 ax ( x)2
28
Example Problem:
A Mustang GT500 goes from 0 to 60 mph in 4.1 seconds.
What is the average acceleration over this period?
The stopping distance from 60 mph has been measured to
be 120 ft. Assuming constant deceleration, find its
magnitude. [all answers should be computed in m/s2]
UF PHY2053, Lecture 3: Motion in One Dimension
29
Free Fall
UF PHY2053, Lecture 3: Motion in One Dimension
30
Free Fall
•  Special case of motion with
constant acceleration,
acceleration of Earth’s
gravitational field g= 9.80 m/s2
•  Important: as much as possible
try to keep a fixed convention
about what is the positive y
direction
•  Recommend: positive y is “up”.
UF PHY2053, Lecture 3: Motion in One Dimension
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Free Fall Height after 1, 2, 3 .. s
•  Acceleration of Earth’s
gravitational field g = 9.80 m/s2
•  Recall
x = vix +
1
2
ax ( t)
2
•  Let us consider viy = 0, yi=0:
start from standing still at origin
t
1s
2s
3s
4s
y
– ½gs2
– 2gs2
– 4.5gs2
– 8gs2
y/y(1s)
1
4
9
16
UF PHY2053, Lecture 3: Motion in One Dimension
32
Summary
•  Graph of velocity vs. time: very useful in describing the
motion of an object
•  Average Acceleration
Slope of the chord
•  Instantaneous Acceleration
Slope of the tangent
•  Distance traveled
Area under graph
•  Motion with constant acceleration:
•  Distance traveled during Δt:
1
x = vix +
•  Velocity change over a traversed distance
(“stopping distance”):
2
vf x
2
vix
2
ax ( t)
= 2ax
2
x
•  Free fall: special case of constant acceleration:
g = 9.81 m/s2 [convention: in the negative y direction]
UF PHY2053, Lecture 3: Motion in One Dimension
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