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Physics 3A: 1-D Motion
Kinematics
motion
of objects, without concern
about what causes that motion.
Dynamics
- study of relation between motion and its causes.
Start with one dimensional (straight-line) motion Average Velocity
Consider the motion of the car in the following:
Positions of the Carat Various Times
t^
x
Position (s)
(m)
A^
0
30
B^
10
52
C^
20
38
D
30
0
E^
40
F^
50
-53^ Shoup – 33
Physics 3A: 1-D Motion
We
define
the average speed (scalar)
as:
where d is the distance traveled and
t is the time interval during
the movement.^ We
define
the average velocity (vector) as "
displacement vector
divided by the change in time
note that d depends on path, but
and thus
does not!
v
d
t
x
x^ f
x^ i
x^ f
x
i
i
v x
x^ ^ t
x
f
x i
t^ f
t^ i
i v^ x
x
Shoup – 34
Physics 3A: 1-D Motion
What are the dimensions of v? What are the units of v in SI system? How do you specify the direction of v in 1-D
motion?
with plus or minus sign.
if (x
> xf
) then v is positive,i
if (x
< xf
) then v is negative.i
m^ s
Shoup – 35
v
length^ time
L T
Physics 3A: 1-D Motion
Shoup – 36
Plotting the car's data gives us: Slope between to points on this graph
gives us the average velocity (v),for example between A & B: In addition, total displacement is the "area under the curve" of a
velocity versus time graph:
rise
run
v^ x
52 m
30 m
10 s
0s
m^ s
^ x
n
x^ n
n
v^ n
t^ n
^ x
lim
^ t
0
n
x^ n
lim
t
0
n
v^ n
t^ n
(Do Example 2.1)
Physics 3A: 1-D Motion
What if we want to know an objects velocity at an exact time?
compute object's "instantaneous velocity"
Instantaneous velocity
v ) – the velocity of an object at a specific
instant of time.
Consider: Take the limit: In calculus:
v^ x
lim
t
0
x^ t
v^ x
lim
^ t
0
x^ ^ t
dx^ dt
Shoup – 37
Physics 3A: 1-D Motion
Summary for
constant velocity, 1-D
Displacement (
vector
Average velocity (
vector
Instantaneous velocity (
vector
Speed (
scalar
Model
for constant velocity:
^ x
x^ f
x^ i
v^ x
x^ ^ t
x^ f
x^ i t^ f
t^ i
i
v^ x
t
x^
x^ f
x^ i
x^ f
x^ i
v^ x
t
x^ f
x^ i
v^ x
t
t^ i^
0
setting
v x
d
x dt
v
d t
Shoup – 38
Physics 3A: 1-D Motion
What if the velocity is not constant?
Acceleration!
We define average acceleration as the"time-rate-of-change"
of velocity, averaged over some time interval: What are the dimensions of acceleration?
[a] = length / time
2
Measures how rapidly the velocity is changing, i.e.a = 2 m/s
2
means velocity is changing 2 m/s in each second
a^ x
v^ xf
v^ xi t^ f
t^ i
v^ x ^ t^
Shoup – 39
Physics 3A: 1-D Motion
Shoup – 40
Just like instantaneous velocity, we can define instantaneous
acceleration which is the
acceleration at one instant in time
Properties of acceleration:
Since its a derivative of velocity wrt time, graphically its theslope of a velocity versus time plot at any point:
0
1
2
3
4
(^40 3020100) -10 -20 -
Slope => acceleration
Velocity (m/s)
Time (s) a^ x
lim
^ t
0
v^ x ^ t
d v^ x dt
Physics 3A: 1-D Motion
Shoup – 45
Now put in value of acceleration:
x f
x^ i
(^1) 2
v
xi
v^ xf
t
v^ xf
v^ xi
a^ x
t
x f
x^ i
^12
v
xi
v^ xi
at
t
x^ f
x^ i
v^ xi^ t^
12
a^
tx (^2)
x
t t
parabola slope = v
xi
slope = v
xf
Physics 3A: 1-D Motion
Shoup – 46
Now we can pull out value of time:
v^ xf
v^ xi
a^
tx
t
v^ xf^
v^ xi a^ x
x f
x^ i
^12
v xi
v^ xf
t
x^ f
x i
^12
v
xi
v^ xf
v xf
v xi a^ x
^ x
i
v 2 xf
v 2 xi 2 a
x
v^ 2 xf
v^ 2 xi
2 a
x
x f^
x^ i
Physics 3A: 1-D Motion
Shoup – 47
Summary:
Equation
Contains x
v i^
vf
a
t
X X
X
X
X
X
X
v^ xf^
v^ xi^
a x^ t
v x
^ v
xi
v^ xf
x f^
x i
^ v
xi
v^ xf
t
x^ f^
x^ i
v^ xi^
t^
a tx^
(^2)
v^ 2 xf
v^ 2 xi
2 a
x
x f^
x^
i
Physics 3A: 1-D Motion
Shoup – 48
Your Mission?
Problem solving
- Draw a picture (Model) 2. Extract from "words" useful information 3. Decide on concept (i.e. constant acceleration?) 4. select equation(s) which use info you have and
also
info you want
- Solve for unknown. 6. Adjust for
significant figures
and include
units
Example of constant acceleration ==>
Free-fall Motion
We all know that objects released near earth fall freelytoward earth's center. Due to gravity (see later lectures) free-falling objects undergoa constant acceleration (vector), pointed toward earth's centerwith magnitude:
a^ y^
g^
9.80 m
^2 s^
32 ft
^2 s^
Physics 3A: 1-D Motion
B A
C
D E F
What's happening? A.
v?, a?
v>0, a= -g
B.
v?, a?
v>0, v
<vb
a
C.
v?, a?
v=
D.
v?, a?
v= -v
a
E.
v?, a?
v< 0, |v
| > |ve
|d
F.
v?, a?
|v
| > |vf^
|e (^) Shoup – 49
Physics 3A: 1-D Motion
Shoup – 50
Converting our constant acceleration equations for
free-fall
motion gives: With g = 9.80 m/s
2 or 32 ft/s
2
v^ 2 yf
v 2 yi
2 g
y
f
y^
i
y^ f
y^ i^
v^ yi
t^
12
g t
(^2)
v^ yf
v^ yi^
g t
v^ y
(^1) 2
v yi
v^ yf
y^ f
y^ i
1 2
v yi^
v^ yf
t
