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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
s 2
32 ft
^2 s^
(Tony: Do Example 2.4)
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
2g
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
12
v^ yi
v^ yf
t
Physics 3A: 2-D Motion
Shoup – 51
Motion in two dimensions
Useful vectors which describe motion:
Position vector in 2-D (
Draw vector with tail at origin, and head at position ofparticle: in unit vector notation: if particle is moving, x= x(t) and y = y(t):
y
x
r
path of particle
r
x
i
y
j
^ r t
x
t
i
y
t
j
r
Physics 3A: 2-D Motion
Shoup – 52
Displacement vector in 2-D
Draw vector from initial position of particle to final positionof particle: in unit vector notation: Average velocity vector (
v
v
r t
^ x
x^ f
x^ i
y^
y^ f^
y^ i
^ t
t^ f
t^ i
r
x
ri
path of particle
∆ r^ rf
y
with:
r
r^ f
r^ i
r
x
i
y
j
Physics 3A: 2-D Motion
Shoup – 53
Instantaneous velocity vector in 2-D (
Average acceleration vector (
v
v
lim
t
0
r^ t
d
r d t
v
! d
" r d t
! d d t
x
$ i
% y
& j
'
! d x^ d t
$ i
% d y^ d t
& j
v
! v^ x
$ i
% v^
y
& j
( a
) a
v f
,
t^ f
, t^ i
./ v^
/ t^
y
x
vf
vi^
-v i vf ∆v t^ f
t^ i
∆v
Physics 3A: 2-D Motion
Shoup – 54
Instantaneous acceleration vector in 2-D (
If acceleration is constant, can use results from chapter 2 for x
y components separately:
0 a
1
a
2
lim
3 t
4 0
576
v
(^6) t
2
d
8
v d t
v xf
9 v xi
: a x^ t^
v yf
9 v yi
: a y^
t
; v f
< v xf
= i
v yf
? j
@ v f
A < v xi
a
tx^
= B i
A > v yi
a y^ t
?B j
C v f
E D v^ xi
F i
G v yi
H j I
EG a x
F i
G a^ y
H^ Ij^ t^ J v^ f
9
K v i
:
L a t
Physics 3A: 2-D Motion
Shoup – 55
Constant acceleration, do same for position: x f
M x i
N v xi
t
N 1 2
a x^
(^2) t
y f
M y i
N v^ yi
t
N 1 2
a y^
(^2) t
O r^ f
< x f
= i
y f
? j
P
r^ f
R Q
x
i
S
v
xi
t
S
a
x^
2 t
U T i
R S
y
i
S
v
yi
t
S
a
y^
2 t
V T j
P
r^
f
R Q
x
i
U i
S
y^
i
VT j
RS
v
xi
t
U i
S
v
yi
t
V
j
T
S
R a
x
U i
S
a
y
VT j^
2 t
W r^ f
X
Y r^ i
Z
[ v ti^
Z 1 2
a t
2
Physics 3A: 2-D Motion
Shoup – 60
Can also derive equation for path (y(x)):
y^ f
tan
i x
f
g
2 v
2 i cos
2
i
x^ 2 f
Physics 3A: 2-D Motion
Shoup – 61
The two dimensional vector equations for projectile motion are: The two most useful quantities to compute when dealing with
projectile motion are the maximum height, h, and the range(horizontal distance), R
r^ f
r^ i
v^ i
t
g t
(^2)
v^
f^
v^
i
g t
y
x
vti^
½ gt
2
rf O
(x,y)
y
x
O
h^ R
θ^ i vi
vy
=
A
B
Why?
Physics 3A: 2-D Motion
Shoup – 62
Two ways to compute h:
or
v^ 2 f
(^2) v i
2a
y f
y^
i
v^ 2 i
2g
h
h
v 2 yi
v^ 2 yf
2 g
v sini^
i ^2
2g
h
(^2) v i sin
2
i
2 g
v^
yf
v^
yi
at
v^
sini
!#" (^) i
g t
t^ A
$ v^
sini
%&
' i
g
(time to max y)
y^
f
(
h
) (
v^
sini
)*
t^ A
,
g t
2 A
h
. - v^ i
sin
./
(^0) i 0 v
sini^
./
(^0) i
g
11 2
g
2 v
sini^
./
(^0) i
g
32
h
v 2 sini
(^2)
i
2g
Physics 3A: 2-D Motion
Shoup – 63
Now to compute R (note t
B^
= 2 t
):A
we can simplify this by using the trigonometry identity of
sin(
θ ) = 2 sin(
θ ) cos(
θ
R
4
x^
f
4
x
i
5
v
xi
t
4
v
cosi^
(^67)
(^8) i
t^ B
R
9
v
cosi^
:; i <^2
t^
A
9
v^
cosi
:;
< i
v^
sini
:; i <
g
R
9
v^
2 cosi
:;
< i
sin
:;
< i
g
R
9
v^
2 sini
: 2
;
< i
g^
Physics 3A: 2-D Motion
Shoup – 64
Summary:
How do you maximize h? R?
R
9
v^
2 sini
: 2
;
< i
g
h
v^
2 sini
2
i
g
y(m) 200 150 100 50
50
100
150
200
250
300
x(m)
75
o
60
o^45
o o 30 o 15
v = 50 m/si^
Physics 3A: 2-D Motion
Shoup – 65
vv^ atxf xi^ x
Uniform Circular Motion
example of where velocity's direction changes, but not itsmagnitude Consider: Acceleration is always perpendicular to velocity and velocity isalways tangent to circular path
If there was component parallel to velocity, speed wouldincrease acceleration is parallel to
v in above figure.
∆θ
rf
ri
∆r
vi
vf
A
B
vi v
f
∆v ri
rf
∆θ
∆θ
v^ i
v^ f
v^
v
Physics 3A: 2-D Motion
Shoup – 66
So direction of acceleration constantly changing, what about
magnitude? from previous diagram, the two triangles are similar and thus:
v v
r r
^ a
(
v t
v
r
v
r
a
v r
r^ t
but:
v
v
lim
t^
0
!
"! r^ " t
so if take limit:
ac
v
2 r^
centripetal acceleration (center seeking)
Physics 3A: 2-D Motion
Shoup – 67
Is uniform circular motion "constant accelerations motion"?
NO!
acceleration's direction is constantly changing
In uniform circular motion, we define time object takes to
complete one revolution (lap) as the motion's period (
T
d
$
v t
%
r
&
v T
T
&
%
r
v^
Do Example
Physics 3A: 2-D Motion
Shoup – 72
To compute velocity take the derivative:
this relates ("converts") the LGM' measured velocity of P tothe LGM measured velocity can go the other way as well: warning: remember this is a vector equation!
r^ PO
r^ PO '
v^ O ' O
t
d d t
r^ PO
d d t
r^ PO '
v^
O' O
t
v^
PO
v^ PO '
v^ O' O
v^
PO '
v^
PO
v^ O 'O
Physics 3A: 2-D Motion
Shoup – 73
Review example 3.6 on page 93
River is flowing up
