2. February 5-8, 2013
Motion in Two Dimensions
Reminder of vectors and vector algebra
Displacement and position in 2-D
Average and instantaneous velocity in 2-D
Average and instantaneous acceleration in 2-D
Projectile motion
Uniform circular motion
Relative velocity*
3. February 5-8, 2013
Vector and its components
The components are the
legs of the right triangle
whose hypotenuse is A
y
x A
A
A
2 2 1
tan y
x y
x
A
A A A and
A
)
sin(
)
cos(
A
A
A
A
y
x
Or,
x
y
x
y
y
x
A
A
A
A
A
A
A
1
2
2
tan
or
tan
4. February 5-8, 2013
Which diagram can represent ?
A) B)
C) D)
Vector Algebra
1
2 r
r
r
r
2
r
1
r
r
2
r
1
r
r
2
r
1
r
r
2
r
1
r
1
r
5. February 5-8, 2013
Kinematic variables in one dimension
Position: x(t) m
Velocity: v(t) m/s
Acceleration: a(t) m/s2
Kinematic variables in three dimensions
Position: m
Velocity: m/s
Acceleration: m/s2
All are vectors: have direction and
magnitudes
Motion in two dimensions
k
v
j
v
i
v
t
v z
y
x
ˆ
ˆ
ˆ
)
(
y
x
z
i
j
k
x
k
z
j
y
i
x
t
r ˆ
ˆ
ˆ
)
(
k
a
j
a
i
a
t
a z
y
x
ˆ
ˆ
ˆ
)
(
6. February 5-8, 2013
In one dimension
In two dimensions
Position: the position of an object is
described by its position vector
--always points to particle from origin.
Displacement:
x1 (t1) = - 3.0 m, x2 (t2) = + 1.0 m
Δx = +1.0 m + 3.0 m = +4.0 m
Position and Displacement
)
(t
r
1
2 r
r
r
j
y
i
x
j
y
y
i
x
x
j
y
i
x
j
y
i
x
r
ˆ
ˆ
ˆ
)
(
ˆ
)
(
)
ˆ
ˆ
(
)
ˆ
ˆ
(
1
2
1
2
1
1
2
2
)
(
)
( 1
1
2
2 t
x
t
x
x
1
2 r
r
r
7. February 5-8, 2013
Average velocity
Instantaneous velocity
v is tangent to the path in x-y graph;
Average & Instantaneous Velocity
dt
r
d
t
r
v
v
t
avg
0
0
t
lim
lim
j
v
i
v
j
t
y
i
t
x
v y
avg
x
avg
avg
ˆ
ˆ
ˆ
ˆ ,
,
t
r
vavg
j
v
i
v
j
dt
dy
i
dt
dx
dt
r
d
v y
x
ˆ
ˆ
ˆ
ˆ
8. February 5-8, 2013
Motion of a Turtle
A turtle starts at the origin and moves with the speed of v0=10 cm/s in
the direction of 25° to the horizontal.
(a) Find the coordinates of a turtle 10 seconds later.
(b) How far did the turtle walk in 10 seconds?
9. February 5-8, 2013
Motion of a Turtle
Notice, you can solve the
equations independently for the
horizontal (x) and vertical (y)
components of motion and then
combine them!
y
x v
v
v
0
0 0 cos25 9.06 cm/s
x
v v
X components:
Y components:
Distance from the origin:
0 90.6 cm
x
x v t
0 0 sin 25 4.23 cm/s
y
v v
0 42.3 cm
y
y v t
cm
0
.
100
2
2
y
x
d
10. February 5-8, 2013
Average acceleration
Instantaneous acceleration
The magnitude of the velocity (the speed) can change
The direction of the velocity can change, even though the
magnitude is constant
Both the magnitude and the direction can change
Average & Instantaneous Acceleration
dt
v
d
t
v
a
a
t
avg
0
0
t
lim
lim
j
a
i
a
j
t
v
i
t
v
a y
avg
x
avg
y
x
avg
ˆ
ˆ
ˆ
ˆ ,
,
t
v
aavg
j
a
i
a
j
dt
dv
i
dt
dv
dt
v
d
a y
x
y
x ˆ
ˆ
ˆ
ˆ
11. February 5-8, 2013
Position
Average velocity
Instantaneous velocity
Acceleration
are not necessarily same direction.
Summary in two dimension
j
y
i
x
t
r ˆ
ˆ
)
(
j
a
i
a
j
dt
dv
i
dt
dv
dt
v
d
t
v
t
a y
x
y
x
t
ˆ
ˆ
ˆ
ˆ
lim
)
(
0
j
v
i
v
j
t
y
i
t
x
t
r
v y
avg
x
avg
avg
ˆ
ˆ
ˆ
ˆ ,
,
j
v
i
v
j
dt
dy
i
dt
dx
dt
r
d
t
r
t
v y
x
t
ˆ
ˆ
ˆ
ˆ
lim
)
(
0
dt
dx
vx
dt
dy
vy
2
2
dt
x
d
dt
dv
a x
x
2
2
dt
y
d
dt
dv
a
y
y
)
(
and
),
(
, t
a
t
v
(t)
r
12. February 5-8, 2013
Motion in two dimensions
t
a
v
v
0
Motions in each dimension are independent components
Constant acceleration equations
Constant acceleration equations hold in each dimension
t = 0 beginning of the process;
where ax and ay are constant;
Initial velocity initial displacement ;
2
2
1
0 t
a
t
v
r
r
t
a
v
v y
y
y
0
2
2
1
0
0 t
a
t
v
y
y y
y
)
(
2 0
2
0
2
y
y
a
v
v y
y
y
t
a
v
v x
x
x
0
2
2
1
0
0 t
a
t
v
x
x x
x
)
(
2 0
2
0
2
x
x
a
v
v x
x
x
j
a
i
a
a y
x
ˆ
ˆ
j
v
i
v
v y
x
ˆ
ˆ 0
0
0
j
y
i
x
r ˆ
ˆ 0
0
0
13. February 5-8, 2013
Define coordinate system. Make sketch showing axes, origin.
List known quantities. Find v0x , v0y , ax , ay , etc. Show initial
conditions on sketch.
List equations of motion to see which ones to use.
Time t is the same for x and y directions.
x0 = x(t = 0), y0 = y(t = 0), v0x = vx(t = 0), v0y = vy(t = 0).
Have an axis point along the direction of a if it is constant.
Hints for solving problems
t
a
v
v y
y
y
0
2
2
1
0
0 t
a
t
v
y
y y
y
)
(
2 0
2
0
2
y
y
a
v
v y
y
y
t
a
v
v x
x
x
0
2
2
1
0
0 t
a
t
v
x
x x
x
)
(
2 0
2
0
2
x
x
a
v
v x
x
x
14. February 5-8, 2013
2-D problem and define a coordinate
system: x- horizontal, y- vertical (up +)
Try to pick x0 = 0, y0 = 0 at t = 0
Horizontal motion + Vertical motion
Horizontal: ax = 0 , constant velocity motion
Vertical: ay = -g = -9.8 m/s2, v0y = 0
Equations:
Projectile Motion
2
2
1
gt
t
v
y
y iy
i
f
t
a
v
v y
y
y
0
2
2
1
0
0 t
a
t
v
y
y y
y
)
(
2 0
2
0
2
y
y
a
v
v y
y
y
t
a
v
v x
x
x
0
2
2
1
0
0 t
a
t
v
x
x x
x
)
(
2 0
2
0
2
x
x
a
v
v x
x
x
Horizontal Vertical
15. February 5-8, 2013
X and Y motions happen independently, so
we can treat them separately
Try to pick x0 = 0, y0 = 0 at t = 0
Horizontal motion + Vertical motion
Horizontal: ax = 0 , constant velocity motion
Vertical: ay = -g = -9.8 m/s2
x and y are connected by time t
y(x) is a parabola
Projectile Motion
gt
v
v y
y
0
2
2
1
0
0 gt
t
v
y
y y
x
x v
v 0
t
v
x
x x
0
0
Horizontal Vertical
16. February 5-8, 2013
2-D problem and define a coordinate system.
Horizontal: ax = 0 and vertical: ay = -g.
Try to pick x0 = 0, y0 = 0 at t = 0.
Velocity initial conditions:
v0 can have x, y components.
v0x is constant usually.
v0y changes continuously.
Equations:
Projectile Motion
0
0
0 cos
v
v x
Horizontal Vertical
0
0
0 sin
v
v x
gt
v
v y
y
0
2
2
1
0
0 gt
t
v
y
y y
x
x v
v 0
t
v
x
x x
0
0
17. February 5-8, 2013
Initial conditions (t = 0): x0 = 0, y0 = 0
v0x = v0 cosθ0 and v0y = v0 sinθ0
Horizontal motion:
Vertical motion:
Parabola;
θ0 = 0 and θ0 = 90 ?
Trajectory of Projectile Motion
2
2
1
0
0 gt
t
v
y y
x
x
v
x
t
t
v
x
0
0
0
2
0
0
0
2
x
x
y
v
x
g
v
x
v
y
2
0
2
2
0
0
cos
2
tan x
v
g
x
y
18. February 5-8, 2013
Initial conditions (t = 0): x0 = 0, y0 = 0
v0x = v0 cosθ0 and v0x = v0 sinθ0, then
What is R and h ?
Horizontal Vertical
2
2
1
0
0
0 gt
t
v y
t
v
x x
0
0
g
v
g
v
v
t
v
x
x
R x
0
2
0
0
0
0
0
0
0
2
sin
sin
cos
2
g
v
g
v
t
y 0
0
0 sin
2
2
2
0
2
2
1
0
0
2
2
2
t
g
t
v
gt
t
v
y
y
h y
h
h
y
g
v
h
2
sin 0
2
2
0
y
y
y
y
y v
g
v
g
v
gt
v
v 0
0
0
0
2
h
gt
v
v y
y
0
2
2
1
0
0 gt
t
v
y
y y
x
x v
v 0
t
v
x
x x
0
0
19. February 5-8, 2013
Projectile Motion
at Various Initial Angles
Complementary
values of the initial
angle result in the
same range
The heights will be
different
The maximum range
occurs at a projection
angle of 45o
g
v
R
2
sin
2
0
20. February 5-8, 2013
Uniform circular motion
Constant speed, or,
constant magnitude of velocity
Motion along a circle:
Changing direction of velocity
21. February 5-8, 2013
Circular Motion: Observations
Object moving along a
curved path with constant
speed
Magnitude of velocity: same
Direction of velocity: changing
Velocity: changing
Acceleration is NOT zero!
Net force acting on the
object is NOT zero
“Centripetal force” a
m
Fnet
22. February 5-8, 2013
Centripetal acceleration
Direction: Centripetal
Uniform Circular Motion
r
v
t
v
a
r
v
r
v
t
r
t
v
r
r
v
v
r
r
v
v
r
2
2
so,
O
x
y
ri
R
A B
vi
rf
vf
Δr
vi
vf
Δv = vf - vi
23. February 5-8, 2013
Uniform Circular Motion
Velocity:
Magnitude: constant v
The direction of the velocity is
tangent to the circle
Acceleration:
Magnitude:
directed toward the center of
the circle of motion
Period:
time interval required for one
complete revolution of the
particle
r
v
ac
2
r
v
ac
2
v
r
T
2
v
ac
24. February 5-8, 2013
Position
Average velocity
Instantaneous velocity
Acceleration
are not necessarily in the same direction.
Summary
j
y
i
x
t
r ˆ
ˆ
)
(
j
a
i
a
j
dt
dv
i
dt
dv
dt
v
d
t
v
t
a y
x
y
x
t
ˆ
ˆ
ˆ
ˆ
lim
)
(
0
j
v
i
v
j
t
y
i
t
x
t
r
v y
avg
x
avg
avg
ˆ
ˆ
ˆ
ˆ ,
,
j
v
i
v
j
dt
dy
i
dt
dx
dt
r
d
t
r
t
v y
x
t
ˆ
ˆ
ˆ
ˆ
lim
)
(
0
dt
dx
vx
dt
dy
vy
2
2
dt
x
d
dt
dv
a x
x
2
2
dt
y
d
dt
dv
a
y
y
)
(
and
),
(
, t
a
t
v
(t)
r
25. February 5-8, 2013
If a particle moves with constant acceleration a, motion
equations are
Projectile motion is one type of 2-D motion under constant
acceleration, where ax = 0, ay = -g.
Summary
j
t
a
t
v
y
i
t
a
t
v
x
j
y
i
x
r yi
yi
i
xi
xi
i
f
f
f
ˆ
)
(
ˆ
)
(
ˆ
ˆ 2
2
1
2
2
1
j
t
a
v
i
t
a
v
j
v
i
v
t
v y
iy
x
ix
fy
fx
f
ˆ
)
(
ˆ
)
(
ˆ
ˆ
)
(
t
a
v
v i
2
2
1
t
a
t
v
r
r i
i
f
