The Aspects of Motion in a Twoand Three- Dimension

1. 3 Motion in Two & Three Dimensions
• Displacement, velocity, acceleration
• Case 1: Projectile Motion
• Case 2: Circular Motion
• Hk: 51, 55, 69, 77, 85, 91.

2. Position & Displacement Vectors
j
y
i
x
r ˆ
ˆ 


1
2 r
r
r







3. Velocity Vectors
t
r
vav





dt
r
d
v



2
2
2
)
(
)
(
)
( z
y
x v
v
v
v 



4. Relative Velocity
• Examples:
• people-mover at airport
• airplane flying in wind
• passing velocity (difference in velocities)
• notation used:
velocity “BA” = velocity of B with respect to A

5. Example:

6. Acceleration Vectors
t
v
aav





dt
v
d
a




7. Direction of Acceleration
• Direction of a = direction of velocity
change (by definition)
• Examples: rounding a corner, bungee
jumper, cannonball (Tipler),
Projectile (29, 30 below)

8. Projectile Motion
• begins when projecting force ends
• ends when object hits something
• gravity alone acts on object

9. Horizontal V Constant

10. Two Dimensional Motion
(constant acceleration)
t
a
v
v x
ox
x 

2
2
1
t
a
t
v
x x
ox 


t
a
v
v y
oy
y 

2
2
1
t
a
t
v
y y
oy 



11. Range vs. Angle

12. Example 1: Calculate Range
(R)
vo = 6.00m/s qo = 30°
xo = 0, yo = 1.6m; x = R, y = 0
s
m
v
v o
o
ox /
20
.
5
30
cos
00
.
6
cos 

 q
s
m
v
v o
o
oy /
00
.
3
30
sin
00
.
6
sin 

 q

13. Example 1 (cont.)
2
2
2
1
2
2
1
9
.
4
3
6
.
1
)
8
.
9
(
30
sin
6
6
.
1
t
t
t
t
t
a
t
v
y y
oy










0
6
.
1
3
9
.
4 2


 t
t
Step 1

14. Quadratic Equation
0
2


 c
bx
ax
a
ac
b
b
x
2
4
2




0
6
.
1
3
9
.
4 2


 t
t
6
.
1
3
9
.
4





c
b
a

15. a
ac
b
b
x
2
4
2




Example 1 (cont.)
6
.
1
3
9
.
4





c
b
a
)
9
.
4
(
2
)
6
.
1
)(
9
.
4
(
4
)
3
(
3 2





t
)
9
.
4
(
2
353
.
6
3

t
954
.
0
342
.
0




t
t
End of Step 1

16. Example 1 (cont.)
t
v
t
a
t
v
x ox
x
ox 


 2
2
1
Step 2
(ax = 0)
m
t
v
x o
o 96
.
4
)
954
.
0
(
30
cos
6
cos 


 q
“Range” = 4.96m
End of Example

17. Circular Motion
• Uniform
• Non-uniform
• Acceleration of Circular Motion

18. 18
Centripetal Acceleration
• Turning is an acceleration toward center of
turn-radius and is called Centripetal
Acceleration
• Centripetal is left/right direction
• a(centripetal) = v2/r
• (v = speed, r = radius of turn)
• Ex. V = 6m/s, r = 4m.
a(centripetal) = 6^2/4 = 9 m/s/s

19. Tangential Acceleration
• Direction = forward along path (speed
increasing)
• Direction = backward along path (speed
decreasing)
t
dv
at



20. Total Acceleration
• Total acceleration
= tangential + centripetal
• = forward/backward + left/right
• a(total) = dv/dt (F/B) + v2/r (L/R)
• Ex. Accelerating out of a turn;
4.0 m/s/s (F) + 3.0 m/s/s (L)
• a(total) = 5.0 m/s/s

21. Summary
• Two dimensional velocity, acceleration
• Projectile motion (downward pointing
acceleration)
• Circular Motion (acceleration in any
direction within plane of motion)

22. Ex. A Plane has an air speed vpa = 75m/s. The wind has a
velocity with respect to the ground of vag = 8 m/s @ 330°.
The plane’s path is due North relative to ground. a) Draw
a vector diagram showing the relationship between the air
speed and the ground speed. b) Find the ground speed
and the compass heading of the plane.
(similar
situation)

23. )
v
,
0
(
)
330
sin
8
,
330
cos
8
(
)
sin
75
,
cos
75
(
v
v
v
pg
ag
pa
pg







q
q



0
330
cos
8
cos
75 


q


 3
.
95
q
pg
v
m/s
70.7
330
sin
8
3
.
95
sin
75 





25. PM Example 2:
vo = 6.00m/s qo = 0°
xo = 0, yo = 1.6m; x = R, y = 0
s
m
v
v o
o
ox /
00
.
6
0
cos
00
.
6
cos 

 q
s
m
v
v o
o
oy /
0
0
sin
00
.
6
sin 

 q

26. PM Example 2 (cont.)
2
2
2
1
2
2
1
9
.
4
0
6
.
1
)
8
.
9
(
0
sin
6
6
.
1
t
t
t
t
a
t
v
y y
oy










571
.
0
9
.
4
6
.
1
6
.
1
9
.
4 2





t
t
Step 1

27. PM Example 2 (cont.)
t
v
t
a
t
v
x ox
x
ox 


 2
2
1
Step 2
(ax = 0)
m
t
v
x o
o 43
.
3
)
571
.
0
(
0
cos
6
cos 


 q
“Range” = 3.43m
End of Step 2

28. PM Example 2: Speed at
Impact
s
t 571
.
0

t
a
v
v x
ox
x 
 t
a
v
v y
oy
y 

s
m
t
vx /
6
)
0
(
6 


s
m
vy
/
59
.
5
571
.
0
)
8
.
9
(
)
0
(




s
m
v
v
v y
x /
20
.
8
)
59
.
5
(
)
6
( 2
2
2
2






29. v1
0
1
2
3
4
5
6
0 2 4 6 8 10 12 14
x(m)
y(m)
1. v1 and v2 are located on trajectory.
1
v
2
v
v

a

30. Q1. Given 1
v
2
v
v

locate these on the
trajectory and form v.
0
1
2
3
4
5
6
0 2 4 6 8 10 12 14
x(m)
y(m)
1
v
2
v

31. Velocity in Two Dimensions
• vavg // r
• instantaneous “v” is limit of “vavg” as t  0
t
r
vavg






32. Acceleration in Two Dimensions
t
v
aavg





• aavg // v
• instantaneous “a” is limit of “aavg” as t  0

33. Displacement in Two Dimensions
ro
r
r
o
r
r
r






r
r
r o







34. v1
0
1
2
3
4
5
6
0 2 4 6 8 10 12 14
x(m)
y(m)
1. v1 and v2 are located on trajectory.
1
v
2
v
v

a

35. Ex. If v1(0.00s) = 12m/s, +60° and v2(0.65s) = 7.223 @
+33.83°, find aave.
)
39
.
10
,
00
.
6
(
))
60
sin(
0
.
12
),
60
cos(
0
.
12
(
1 

v
)
02
.
4
,
00
.
6
(
))
83
.
33
sin(
223
.
7
),
83
.
33
cos(
223
.
7
(
2 

v
s
m
v
v
v /
)
37
.
6
,
0
(
)
39
.
10
,
00
.
6
(
)
02
.
4
,
00
.
6
(
1
2 






s
s
m
s
s
m
t
v
a /
/
)
8
.
9
,
0
(
00
.
0
65
.
0
/
)
37
.
6
,
0
(









36. Q1. Given 1
v
2
v
v

locate these on the
trajectory and form v.
0
1
2
3
4
5
6
0 2 4 6 8 10 12 14
x(m)
y(m)
1
v
2
v

37. Q2. If v3(1.15s) = 6.06m/s, -8.32° and v4(1.60s) = 7.997, -41.389°, write the
coordinate-forms of these vectors and calculate the average acceleration.
)
8777
.
0
,
00
.
6
(
))
32
.
8
sin(
06
.
6
),
32
.
8
cos(
06
.
6
(
3 




v
)
2877
.
5
,
00
.
6
(
))
39
.
41
sin(
997
.
7
),
39
.
41
cos(
997
.
7
(
4 




v
s
m
v
v
v /
)
41
.
4
,
0
(
)
8777
.
0
,
00
.
6
(
)
2877
.
5
,
00
.
6
(
1
2 








s
s
m
s
s
m
t
v
a /
/
)
8
.
9
,
0
(
15
.
1
60
.
1
/
)
41
.
4
,
0
(








v3
v4
v a