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3 Motion in Two & Three Dimensions. Displacement, velocity, acceleration Case 1: Projectile Motion Case 2: Circular Motion Hk: 51, 55, 69, 77, 85, 91. Position & Displacement Vectors. Velocity Vectors. Relative Velocity. Examples: people-mover at airport airplane flying in wind - PowerPoint PPT Presentation
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3 Motion in Two & Three Dimensions
• Displacement, velocity, acceleration
• Case 1: Projectile Motion• Case 2: Circular Motion
• Hk: 51, 55, 69, 77, 85, 91.
Position & Displacement Vectors
jyixr ˆˆ
12 rrr
Velocity Vectors
t
rvav
dt
rdv
222 )()()( zyx vvvv
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
Example:
Acceleration Vectors
t
vaav
dt
vda
Direction of Acceleration
• Direction of a = direction of velocity change (by definition)
• Examples: rounding a corner, bungee jumper, cannonball (Tipler), Projectile (29, 30 below)
Projectile Motion
• begins when projecting force ends
• ends when object hits something
• gravity alone acts on object
Horizontal V Constant
Two Dimensional Motion (constant acceleration)
tavv xoxx
221 tatvx xox
tavv yoyy
221 tatvy yoy
Range vs. Angle
Example 1: Calculate Range (R)
vo = 6.00m/s o = 30°
xo = 0, yo = 1.6m; x = R, y = 0
smvv ooox /20.530cos00.6cos
smvv oooy /00.330sin00.6sin
Example 1 (cont.)
2
221
221
9.436.1
)8.9(30sin66.1
tt
tt
tatvy yoy
06.139.4 2 tt
Step 1
Quadratic Equation
02 cbxaxa
acbbx
2
42
06.139.4 2 tt
6.1
3
9.4
c
b
a
a
acbbx
2
42
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
353.63t
954.0
342.0
t
t
End of Step 1
Example 1 (cont.)
tvtatvx oxxox 221Step 2
(ax = 0)
mtvx oo 96.4)954.0(30cos6cos
“Range” = 4.96m
End of Example
Circular Motion
• Uniform
• Non-uniform
• Acceleration of Circular Motion
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
Tangential Acceleration
• Direction = forward along path (speed increasing)
• Direction = backward along path (speed decreasing)
t
dvat
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
Summary
• Two dimensional velocity, acceleration
• Projectile motion (downward pointing acceleration)
• Circular Motion (acceleration in any direction within plane of motion)
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)
)v,0()330sin8,330cos8()sin75,cos75(
vvv
pg
agpapg
0330cos8cos75 3.95
pgvm/s 70.7330sin83.95sin75
PM Example 2:
vo = 6.00m/s o = 0°
xo = 0, yo = 1.6m; x = R, y = 0
smvv ooox /00.60cos00.6cos
smvv oooy /00sin00.6sin
PM Example 2 (cont.)
2
221
221
9.406.1
)8.9(0sin66.1
t
tt
tatvy yoy
571.09.4
6.1
6.19.4 2
t
t
Step 1
PM Example 2 (cont.)
tvtatvx oxxox 221Step 2
(ax = 0)
mtvx oo 43.3)571.0(0cos6cos
“Range” = 3.43m
End of Step 2
PM Example 2: Speed at Impact
st 571.0
tavv xoxx tavv yoyy
smtvx /6)0(6 sm
vy
/59.5
571.0)8.9()0(
smvvv yx /20.8)59.5()6( 2222
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.
1v
2v
va
Q1. Given 1v2v
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
)
1v
2v
Velocity in Two Dimensions
• vavg // r
• instantaneous “v” is limit of “vavg” as t 0
t
rvavg
Acceleration in Two Dimensions
t
vaavg
• aavg // v
• instantaneous “a” is limit of “aavg” as t 0
Displacement in Two Dimensions
ro
r
r
orrr
rrr o
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.
1v
2v
va
Ex. If v1(0.00s) = 12m/s, +60° and v2(0.65s) = 7.223 @ +33.83°, find aave.
)39.10,00.6())60sin(0.12),60cos(0.12(1 v
)02.4,00.6())83.33sin(223.7),83.33cos(223.7(2 v
smvvv /)37.6,0()39.10,00.6()02.4,00.6(12
ssms
sm
t
va //)8.9,0(
00.065.0
/)37.6,0(
Q1. Given 1v2v
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
)
1v
2v
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())32.8sin(06.6),32.8cos(06.6(3 v
)2877.5,00.6())39.41sin(997.7),39.41cos(997.7(4 v
smvvv /)41.4,0()8777.0,00.6()2877.5,00.6(12
ssms
sm
t
va //)8.9,0(
15.160.1
/)41.4,0(
v3
v4v a