Transcript Slide 1

Vector Addition
Recall that for vectors in one dimension
(parallel vectors), the vectors are added algebraically.
Vectors in 2 dimensions are add geometrically
*X = (R) (cos θ)
*Y = (R) (sin θ)
Y
R
θ
X
Example
3A:Section Review Pg. 87 1, 2
2.) 126m 10o= θ
Y
θ– theta:
means the
measure of the
angle
R
θ
X
Vector Resultants with Right Triangles
c2 = a2 + b2
[R2 = x2 + y2 ]
7 km
h
R2 = (6km/h) 2 + (7km/h)2
R2 = 36km2/h2 + 49km2/h2
R = 85 km2/h2
R
θ
6km/h
R = ___________ km/h
Solving Problems with 2 vectors in different directions:
We will not do this vector addition graphically.
We will not do it using the Law of Cosines
→We will do it trigonometrically or as the books says, analytically.
All the Trig. You need to know for Physics I
*SOH, CAH, TOA
θ
Side adjacent to θ
Side
opposite
of θ
Sin θ =
Opp. Side
Hypotenuse
Cos θ =
Adjacent side
Hypotenuse
Tan θ =
Opposite side
Adjacent side
Example
Sample 3B page 93
*Add to your notes
Vr2 = Vx2 + Vy2
Angles are in Standard Position
What you learned in math is used in Physics!
900
Example:
0o
1800
50km/h
?
θ
270o
90 km/h
Finding the resultant without R 2 = X 2 + Y 2
R = Resultant
I
Tan θ = 50 km/hr = .5556
90 km/hr
θ = 29.05o
II
Cos 29.05o = 90km/hr
R
R = 90 km/hr
.8742
R = 102.9 km/hr
R = 102.9 km/hr @ 29.05o STD position
Motion in 2 dimensions
I. Projectiles: Objects thrown or projected into the air are projectiles
whose parabolic paths are called trajectories.
Motion of a projectile along a trajectory is called projectile motion.
Max
Height
*Vx = (Vi) (cos θ)
Vi
θ
Vy
*Vy = (Vi) (sin θ)
Vx
Once a projectile loses contact with the hand, bat, or gun barrel,
it is only accelerated by g, in the X direction, a = 0
Motion in 2 dimensions
*Vx = (Vi) (cos θ)
Max
Height
Y
*Vy = (Vi) (sin θ)
θ
X
We can analyze the motion separately along each axis:
X direction
X = (Vx)(t) + (½)at2 = (Vx)(t)
X = (Vx)(t) = (Vi cos θ)t
Y direction
Motion in 2 dimensions
*Vx = (Vi) (cos θ)
Max
Height
Y
*Vy = (Vi) (sin θ)
θ
X
We can analyze the motion separately along each axis:
X direction
X = (Vx)(t) + (½)at2 = (Vx)(t)
Y direction
Motion in 2 dimensions
*Vx = (Vi) (cos θ)
Max
Height
Y
*Vy = (Vi) (sin θ)
θ
X
We can analyze the motion separately along each axis:
X direction
X = (Vx)(t) + (½)at2 = (Vx)(t)
X = (Vx)(t) = (Vi cos θ)t
Y direction
Y = (Vy)(t) + ½gt2
Motion in 2 dimensions
*Vx = (Vi) (cos θ)
Max
Height
Y
*Vy = (Vi) (sin θ)
θ
X
We can analyze the motion separately along each axis:
X direction
X = (Vx)(t) + (½)at2 = (Vx)(t)
X = (Vx)(t) = (Vi cos θ)t
Y direction
Y = (Vy)(t) + ½gt2
Y = (Vy)(t) + ½gt2 = (Vi sin θ)t + ½gt2
* t = ½ time of flight for Y
*Time of Flight
Equation:
(derived from the Y equation)
Y = vi sinθ (t) + ½ gt2
2vi sin θ = t
g
or
2vy = t
g
Calculating Xmax without time:
X max = 2 Vi 2 sinθ cos θ
g
X max = 2 Vi 2 sin 2 θ
g
Example : A stone is thrown horizontally at 15m/s from the top of a cliff
that is 44 m high.
A. How long does the stone take to reach the bottom of the cliff?
B. How far from the base of the cliff does the stone strike the ground?
C. Sketch the trajectory of the stone.
Acceleration only vertically means:
the time aloft is independent of the horizontal component
of the trajectory.
Max height = max Y
Pop up
Y
Same t: time
*if vi is the same
Fly out
Homework #1 BK & WKBK A-C Chp. 3 (15)
Book Sections 3A,3B & 3C
3A: Pg. 91 1-4
2.) 45.6m at 9.5o East of North
4.) 1.8m at 49o below the horizontal
3B: pg. 94 1,3,5 (answers back of the book)
3C: pg. 97
1,2,4
2.) 7.5km at 26o above the horizontal
4.) 171 km at 34o East of North
WKBK 3A & 3B
3A:
3.) 7 jumps at 36o West of North
5.) 65o below the water 533m
3B
4.) x= 29m y=21m
5.) Vx=335km/h Vy=89.8km/h
6.) d= 900m
x=450 m East y= 780m North
Homework #2 Chp. 3 Book & WKBK 3D-3E (11)
BOOK-3D Problems pg. 102
2,3,4
2.) 4.9m/s
3.)7.6m/s
4.) 5.6m
3E problems pg. 104
4.) 7m/s
Workbook 3 D-E
3D 1. 5.98 m = X
3. 170m = Y
4. y= -.735m
3E 1. Vi = 45.8 m/s
2. V=Vx= 68.2 m/s
4. Vi=20.8 m/s
Ymax = 11m
Ymax = 22.1m
3&4
*Scalars – has magnitude with units
*Vectors – have magnitude with direction
→ Vectors can be added and subtracted creating a Resultant ( R )
3N
3N
4N
6N
7N
3N