Vector Addition

Recall that for vectors in one dimension (parallel vectors), the vectors are added algebraically.

Vectors in 2 dimensions are add geometrically RYX*X = (R) (cos )

*Y = (R) (sin )

Example

3A:Section Review Pg. 87 1, 2 2.) 126m 10o=

RYVector Resultants with Right Triangles

c2 = a2 + b2 [R2 = x2 + y2 ]

RX7 km h6km/hR2 = (6km/h) 2 + (7km/h)2R2 = 36km2/h2 + 49km2/h2R = 85 km2/h2

R = ___________ km/h theta: means the measure of the angle

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, TOAhypotenuseSide adjacent to Side opposite of Sin = Opp. Side Hypotenuse

Cos = Adjacent side Hypotenuse

Tan = Opposite side Adjacent sideSolving Problems with 2 vectors in different directions:

ExampleSample 3B page 93*Add to your notes

Vr2 = Vx2 + Vy2

Angles are in Standard PositionWhat you learned in math is used in Physics!0o270o900180050km/h90 km/hExample:?

Finding the resultant without R 2 = X 2 + Y 2 R = ResultantTan = 50 km/hr = .5556 90 km/hr

= 29.05o

Cos 29.05o = 90km/hrRR = 90 km/hr .8742R = 102.9 km/hr

R = 102.9 km/hr @ 29.05o STD positionIII

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 )

*Vy = (Vi) (sin )

Once a projectile loses contact with the hand, bat, or gun barrel, it is only accelerated by g, in the X direction, a = 0

ViVxVy

Motion in 2 dimensions

Max Height *Vx = (Vi) (cos )

*Vy = (Vi) (sin )

We can analyze the motion separately along each axis:

X directionY directionX = (Vx)(t) + ()at2 = (Vx)(t) X = (Vx)(t) = (Vi cos )t

YX

Motion in 2 dimensions

Max Height *Vx = (Vi) (cos )

*Vy = (Vi) (sin )

We can analyze the motion separately along each axis:

X directionY directionX = (Vx)(t) + ()at2 = (Vx)(t)

YX

Motion in 2 dimensions

Max Height *Vx = (Vi) (cos )

*Vy = (Vi) (sin )

We can analyze the motion separately along each axis:

X directionY directionX = (Vx)(t) + ()at2 = (Vx)(t) Y = (Vy)(t) + gt2 X = (Vx)(t) = (Vi cos )t

YX

Motion in 2 dimensions

Max Height *Vx = (Vi) (cos )

*Vy = (Vi) (sin )

We can analyze the motion separately along each axis:

X directionY directionX = (Vx)(t) + ()at2 = (Vx)(t) Y = (Vy)(t) + gt2 X = (Vx)(t) = (Vi cos )t Y = (Vy)(t) + gt2 = (Vi sin )t + gt2 * t = time of flight for YYX

*Time of Flight Equation: (derived from the Y equation) Y = vi sin (t) + gt2

2vi sin = tg

or

2vy = t g

Calculating Xmax without time:

X max = 2 Vi 2 sin cos g

X max = 2 Vi 2 sin 2 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 YSame t: time *if vi is the sameFly outPop upY

Homework #1 BK & WKBK A-C Chp. 3 (15)

Book Sections 3A,3B& 3C

3A: Pg. 91 1-42.) 45.6m at 9.5o East of North4.) 1.8m at 49o below the horizontal

3B: pg. 94 1,3,5 (answers back of the book)

3C: pg. 97 1,2,42.) 7.5km at 26o above the horizontal4.) 171 km at 34o East of NorthWKBK 3A & 3B3A: 3.) 7 jumps at 36o West of North 5.) 65o below the water 533m

3B 4.) x= 29m y=21m5.) Vx=335km/h Vy=89.8km/h6.) d= 900mx=450 m East y= 780m North

Workbook 3 D-E

3D 1. 5.98 m = X3. 170m = Y4. y= -.735m

3E 1. Vi = 45.8 m/s2. V=Vx= 68.2 m/s4. Vi=20.8 m/s Ymax = 11m Ymax = 22.1mHomework #2 Chp. 3 Book & WKBK 3D-3E (11)

BOOK-3D Problems pg. 102 2,3,42.) 4.9m/s3.)7.6m/s4.) 5.6m

3E problems pg. 1043 & 4 4.) 7m/s

*Scalars has magnitude with units

*Vectors have magnitude with direction

Vectors can be added and subtracted creating a Resultant ( R ) 3N4N7N6N3N3N

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