Holt Physics Chapter 3—Two-dimensional Motion and VectorsI. Section 3-1—Introduction to Vectors

A. Scalars and Vectors1. Magnitude and Direction

a. In chapter 2 we discussed velocity. Velocity has both a magnitude and a direction.

b. Magnitude: a measurement represented by a number (e.g. -12 meters/second)

c. Direction: an indication of orientation. For velocities we use a positive number for right or up and a negative number for left or down.

2. Velocity is an example of a VECTOR. Vectors always have both a direction and a magnitude

a. Vectors are differentiated from scalars in your book by using BOLD type.

b. Vectors are also represented by arrows. The longer the arrow the more magnitude the vector has.

3. Speed is an example of a SCALAR. Scalars only have a magnitude.

B. Properties of vectors1. Vectors can be added graphically

a. When adding vectors, you must make sure that they have the same units and describe similar quantities.

b. The answer found by adding vectors together is called the resultant.

(insert fig 3-2 here)

c. Vectors can only be moved parallel to themselves in a diagram. (see fig 3.3 on p.86)

II. Section 3-2—Vector OperationsA. Coordinate Systems in Two Dimensions

1. We will always orient the coordinate plane so that the y-axis lies north and south, and the x-axis lies east-west.

2. For objects flying or falling through the air we orient the y-axis vertically (up and down) and the x-axis horizontally.

B. Determining Resultant Magnitude and Direction1. We can use right-triangle trigonometry to

calculate both the magnitude and the direction of a resultant.

2. Pythagorean Theorem: a2 + b2 = c2

3. Tangent θ = opposite/adjacentC. Resolving Vectors Into Components

1. Vectors that are added together are called components

2. We can use the sine and cosine functions to calculate missing sides of a right triangle.

3. Sine θ = opposite/hypotenuse4. Cosine θ = adjacent/hypotenuse

III. Section 3-3—Projectile MotionA. Two-dimensional Motion

1. Projectiles are objects thrown or launched into the air.

2. Projectile motion is a form of two-dimensional motion because the projectile has velocities along both the x- and y -axes.

3. Projectiles always follow a curved path called a “trajectory”

4. Trajectories are always shaped like a parabola*5. Projectile motion can be thought of as free fall with

and initial horizontal velocity6. By breaking the projectile’s trajectory into its

component vertical and horizontal component vectors we can use right triangles (again!) to solve problems involving trajectories like the sample problem on p.101

*in a vacuum ignoring air resistance

Projectile Motion Price

VERTICAL MOTION OF A PROJECTILE THAT FALLS FROM REST

vf = -gΔt

vf2 = -2g∆y

Δy = -½g(∆t)2

HORIZONTAL MOTION OF A PROJECTILE

vx = vi = constant

Δx = vx Δt

Sample Problem 3D (p.101)Given: ∆y = -321m ∆x = 45m g = 9.81m/s2

Determine: Initial velocity (vi) Remember that velocity equals displacement divided by time or

v = ∆x/∆t

We know ∆x = 45 but we don’t know ∆t. Using the equations above we can determine ∆t.

The only equation above that we can solve for ∆t is the 3rd one

∆y = -½g(Δt)2

If we substitute and solve for Δt we get-321 = (-½)(9.81)(Δt)2

65.44 = (Δt)2

Δt = 8.1s

Now we simply divide ∆x by ∆t to get vi = 5.5m/s

B. Projectiles Launched at an Angle1. Now let’s considered objects launched into the air

at an angle like balls and bullets.2. Again we will break the vector up into its vertical

(y) and horizontal (x) components and use trigonometry to make right triangles to solve problems.

(Insert Fig. 3-20 here)

vx,i = vi (cos θ) and vy,i = vi (sin θ)

IV. Section 3-4—Relative MotionA. Frames of Reference

1. Velocity measurements depend on the observer’s frame of reference.

2. Albert Einstein called this phenomenon “Relativity”3. We will use subscripts to denote observers’ frames

of reference4. “There is no general equation to work out relative

velocity problems; instead you should develop the necessary equations on your own…” (p.107)

Projectiles Launched at an AnglePrice

PROJECTILES LAUNCHED AT AN ANGLE

vx = vi (cos θ) = constant*∆x = vi (cos θ) ∆t

vy,f = vi (sin θ) – gΔtvy,f

2 = vi2 (sin θ)2 – 2gΔy

Δy = vi (sin θ) Δt - ½g(Δt)2

*we will again ignore air resistance

Practice 3E #33. A baseball is thrown at an angle of 25o relative to the ground at a

speed of 23m/s. If the ball is caught 42m from the thrower, how long was it in the air? How high was the tallest spot in the ball’s path?

DRAW AND LABEL A DIAGRAM

Given: θ = 25o Δx = 42m vi = 23m/s

SOLVE FOR ΔtUse this equation: Δx = vi (cos θ) Δt

42 = (23)(cos 25)(Δt) 42 = (23)(.91)(Δt)

2.01s = Δt

SOLVE FOR ΔyRemember: Δy = vy/ΔtSolve for vy: vy = vi (sin θ)

vy = (23)(.422) vy = 9.72m/s

Solve for Δy: Δy = 9.72/2.01 Δy = 4.83m