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Lesson

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Lesson

Vectors Review

R

head

tail

- Scalars have magnitude only
- Distance, speed, time, mass

- Vectors have both magnitude and direction
- displacement, velocity, acceleration

x

A

- The direction of a vector is represented by the direction in which the ray points.
- This is typically given by an angle.

If vector A represents a displacement of three miles to the north…

A

B

Then vector B, which is twice as long, would represent a displacement of six miles to the north!

- The magnitude of a vector is the size of whatever the vector represents.
- The magnitude is represented by the length of the vector.
- Symbolically, the magnitude is often represented as │A │

- Equal vectors have the same length and direction, and represent the same quantity (such as force or velocity).

A

-A

- Inverse vectors have the same length, but opposite direction.

B

A

R

- Vectors are added graphically together head-to-tail.
- The sum is called the resultant.
- The inverse of the sum is called the equilibrant

A + B = R

- Resolve each vector into its x- and y-components.
Ax = AcosAy = Asin

Bx = BcosBy = Bsinetc.

- Add the x-components together to get Rx and the y-components to get Ry.
- Use the Pythagorean Theorem to get the magnitude of the resultant.
- Use the inverse tangent function to get the angle.

- Sample problem: Add together the following graphically and by component, giving the magnitude and direction of the resultant and the equilibrant.
- Vector A: 300 m @ 60o
- Vector B: 450 m @ 100o
- Vector C: 120 m @ -120o

Lesson

Unit Vectors

a

Polar Angle

z

Azimuthal Angle

az

q

ay

y

f

ax

xy Projection

x

- Unit vectors are quantities that specify direction only. They have a magnitude of exactly one, and typically point in the x, y, or z directions.

z

k

j

i

y

x

- Instead of using magnitudes and directions, vectors can be represented by their components combined with their unit vectors.
- Example: displacement of 30 meters in the +x direction added to a displacement of 60 meters in the –y direction added to a displacement of 40 meters in the +z direction yields a displacement of:

- Simply add all the i components together, all the j components together, and all the k components together.

Sample problem: Consider two vectors, A = 3.00 i + 7.50 j and B = -5.20 i + 2.40 j. Calculate C where C = A + B.

Sample problem: You move 10 meters north and 6 meters east. You then climb a 3 meter platform, and move 1 meter west on the platform. What is your displacement vector? (Assume East is in the +x direction).

- Given the vector

Sample problem: You move 10 meters north and 6 meters east. You then climb a 3 meter platform, and move 1 meter west on the platform. How far are you from your starting point?

Lesson

Position, Velocity, and Acceleration Vectors in Multiple Dimensions

x: position

x: displacement

v: velocity

a: acceleration

r: position

r: displacement

v: velocity

a: acceleration

In Unit Vector

Notation

- r = x i + y j + z k
- r = x i + y j + z k
- v = vxi + vyj + vzk
- a = axi + ayj + azk

Sample problem: The position of a particle is given byr = (80 + 2t)i – 40j - 5t2k. Derive the velocity and acceleration vectors for this particle. What does motion “look like”?

Sample problem: A position function has the form r = x i + y j with x = t3 – 6 and y = 5t - 3.

a) Determine the velocity and acceleration functions.

b) Determine the velocity and speed at 2 seconds.

- Let’s look at some video analysis.
- Let’s look at a documentary.
- Homework questions?

Lesson

Multi-Dimensional Motion with Constant (or Uniform) Acceleration

Sample Problem: A baseball outfielder throws a long ball. The components of the position are x = (30 t) m and y = (10 t – 4.9t2) m

a) Write vector expressions for the ball’s position, velocity, and acceleration as functions of time. Use unit vector notation!

b) Write vector expressions for the ball’s position, velocity, and acceleration at 2.0 seconds.

Sample problem: A particle undergoing constant acceleration changes from a velocity of 4i – 3j to a velocity of 5i + j in 4.0 seconds. What is the acceleration of the particle during this time period? What is its displacement during this time period?

g

g

g

g

g

- This shows the parabolic trajectory of a projectile fired over level ground.
- Acceleration points down at 9.8 m/s2 for the entire trajectory.

- The velocity can be resolved into components all along its path. Horizontal velocity remains constant; vertical velocity is accelerated.

vx

vx

vy

vy

vx

vy

vx

vy

vx

y

y

x

x

t

t

Vy

Vx

t

t

ay

ax

t

t

Vo,y = Vo sin

Vo,x = Vo cos

- …resolve the initial velocity into components.

Vo

Sample problem: A soccer player kicks a ball at 15 m/s at an angle of 35o above the horizontal over level ground. How far horizontally will the ball travel until it strikes the ground?

Sample problem: A cannon is fired at a 15o angle above the horizontal from the top of a 120 m high cliff. How long will it take the cannonball to strike the plane below the cliff? How far from the base of the cliff will it strike?

Lesson

Monkey Gun Experiment – shooting on an angle

Lesson

A day of derivations

Sample problem: derive the trajectory equation.

Sample problem: Derive the range equation for a projectile fired over level ground.

Sample problem: Show that maximum range is obtained for a firing angle of 45o.

Lesson

Relative Motion

- When observers are moving at constant velocity relative to each other, we have a case of relative motion.
- The moving observers can agree about some things, but not about everything, regarding an object they are both observing.

P

particle

vrel

A

B

observer

observer

vA

P

particle

vrel

A

B

observer

observer

In this case, it looks to A like P is moving to the right at twice the speed that B is moving in the same direction.

P

particle

-vrel

A

B

observer

observer

vA

vB

it looks like P is moving to the right at the same speed that A is moving in the opposite direction, and this speed is half of what A reports for P.

vrel

P

particle

-vrel

A

B

observer

observer

vA

vB

vB = vA – vrel

vA = vB + vrel

vrel

Sample problem: Now show that although velocity of the observers is different, the acceleration they measure for a third particle is the same provided vrel is constant. Begin with vB = vA - vrel

- If observers are moving but not accelerating relative to each other, they agree on a third object’s acceleration, but not its velocity!

- Frames of reference which may move relative to each other but in which observers find the same value for the acceleration of a third moving particle.
- Inertial reference frames are moving at constant velocity relative to each other. It is impossible to identify which one may be at rest.
- Newton’s Laws hold only in inertial reference frames, and do not hold in reference frames which are accelerating.

Sample problem: How long does it take an automobile traveling in the left lane at 60.0km/h to pull alongside a car traveling in the right lane at 40.0 km/h if the cars’ front bumpers are initially 100 m apart?

Sample problem: A pilot of an airplane notes that the compass indicates a heading due west. The airplane’s speed relative to the air is 150 km/h. If there is a wind of 30.0 km/h toward the north, find the velocity of the airplane relative to the ground.