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Vectors. Vector : quantity that has magnitude and direction Scalar : quantity that has magnitude only Example: 20 mi north (vector) vs. 20 mi (scalar) Arrows represent vectors graphically. (or A ). Magnitude = . (or A ). (arrow length proportional to vector magnitude). A. A. B. B.

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vectors
Vectors
  • Vector: quantity that has magnitude and direction
  • Scalar: quantity that has magnitude only
    • Example: 20 mi north (vector) vs. 20 mi (scalar)
  • Arrows represent vectors graphically

(or A)

Magnitude =

(or A)

(arrow length proportional to vector magnitude)

vectors1

A

A

B

B

Vectors
  • Two vectorsAandBare equal if|A| = |B| ANDA B:
  • For vectorCantiparallel toAandB, with same length asAandB A = B = – C
  • Vector addition: Must take direction into account!
    • Use tail-to-tip rule: place tail of 2nd vector to tip of 1st vector
    • Works for any number of vectors

C

vector addition and subtraction

B

B

A

A

C

B

D

E

–B

A–B

Vector Addition and Subtraction
  • ForA + B = C:
    • Vector addition is commutative (order doesn’t matter) – try it!
  • For multiple vectorsA + B+ D= E:
  • Rules for vector subtraction are the same – just consider adding a negative vector:

Vector Addition and Subtraction Interactive

A

A

A – B= A + (– B)

component vectors

Ax

Ay

Ay

Ax

Component Vectors
  • Imagine the hassle of measuring a triangle each time you want to add vectors!
    • Would need protractor, ruler, etc.
  • Use of vector components provides simple yet accurate method for adding vectors

y

Axand Ay are determined from trigonometry:

Ax = Acosq and Ay = Asinq

Ax(Ay) could be positive or negative depending on whether or not it points in the +x or –x (+y or –y) direction

A = Ax+ Ay

A

q

x

A = (Ax2 + Ay2)1/2

component vectors1

x

y

Ay

Ax

Component Vectors
  • It may be helpful to envision components as “shadows” that are projected along each axis:
    • The length of the “shadow” depends on the orientation of A

y

A

q

x

component vectors2
Component Vectors
  • Any vector can be represented by its x and y component vectors
  • Sometimes it is convenient to determine the components of vectors in a tilted coordinate system:
slide7
CQ1: Consider vector shown below. Which of the following could not be a pair of component vectors for vector ?
component vectors3

In terms of components:

By

Ay

Cy

Ax

Bx

Cx

Component Vectors

y

B

Ax + Bx = Cx

Ay + By = Cy

C

  • Using components to add vectors A + B = C:

A

x

slide9

CQ2: A man entered a cave and walked 100 m north. He then made a sharp turn 150° to the west and walked 87 m straight ahead. How far is the man from where he entered the cave? (Note: sin30° = 0.50; cos30° = 0.87.)

  • 25 m
  • 50 m
  • 100 m
  • 150 m
example problem 3 20
Example Problem #3.20
  • The helicopter view in the figure at right shows two people pulling on a stubborn mule. Find
  • the single force that is equivalent to the two forces shown, and
  • the force that a third person would have to exert on the mule to make the net force equal to zero.
  • The forces are measured in units of Newtons (N).

Solution (details given in class):

(a) 185 N, q = 77.8° from x axis

(b) 185 N, q = 258° from x axis

2 d and 3 d motion

z

y

x

2–D and 3–D Motion
  • Strong similarity between 1–D motion and 2–D (or 3–D) motion
  • Same kinematic quantities (displacement, velocity, acceleration) are used
  • Only difference is vector quantities can have up to three non-zero components
  • First, we describe position:
  • Displacement (change in position):
2 d and 3 d velocity
2–D and 3–D Velocity
  • Average velocity:
    • Or:
  • Instantaneous velocity:
  • Instantaneous speed:
    • Think Pythagorean theorem!
  • Just like in 1–D motion, average velocity only depends on beginning and ending position
  • Instantaneous velocity is tangent to the curve of versus time
2 d and 3 d acceleration
2–D and 3–D Acceleration

(same as 1–D)

  • Average acceleration defined as:
    • are the instantaneous velocities at t2 and t1, respectively
    • points in the same direction as
  • Instantaneous acceleration:
    • points toward the inside of any turn particle is making
  • Remember that an object can have a non-zero acceleration even though its speed remains unchanged
  • Example: uniform circular motion
    • Speed remains the same
    • Velocity keeps changing due to changing direction

(same as 1–D)

slide14

CQ3: A weather balloon travels upward for 6 km while the wind blows it 10 km north and 8 km east. Approximately what is its final displacement from its initial position?

  • 7 km
  • 10 km
  • 14 km
  • 20 km
projectile motion

v0

Projectile Motion
  • Have you ever wondered how to predict how far a batted baseball will travel, or how far a skier will travel after jumping from a ramp?
  • These are examples of projectile motion
    • Projectile (baseball, skier) follows a trajectory (path in space) determined by its initial velocity and effects of gravity
  • Start with simple model
    • Projectile represented by single particle
    • Neglect air resistance, curvature & rotation of the earth

y

(Usually put start of motion at origin of coordinates)

x

projectile motion1

v0y

v0x

Projectile Motion
  • Notice that acceleration is constant (= g) and is in vertical direction (due to gravity) pointing toward earth’s surface
  • This is motion with constant acceleration in 2–D
  • Particle is given an initial velocity v0:
  • We can analyze projectile motion as:
    • Horizontal motion with constant velocity(ax = 0)
    • Vertical motion with constant acceleration (due to gravity)

y

ax = 0  vx = v0x = v0cosq (constant in time)

ay = –g  vy varies with time

v0x

v0y

q

x

slide17

CQ4: Two skydivers are playing catch with a ball while they are falling through the air. Ignoring air resistance, in which direction should one skydiver throw the ball relative to the other if the one wants the other to catch it?

  • above the other since the ball will fall faster
  • above the other since the ball will fall more slowly
  • below the other since the ball will fall more slowly
  • directly at the other since there is no air resistance
projectile motion2
Projectile Motion

Velocity Components Interactive

Position and Time Interactive

  • The equations describing the motion follow directly from our discussion of 1–D motion with constant acceleration:
  • Notice that we treat the x– and y–components separately

y–direction (ay = –g):

x–direction (ax = 0):

projectile motion3

y = ymax

x = xmax

What is the maximum height reached? (or At what value of y is vy = 0?)

What is the maximum range? (or What is the value of x when y = 0 [other than at the origin]?)

Projectile Motion
  • You can answer lots of questions concerning the path the projectile takes by using these equations
  • Many times you need to restate the question to determine what the question is looking for

y

x

example problem 3 23
Example Problem #3.23

A student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of 18.0 m/s. The cliff is 50.0 m above a flat, horizontal beach, as shown in the figure at right. (a) What are the coordinates of the initial position of the stone? (b) What are the components of the initial velocity? (c) Write the equations for the x- and y-components of the velocity of the stone with time. (d) Write the equations for the position of the stone with time. (e) How long after being released does the stone strike the beach below the cliff? (f) With what speed and angle of impact does the stone land?

Partial solution (details given in class):

  • 3.19 s
  • 36.1 m/s, q = 60.1°below +x axis
example problem 3 31
Solution (details given in class):

(b) 1.78 s

(a) 32.5 m

Example Problem #3.31

A car is parked on a cliff overlooking the ocean on an incline that makes an angle of 24.0° below the horizontal. The negligent driver leaves the car in neutral, and the emergency brakes are defective. The car rolls from rest down the incline with a constant acceleration of 4.00 m/s2 for a distance of 50.0 m to the edge of the cliff, which is 30.0 m above the ocean. Find (a) the car’s position relative to the base of the cliff when the car lands in the ocean, and (b) the length of time the car is in the air.

cq5 interactive problem target practice
The airplane cargo needs to be dropped at a horizontal position of:CQ5: Interactive Problem: Target Practice
  • 0 m
  • 0.775 m
  • 1.35 m
  • 4.65 m
  • 6.00 m

ActivPhysics Problem #3.7, Pearson/Addison Wesley (1995–2007)