Scalars & Vectors 1_scalar_vector... Vectors can be represented by words “Take your team 2...

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Transcript of Scalars & Vectors 1_scalar_vector... Vectors can be represented by words “Take your team 2...

  • Scalars & Vectors

  • Vectors are quantities that have both a direction

    and a magnitude (size).

     Ex. 2 km, 30ο north of east

     Examples of Vectors used in Physics

     Displacement

     Velocity

     Acceleration

     Force

     Scalars are quantities that have only a

    magnitude(size) are called.

    Scalar Example Magnitude

    Speed 20 m/s

    Distance 10 m

    Age 15 years

    Heat 1000 calories

  •  Vectors can be represented by words  “Take your team 2 ‘clicks’ (km) north”

     “US Air 45, new course 30o at 500 mph.”

     Vectors can be represented by symbols  In the text, boldface indicates vectors.

     Examples:

     Vectors can be represented graphically using arrows  The direction of the arrow is the direction of the

    vector.  The length of the arrow tells the magnitude

     Vectors can be moved parallel to themselves and still be the same vector

     Vectors only tell amount and direction, so a vector doesn’t care where it starts.

    t  Δx

    VaF av

  •  The sum of two vectors is called the

    resultant.

     To add vectors graphically, draw each vector

    to scale.

     Place the tail of the second vector at the tip

    of the first vector.

     Vectors can be added in any order.

     To subtract a vector, add its opposite.

  • VECTOR ADDITION – If 2 similar vectors point in the SAME direction, add them.

     Example: A man walks 54.5 meters east, then another 30 meters east. Calculate his displacement relative to where he started?

    54.5 m, E 30 m, E +

    84.5 m, E

    Notice that the SIZE

    of the arrow conveys

    MAGNITUDE and the

    way it was drawn

    conveys DIRECTION.

  • VECTOR SUBTRACTION - If 2 vectors are going in

    opposite directions, you SUBTRACT.

     Example: A man walks 54.5 meters east, then 30

    meters west. Calculate his displacement relative to

    where he started?

    54.5 m, E

    30 m, W

    24.5 m, E

  • When 2 vectors are perpendicular, you must use the Pythagorean Theorem.

    kmc

    c

    bacbac

    8.10912050

    255295Resultant

    22222

    

    

    

    95 km,E

    55 km, N

    Start

    Finish The hypotenuse in Physics is

    called the RESULTANT.

    The LEGS of the triangle are called the COMPONENTS

    Horizontal Component

    Vertical

    Component

     Example: A man walks 95 km, East then 55 km, north. Calculate his RESULTANT DISPLACEMENT

  •  In the previous example, DISPLACEMENT was asked for

    and since it is a VECTOR we should include a DIRECTION

    on our final answer.

    NOTE: When drawing a right triangle that

    conveys some type of motion, you MUST

    draw your components HEAD TO TOE.

    N

    S

    E W

    N of E

    E of N

    S of W

    W of S

    N of W

    W of N

    S of E

    E of S

    N of E

  •  Just putting North of East on the answer is NOT specific enough for

    the direction. We MUST find the VALUE of the angle.

    o30)5789.0(1

    5789.0 95 55

    

    

    Tan

    sideadjacent sideoppositeTan

    95 km, E

    To find the value of the

    angle we use a Trig

    function called TANGENT.

    So the COMPLETE final answer = 109.8 km, 30 degrees North of East

    N of E

    55 km, N

    109.8 km

  • Resolve each vector into x and y

    components, using sin and cos.

    Add the x components together to get the

    total x component. Add the y component

    together to get the total y component.

    Find the magnitude of the resultant using

    Pythagorean theorem.

    Find the direction of the resultant using the

    inverse tan function.

  •  Any vector can be resolved, that is, broken

    up, into two vectors, one that lies on the x-

    axis and one on the y-axis.

  •  An arrow is shot from a bow at an angle of 25ο

    above the horizontal, with an initial speed of 45

    m/s. Find the horizontal and vertical components

    of the arrow’s initial velocity.

    25o

    vx

    vy

    ?

    ?

    25

    m/s 45

    y

    x

    v

    v

    v

    o

    m/s 4178.40

    )cos(25m/s) 45(

    cos

    cos

    

    

    

    x

    x

    x

    x

    v

    v

    vv

    v

    v

    m/s 1901.19

    )(25sinm/s) 45(

    sin

    sin

    

    

    

    y

    y

    y

    y

    v

    v

    vv

    v

    v

  •  Suppose a person walked 65 m, 25 degrees East of North.

    What were his horizontal and vertical components?

    EmCHopp

    NmCVadj

    hypopphypadj

    hypotenuse

    sideopposite

    hypotenuse

    sideadjacent

    ,47.2725sin65..

    ,91.5825cos65..

    sincos

    sincos

    

    

    

    

    

    

    65 m 25

    H.C. = ?

    V.C = ?

    The goal: ALWAYS MAKE A RIGHT

    TRIANGLE!

    To solve for components, we often use

    the trig functions since and cosine.

  •  A bear, searching for food wanders 35 meters east then 20 meters north. Frustrated, he wanders another 12 meters west then 6 meters south. Calculate the bear's displacement.

    3.31)6087.0(

    6087. 23

    14

    93.262314

    1

    22

    

    

    

    Tan

    Tan

    mR

    35 m, E

    20 m, N

    12 m, W

    6 m, S

    - = 23 m, E

    - = 14 m, N

    The Final Answer: 26.93 m, 31.3 degrees NORTH or EAST

    23 m, E

    14 m, N R

  •  A boat moves with a velocity of 15 m/s, N in a river which flows with a velocity of 8.0 m/s, west. Calculate the boat's resultant velocity with respect to due north.

    1.28)5333.0(

    5333.0 15

    8

    /17158

    1

    22

    

    

    

    Tan

    Tan

    smRv

    15 m/s, N 8.0 m/s, W

    Rv 

    The Final Answer : 17 m/s, @ 28.1 degrees West of North

  •  A plane moves with a velocity of 63.5 m/s at 32

    degrees South of East. Calculate the plane's

    horizontal and vertical velocity components.

    SsmCVopp

    EsmCHadj

    hypopphypadj

    hypotenuse

    opposite

    hypotenuse

    adjacent

    ,/64.3332sin5.63..

    ,/85.5332cos5.63..

    sincos

    sinecos

    

    

    

    

    

    

    63.5 m/s

    32

    H.C. =?

    V.C. = ?

  •  A storm system moves 5000 km due east, then shifts course at 40 degrees North of East for 1500 km. Calculate the storm's resultant displacement.

    NkmCVopp

    EkmCHadj

    hypopphypadj

    hypotenuse

    opp

    hypotenuse

    adj

    ,2.96440sin1500..

    ,1.114940cos1500..

    sincos

    sinecosine

    

    

    

    

    

    

    0.20)364.0(

    364.0 1.2649

    2.964

    1.28192.9641.2649

    1

    22

    

    

    

    Tan

    Tan

    kmR

    5000 km, E

    40

    1500 km

    H.C.

    V.C.

    1500 km + 1149.1 km = 2649.1 km

    2649.1 km

    964.2 km R

    The Final Answer: 2819.1 km @ 20

    degrees, East of North

  • We use the term VECTOR RESOLUTION to suggest

    that any vector which IS NOT on an axis MUST be

    broken down into horizontal and vertical

    components.

    BUT --- the ultimate and

    recurring themes in

    physics is take any and all

    vectors and turn

    them all into ONE BIG

    RIGHT TRIANGLE.

  • 1. Make a drawing showing all the vectors, angles, and given directions.

    2. Make a chart with all the horizontal components in one column and all the vertical components on the other.

    3. Make sure you assign a negative sign to any vector which is moving WEST or SOUTH.

    4. Add all the horizontal components to get ONE value for the horizontal. Do the same for the vertical.

    5. Use the Pythagorean Theorem to find the resultant and Tangent to find the direction.

  • A search and rescue operation produced the

    following search patterns in order:

    1: 30 meters, west

    2: 65 meters, 32 degrees E