Vectors - Download Now PowerPoint by 6lVV7MM



Magnitude + Direction
I. Scalars and Vectors

       A. Scalars – Number and unit. No direction.
               Ex: Mass, time, temperature

       B. Vectors – Quantities that have a direction associated with them.

               Ex. Velocity, force, momentum

            1. Each separate vector is called a COMPONENT.

            2. The single vector that produces the same result as the combined components
            is called the RESULTANT.

                 V1        V2                                    V1 + V2

                COMPONENTS                                   RESULTANT
       Each vector has two ends

TAIL                              HEAD;
    Vectors can be used to represent the
    increasing speed of a ball in free fall.

The distance the ball falls each second increases
because the ball is accelerating.

This is shown by drawing a longer
vector arrow for each time interval.
    constant velocity (v) – motion in a straight line at a constant speed; vf = vo.

    Vectors can also be used to represent a ball rolling horizontally
    on a table at a constant velocity.


    Each vector arrow is drawn the same length to represent the constant velocity.

    The velocity would remain constant but friction makes it slow down
    and eventually stop.
II. Vector Addition and Subtraction
    A. Addition of COLINEAR Vectors – when vectors point along the same direction.

        Ex: If a person walks 10.0 meters, east, stops, and continues 5.0 meters, east.
        What is the resultant?
                                                         Tail to Tip Method of Drawing

               10 m, E                5 m, E             -Tail of the second arrow
              Vector A                Vector B           located At the head of the
                                                         first arrow, the two lengths
                                                         add to give the length of the
              10 m, E + 5 m E = 15 m, E                  total displacement.
                                                         - Can only be done when the
                         Vector R
                                                         vectors point along the same

    Ex: If a person walks 10.0 meters, east, stops, and continues 5.0 meters, west.
    What is the resultant?
                                                    Colinear Vector Addition in
                                                    Opposite Directions:
                                                    -Draw both vector components in
           10 m, E                   5 m, W         their respective direction.
          Vector A                   Vector B       - Find the difference between the
                                                    two vectors and select the direction
                                                    of the vector with the greater
          10 m, E - 5 m E = 5 m, E                  magnitude.
                     Vector R

Ex: What is the velocity of a swimmer who is swimming 10 m/s upstream if the
current downstream is 5 m/s?

                     5 m/s;                              10 m/s up – 5 m/s; down
10 m/s;              downstream                          = 5 m/s; up
B. Addition of Vectors that are Neither Colinear Nor Perpendicular

    Graphical Technique – A diagram is constructed in which the arrow are drawn
    tail to tip. The lengths of the vector arrows are drawn to scale, and the angles
    are drawn accurately (with a protractor). Then, the length of the arrow
    representing the resultant vector is measured with a ruler. This length is
    converted into the magnitude of the resultant vector by using the scale
    factor with which the drawing was constructed.
                                 Ex: A car moves with a displacement A
                                 of 275 m, due west and then with a
                R                displacement B of 125 m in a direction
         B                       55.0° north of west.


                                 The two displacement vectors A and
Vector A = 275 m; W              B are neither colinear nor perpendicular
                                 but add to give the resultant vector R.
Vector B = 125 m; @ 55° N of W
Ex: I ran 10.0 km due east but was pushed 5.0 km, north by a crosswind.
How far did I actually run? What was my angle from the start line?

                                  5.0 km; north


         10.0 km; east

                                      - The resultant, R can be found by using
                                      a ruler to measure the magnitude, and a
                                      protractor to measure, the angle Θ.
C. Parallelogram Method to Vector Addition
   Two vectors are drawn to scale and joined at their tails. Parallel lines
   are drawn to complete the parallelogram.

    The resultant is drawn from where the tails touch to the opposite corner.

   Ex: A boat travels 2 km north, then 4 km north of east at an angle of 30°.
   Determine the magnitude and direction of the resultant.

                                        Two ways to solve for R

                                        - Use a ruler and a protractor

                R                       -Use vector resolution
2 km; n                   4 km, e        (math: Pythagorean’s Theorem;
                    30°                   Trig functions)
D. Vector Component Method (Vector Resolution)

       1. Resolve each vector into x and y components.

       2. Assign positive or negative values to each component based on
       which quadrant it falls in.

       3. Reduce the problem to the sum of the two vectors.

             Σ x = v1x + v2x + any others
             Σ y = v1y + v2y + any others

       4. Determine the magnitude and direction of the resultant. Since Σ x and Σ y
       are at right angles, the Pythagorean Theorem can be used to determine the
       magnitude of the resultant.

        The definition of the tangent of the angle can be used to determine the
        direction of the resultant.
             tan Θ = vy                     Θ = tan-1
                     vx                              (v)
               4 km, e                                                  4 km, e

                   28°                                                         28°
                         R                                                           R
2 km; n                                                   2 km; n

  Vector 1                                4 km, e                                  Vector 2
  (2 km; n)                                                  v2y               (4 km; e @ 28°)
 V1x = 0 km                                        v2x                          V2x = 4 km

  V1y = 2 km                 V2y = sin θ = o/h           V2x = cos θ = a/h       V2y = 2 km
                             V2y = o = sin θ (h)         V2x = a = cos θ (h)
                             V2y = sin (28°) ( 4 km)     V2x = cos (28°) ( 4 km)
                             V2y = 2 km                  V2x = 4 km
            v2x                             Σx

                  R                              R

Σ x = v1x + v2x = 0 km + 4 km = 4 km

Σ y = v1y + v2y = 2 km + 2 km = 4 km
             Σ x = 4 km
                                           Θ = tan-1       vy
                                                   ( )     vx
Σ y = 4 km
                                           Θ = tan-1 Σ x
                                                   ( Σy         )
                                           Θ = tan-1 4 km
                                                     4 km           )
                                           Θ = 45º; East of North

sin Θ = o                 sin Θ = Σ x
R=     Σx                  R=      4
     (sin Θ)                    (sin 37)

R = 7 km
       Rule for Adding Vectors
When adding vectors, it doesn’t matter which
 order they are added in. The resultant will
 always be the same.
Ex: A man walks 100 ft at 30º north of east; 200 ft at 40º south of east; and 400 ft
at 90º.
                                           ΣRx = V1x + V2x + V3x
                                           ΣRx = V1cosΘ + V2cosΘ + V3cosΘ
                                           ΣRx = 100(cos30) + 200(cos40) + 0
                                V3         ΣRx = 240 ft
         V1              40º

   ΣRy = V1y + (-V2y) + V3y
   ΣRy = V1sinΘ - V2sinΘ + V3sinΘ

   ΣRy = 100(sin30) - 200(sin40) + 400

   ΣRy = 321 ft
                                     Θ = tan-1 Ry
                                             ( )

                                     Θ = tan-1 321 ft
                       Ry = 321 ft
                                             ( 240 ft   )
   Θ                                 Θ = 53º; North of East
       Rx = 240 ft

                     sin Θ = Ry
sin Θ = o
R=      321
     (sin 53)

R = 402 ft
III. Relative Motion/Velocity

    The concept of a frame of reference is important when solving for
    these problems. It is important to always keep in mind the relationship
    between each variable involved.

   A. Each velocity vector is labeled by two subscripts:
         First subscript refers to the object
         Second subscript refers to the reference frame in which it has a velocity

  Ex: Boat goes across a river
                           VBW = velocity of the boat with respect to the water:
                           how fast the boat travels in the water across the river.
                           VWS = velocity of the water with respect to the shore:
         vBW               how fast the water in the river travels when compared
                           to the shore.

                           VBS = velocity of the boat with respect to the shore:
                           how fast the boat (that’s in the water) is traveling
       Current             when compared to the shore).
                               vBS = vBW + vSW

                                vBW = -vWB

                     Ex: A passenger is walking towards the front of a
                     moving train. The people sitting on the train see
                     the passenger walking with a velocity of +2.0 m/s,
                     where the plus sign denotes a direction to the right.
                     Suppose the train is moving with a velocity of
                     +9.0 m/s relative to an observer standing on
                     the ground.
                     What would the ground-based observer see the
                     passenger moving with?

vPT = velocity of the PASSENGER relative to the TRAIN = + 2.0 m/s
vTG = velocity of the TRAIN relative to the GROUND = + 9.0 m/s
                            vPG = vPT + vTG
vPG = velocity of the PASSENGER relative to the GROUND = + 11 m/s
                                  =               vPG
        vPT        vTG
Ex: The engine of a boat drives it across a river that is 1800 m wide. The
velocity VBW of the boat relative to the water is 4.0 m/s, directed perpendicular
to the current, as shown below. The velocity VWS of the water relative to the shore
is 2.0 m/s.
a.) What is the velocity VBS of the boat relative to the shore?
b.) How long does it take for the boat to cross?

a.) The velocity VBS of the boat relative to the shore is the vector sum of the velocity
VBW of the boat relative to the water and the velocity VWS of the water
relative to the shore: VBS = VBW + VWS.

       VBS = √(VBW 2 + VWS2)                         Θ = tan-1 VBW
                                                               VWS   )
       VBS = √( (4.0 m/s)2 + (2.0 m/s)2 )
                                                      Θ = tan-1 4.0
        = 4.5 m/s                                              (2.0   )
                                                       = 63º
b.) The component of VBS that is parallel to the width of the river determines
how fast the boat is moving across the river; this parallel component is
VBS (sin Θ) = VBW = 4.0 m/s. The time for the boat to cross the river is equal
to the width of the river divided by the magnitude of this velocity component.

   t=    width        = 1800 m     = 450 s
        VBS (sin Θ)     4.0 m/s
IV. Projectile Motion
   The motion of an object that is projected into the air at an angle.

   Solve these problems in the same way as you would velocity problems.
   Use the kinematic formulas for constant acceleration to get formulas for
   projectile motion.

   Horizontal Motion                              Vertical Motion

    vfx = vix + axΔt                                vfy = viy + ayΔt

    Δx = xi + vixt + ½ axΔt2                        Δy = yi + viyt + ½ ayΔt2
    vfx2 = vix2 + 2axΔx                             vfy2 = viy2 + 2ayΔy
     Idea behind trajectory/projectile motion:
Observe the motion of the ball as it falls focusing on the path followed (trajectory).
In projectile motion, we are combining the motion of the ball in free fall
with the motion of the ball rolling on the table at a constant velocity.

This is seen when rolling the ball off of the table.
The ball rolling on the table would continue forever in a straight line if
 gravity is ignored. The ball
in free fall would continue to
increase its speed if air
resistance is ignored.
Since the ball is moving at a constant velocity and in free fall at the same
time, the horizontal and vertical vectors are added together during
equal time intervals. This is done for each time interval until the ball
hits the ground.
The path the ball follows can be seen by connecting the
intermittent resultant vectors.
The two-dimensional motion of the spacecraft can be viewed as the
combination of the separate x and y motions.
                       Kinematic Variables

                      y, x, ax, ay, vy, viy, vx, vix
Ex: In the x direction, a spacecraft has an initial velocity component of vix = +22 m/s
and an acceleration component ax = +24 m/s2. In the y direction, the analogous
quantities are viy = +14 m/s and ay = +12 m/s2. The directions to the right and
upward have been chosen as the positive directions. After a time of 7.0 s, find the:
a.) x and vx
b.) y and vy
c.) final velocity (magnitude and direction of the aircraft)

             Treat the motion in the x and y direction separately

a.) data for motion in the x direction
    ax = +24 m/s2            x = vixt + ½ axt2
    vix = +22 m/s            x = (22 m/s)(7.0 s) + ½ (24 m/s2)(7.0 s) 2
    t = 7.0 s
    x=?                      x = +740 m
    vx = ?
                            vx = vix + axt
                            vx = (22 m/s) + (24 m/s2)(7.0s)
                            vx = +190 m/s
 b.) data for motion in the y direction
     ay = +12 m/s2            y = viyt + ½ ayt2
     viy = +14 m/s            y = (14 m/s)(7.0 s) + ½ (12 m/s2)(7.0 s) 2
     t = 7.0 s
     y=?                      y = +390 m
     vy = ?
                             vy = viy + ayt
                             vy = (14 m/s) + (12 m/s2)(7.0s)
                             vy = +98 m/s

c.) final velocity
 vfy = 98 m/s                             Vf = √(Vfx2 + Vfy2)
                                          Vf = √( (190 m/s)2 + (98 m/s)2 )
                                          Vf = 210 m/s
                      vfx = 190 m/s       Θ = tan-1 vy
                                                 ( )vx
                                          Θ = tan 98
                                                 (190)          = 27° above the positive
                                                                  x axis
Ex: The figure shows an airplane moving horizontally with a constant velocity of
+115 m/s at an altitude of 1050 m. The directions to the right and upward have been
chosen as the positive directions. The plane releases a “care package” that falls to
the ground along a curved trajectory. Ignoring air resistance, determine the time
required for the package to hit the ground.
                                                            The time required for
                                                            the package to hit the
                                                            ground is the time it
                                                            takes for it to move
                                                            1050 m. It moves both
                                                            downward and to the

                                                            These two parts of motion
                                                            occur independently.

                                                            The package is not
                                                            moving in the y direction
                                                            at the instant of release,
                                                            so viy = 0 m/s.
 When the package hits the ground, the y component of its displacement is
 y = -1050 m as the drawing shows. The acceleration is due to gravity,
 so g = - 9.80 m/s2.

 (Note, if you make y = +, then make g = +. I left it as a negative number to
 demonstrate that its just a frame of reference)

  What we know:                       y = viyt + ½ ayt2
  y = -1050 m
  ay = -9.80 m/s2
  viy = 0 m/s                          t = √(2y / ay)
  t=?                                  t = √( 2(-1050 m) / -9.80 m/s2) )
                                       t = 14.6 s

Note: the vertical speed of the package picks up over time, but the horizontal
component stays the same, v0x = vx = 115 cm. This is because the package
is traveling at the same speed as the plane in the x direction. Therefore, the
package does not accelerate in the x direction.
Two packages are dropped from the plane. Package A drops vertically and
package B drops at an angle. They both hit the ground at the same time because
they have the same y components of their velocities. However, package B will
hit the ground with a greater speed due to the magnitude and direction of its x
Ex: Find the speed of the package and the direction of the velocity vector just
before the package in the previous problem hits the ground.

 Image upon impact

                  vfx = 115 m/s                                vfx = 115 m/s

            vf       ground                               vf      ground
vfy = ?                               vfy = -143 m/s

What we know to solve for vy :           So,   vf = √( (115 m/s)2 + (-143 m/s) 2 )
ay = -9.80 m/s2                                vf = 184 m/s
viy = 0 m/s
t = 14.6 s                                Θ = cos-1 vfx
                                                 ( ) vf
vfy = viy + ayt
                                          Θ = cos 115
vfy = 0m/s + (-9.80 m/s2)(14.6 s)                ( 184 )
 vfy = -143 m/s                           Θ = 51.3° below the x axis
Angled projectile motion:
A projectile launched at an angle would continue in a straight line at
a constant velocity if gravity is ignored. However, gravity makes the
projectile accelerate to Earth. This shows us why a projectile launched
at an angle follows a parabolic trajectory.
Since the projectile is launched at an angle, it now has both horizontal
and vertical velocities.

   When dealing with angled projectile problems, we neglect air resistance and
   say that gravity is g = - 9.80 m/s2 (downward) and also say that the vertical
   velocity of any object moving upwards will have a + velocity. This makes
   solving the problems a lot “cleaner.”
The horizontal component of the velocity remains constant.

The vertical component of the velocity changes as the projectile
moves up or down. This is shown when looking at the vector components
of the trajectory at regular time intervals.
   The up and right vectors represent the velocity given to the projectile when
   launched. The vertical vectors decrease in magnitude due to gravity.
   Eventually, the effects of gravity will reduce the upward velocity to zero.
   This occurs at the top of the parabolic trajectory where there is only horizontal

                                          After gravity reduces the upward (vertical)
                                          speed to zero it begins to add a downward
                                          velocity. This velocity increases until the
                                          projectile return to the ground.

When looking at each half of the
trajectory (up and down) you can
determine that the speed of the
projectile going up is equal to the
speed of the projectile coming down.
(Provided air resistance is ignored.)
The only difference is the direction
of the motion.
Problem Solving Projectile Motion:

Since the horizontal and vertical components of a projectile launched at any angle
are still independent of one other, the equations pertaining to linear motion and
free fall can still be used in problem solving.
(As long as we look at them separately.)

Before the projectile is launched, t = 0, the position is zero, y = 0.
The other half of the equation is the part that helps us find time;
when y at the end of the trajectory is zero all the time needed to complete
the trajectory has passed.

       y = viyt + ½ gt2                      x = vixt + ½ at2
When solving problems, the following procedures will be used:

1. Make a sketch of the situation and label all given and unknown quantities.
2. Determine what type of projectile motion is being observed – angled.
3. List all of the equations that apply to the problem.
4. Determine what variables are known and needed.
5. Substitute the given quantities into the equation using proper units.
6. Solve the equation, carrying units throughout the problem.
7. Check your answer to make sure it makes sense.
8. Box or circle your final answer(s).
Ex: A baseball player hits a home run, and the ball lands in the left-field seats,
7.5 m above the point at which the ball was hit. The ball lands with a velocity
of 36 m/s at an angle of 28º below the horizontal. Ignoring air resistance, find the
initial velocity with which the ball leaves the bat.

What we want to find: vi and θ
           To solve for these, we need to find viy and vix

            Vi = √(Vix2 + Viy2)      and          Θ = tan-1 Viy
                                                             ( )
    Since air resistance is ignored, the horizontal component of the velocity, vx
    is constant.
                       vix = vfx = (36 m/s) cos 28° = 32 m/s

What we know about the y component:
y = + 7.5 m
ay = -9.80 m/s2
vfy = (-36 sin 28°) m/s
viy = ?

vfy2 = viy + 2ayy
viy = √(vfy2 – 2ayy)

viy = √( (-36 sin 28°) m/s)2 – 2(-9.80 m/s2)(7.5 m) )
viy = 21 m/s

 Vi = √(Vix2 + Viy2)

 Vi = √( (32 m/s)2 + (21 m/s)2 )

 Vi = 38 m/s

 Θ = tan-1 Viy
          ( )

 Θ = tan-1 21
          ( )

  Θ = 33° above the x axis

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