Introduction to Vectors
Scalars and Vectors
 In Physics, quantities are described
as either scalar quantities or
vector quantities .
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Introduction to Vectors
Scalar Quantities
 Involve only a magnitude, which
includes numbers and units.
 Examples include distance and
speed.
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Introduction to Vectors
Vector Quantities
 Involve a direction, in addition to
numbers and units.
 Can be represented graphically with
arrows.
 The longer the arrow, the greater the
magnitude it represents.
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Vector Operations
Drawing Vectors
In order to draw vectors that
indicate direction, you need to
work within a coordinate
system.
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Vector Operations
Coordinate Systems
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Vector Operations
Coordinate System
When working with the Cartesian
coordinates, adding vectors can
be accomplished with the “tip-totail” method.
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Example: A child walks 2.0 m east, pauses,
and then continues 3.0 m east.
0
The resultant (R) = 5.0 m east.
If the two vectors have different directions, they
are still added tip to tail.
Example: A child walks 2.0 m east, then turns
around and walks 4.0 m west.
0
The resultant is 2.0 m west.
Component vectors are
added “tip-to-tail.”
The resultant vector is
drawn “tail-to-tip.”
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4 m east
3 m north
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Adding vectors graphically, using the “tip-totail” method.
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4 m east
3 m north
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A man walks 3 m north, and then 4 m east.
Find his displacement.
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4 m east
3 m north
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You are allowed to move the vectors, but
don’t change the direction or length.
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tail
tip
4 m east
3 m north
Line up the tip of one vector with the tail
of the other.
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4 m east
3 m
north
Line up the tip of one vector with the tail
of the other.
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4 m east
3 m
north
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Resultant vector
Now, draw the resultant vector from “tailto-tip” as shown above.
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Vector Operations
When we want to add two component
vectors that are perpendicular to
each other, we can use the
Pythagorean theorem to find the
magnitude of the resultant vector,
and the tangent function to find
the direction of the resultant
vector.
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Vector Operations
3.02+4.02 = R2 R =  25m2
= 5.0m
Resultant vector=R
3.0 m north
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4.0 m east
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Vector Operations
tan = opp/adj tan-1(opp/adj) = 
tan-1(3.0m/4.0m)
= 37
5.0 m NE
Full Answer:
5.0m at 37°N of E
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3.0 m north

4.0 m east
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What if you know the resultant vector
angle and need to find a component
vector?
tan = opp/adj
tan(adj) = opp
Tan37(4.0m)=
3.01m
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tan-1(opp/adj) = 
?
?
=37
4.0 m east
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Then use the Pythagorean theorem to
find hypotenuse…
3.01m2+4.0m2 = R2
R= 25m2=5.0m
20
?
3.01m
=37
4.0 m east
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Problems 3A on page 91
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Vector Operations
Resolution of Vectors – The process of
breaking up a vector into two or more
components.
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Vector Operations
Resolution of Vectors
• When a golf ball is hit upwards at
an angle, it will move in two
dimensions.
• The ball moves forward as it
moves upwards and as it falls down.
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Vector Operations
Resolution of Vectors
In order to figure out how far
the ball will travel, you will be
required to resolve the initial
velocity vector into horizontal
velocity (x) and vertical velocity (y)
components.
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Vector Operations
Resolution of Vectors
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
Resolving a
vector into its
components is
the opposite of
combining
component
vectors into a
resultant vector.
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
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If you were given component vectors
A and B, you could find C.
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
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If you are given C, you need to be able
to find Cy and Cx.
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Vector Operations
Vi
55o
55o
Vi = 35 m/s
V
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A cannonball is
fired with an
initial velocity
of 35 m/s at
an angle of 55o
above the
horizontal.
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Vector Operations
V
Vii
o
55
55
o
Vx
Vi = 35 m/s
V
29
Vy
Draw the x and y
components of the
velocity vector to
form a right
triangle.
Use Trigonometry
functions to find the
magnitude of each
component. 3/23/2016 2:21 AM
Vector Operations
V
VVii
hyp
o
55
55
o
Vx
Vi = 35 m/s
V
30
Vy
The original
velocity vector is
the hypotenuse of
the triangle.
IT IS NOT THE
PATH OF THE
PROJECTILE!!!
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Vector Operations
hyp
Vii
V
V
o
55
55
o
opp
Vy
The vertical
component (Vy)
represents the
opposite side.
Vx
Vi = 35 m/s
V
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Vector Operations
hyp
V
VVii
o
55
55
o
Vx adj
opp
Vy
The horizontal
component (Vx)
represents the
adjacent side.
Vi = 35 m/s
V
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Vector Operations
hyp
VVii
V
o
55
55o
Vx adj
Vi = 35 m/s
V
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opp
Vy
From Geometry, we recall
opp
Sin  = ----hyp
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Vector Operations
hyp
V
VVii
55o
55o
Vx adj
Vi = 35 m/s
V
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opp
Vy
We multiply both
sides of the
equation by hyp
opp
hyp x Sin  = ----- x hyp
hyp
hyp x sin  = opp
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Vector Operations
VV
V
i
hyp
i
55o
Vx adj
Vi = 35 m/s
V
35
opp
Vy
hyp x sin  = opp
or
opp = hyp x sin 
Remember, opp is
really Vy and hyp is Vi,
so,
Vy = Visin 
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Vector Operations
hyp
Vi i
VV
55o
Vx adj
Vi = 35 m/s
V
Vy = 29 m/s
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Vy = Visin 
opp
Vy
Vy = (35 m/s)(sin 55o)
Vy = 28.67032155 m/s
Vy = 29 m/s
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Resolution of Vectors
V
VV
ii
hyp
55o
Vx adj
Vi = 35 m/s
V
Vy = 29 m/s
37
opp
Vy
Now, for the
horizontal (X)
component of the
initial velocity.
From our diagram,
we can see that Vx
represents the
adjacent side of
the triangle.
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Resolution of Vectors
V
i
VV
hyp
i
55o
Vx adj
Vi = 35 m/s
V
Vy = 29 m/s
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opp
Vy
We could use either
cosine, tangent or
the Pythagorean
theorem to find
the value of Vx.
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Vector Operations
hyp
Vi i
VV
55
55o o
Vx adj
Vi = 35 m/s
V
Vy = 29 m/s
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opp
Vy
adj
cos  = ------hyp
Multiplying both sides by
hyp, we get
adj
hyp x cos  = ------- x hyp
hyp
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Vector Operations
V
VVii
hyp
55o
Vx adj
VVi = 35 m/s
Vy = 29 m/s
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Vx = 20. m/s
opp
Vy
hyp x cos  = adj, or
adj = hyp x cos 
We recall that adj = Vx
so,
Vx = Vicos 
Vx = (35 m/s)(cos 55o)
Vx = 20.07517527
m/s
1 m/s
Vx = 2.0 X103/23/2016
2:21 AM
One more thing…

If the angle is below horizontal, or it is a ski
slope, etc. draw it like this…
Vx

Vy
Vi
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 Problems
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3B page 94
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Adding Vectors Algebraically
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Example

A car drives 67 km at an angle of 20 degrees
south of east and then drives 78 km at an
angle of 67 degrees north of east. What is
the car’s resultant displacement ?(Give
magnitude and direction – ALWAYS!).
–
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SIX Steps…
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1. Draw a LARGE diagram showing all
the vectors(ADDED TIP TO TAIL), all
angles, the Resultant Vector and
finally all components!
d2=78km
R
dx1
d2
dy2
20
d1
dy1
d1=67km
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67
dx2
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2. Find all the x
components and y
components of each
vector(Remember signs!!).
d2=78km
R
dx1
dy,2
d2
20
d1
dy1
d1=67km
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67
dx2
dy1 = -d1sin 
= -(67km)(sin20)
= -23km
dy2 = d2sin 
(78km)(sin67)
= + 72km
dx1 = d1cos 
= (67km)(cos20)
=+63km
dx2 = d2cos 
=(78km)(cos67)
= + 31km
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3. Find the total x and
total y (Remember signs!!)
R
dx1
dx.Total =
63km+31km=
+94km
d2
dy2
23
d1
dy1
67
dx2
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dy.total =
-23km+72km=
+49km
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4. Draw a NEW triangle showing the total x
and y, then draw in the resultant(this should
look similar to the 1st resultant you drew if
your angles were estimated correctly.)
R
yT=49km
xT=94km
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5. Use the Pythagorean
Theorem to calculate R…
1.1X102km
R
49km
94km
49
xT2 + yT2 = R2
94km2+49km2=R2
R2 = 11237km2
R = 106km
=1.1X102km
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You are not
finished!!
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6. Use the tangent
function to determine 
Tan-1(opp/adj) =
2
1.1X10
R km
49km

94km
51

Tan-1(49/94)
= 28  North of
East
Full Answer is
1.1X102km at
28 North of
East
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You should be able to do this with any
number of vectors…just a pain…
32
53
52
R
25
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Also, if a direction is already straight up
or down or left or right, you do not have
to resolve it into components, it is just
one component already!!

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Ex: If a cat climbs up a tree, it is only a Y
component and there is no X component – so
call X zero for that vector.
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HOMEWORK
 Problems 3C, page 97
 -page 95 has a good example also, if you need
it – basically the same problem as my example
with different numbers.
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Horizontal Projectile Motion
The motion in the x dimension has
no effect on the motion in the y
dimension, when air resistance is
ignored, so “forward” velocity won’t
keep objects in the air longer.
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 The
path of a projectile is a
curve called a parabola.
– Air resistance affects the
path (page 99)
 Projectile motion is freefall,
with an initial horizontal velocity.
 See figure 3-19
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Recall Vertical Freefall Formulas from
Rest (now the negative is in the
formula… don’t ask me why …so put a
positive 9.81 m/s2 in these formulas).
Vy,f = -gt
 Vy,f2 = -2gy
 y = -1/2g(t)2
 Horizontal motion is considered
constant.
 Vx,i = Vx,f = constant
so you can use x = vxt for ANY
horizontal distance calculation

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Example 1
A softball is thrown horizontally off
the top of a 890.0m building. It
hits the ground 90.0 m from the
base of the building. How long was
the ball in the air?
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1. Draw a picture…
Vx (unknown)
890.0 m
90.0 m
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2. Look at possible formulas to find
time…
Vx (unknown)
x = vxt
y = -1/2g(t)2
890.0 m
90.0 m
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3. Solve for time with the second
formula
Vx (unknown)
x = vxt
y = -1/2g(t)2
890.0 m
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90.0 m
t = -(2
t = -2 y/g
·-890.0)/9.81=13.5 s
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Example 2
Dakota is thrown horizontally off the
top of a 25m building. He hits the
ground* 27.0 m from the base of
the building. How fast was Dakota
thrown?
*of course there was a safety net 
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1. Draw a picture…
Vx (unknown)
25 m
27.0 m
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2. Look at possible formulas to find
time…because we can’t get velocity
without time and distance!
Vx (unknown)
x = vxt
y = -1/2g(t)2
25 m
27.0 m
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3. Solve for time with the second
formula
Vx (unknown)
x = vxt
y = -1/2g(t)2
25 m
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27.0 m
t = -(2
t = -2 y/g
·-25m)/9.81m/s=2.3s
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4. Then solve for vx with the first
formula
Vx (unknown)
x = vxt
y = -1/2g(t)2
t = 2.3s
25 m
27.0 m
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Vx (unknown)
x = vxt
y = -1/2g(t)2
25 m
t = 2.3s
x/ t = vx
27.0 m
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Vx=27.0m/ 2.3s = 12m/s
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If you know horizontal velocity and
distance in the x direction, you could
solve for vertical y distance too…


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Find time…using x = vxt
Then find y using y = -1/2g(t)2
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Homework

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Problems 3D page 102
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Projectiles Launched at an Angle
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The cart below moves to the right with a
uniform velocity of 3.0 m/s. As it moves,
it launches a ball straight up with a
velocity of 2.0 m/s. Will the ball land in
front of the cart, behind the cart or on
top of the cart? (ignore air resistance.)
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Answer. In addition to the vertical velocity
of 2.0 m/s, the ball will have the same
velocity in the horizontal as the cart. If
there is no air resistance, the ball will
continue moving forward at 3.0 m/s, until it
lands back in the launch tube on the cart.
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Components to analyze objects
launched at an angle



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If the initial velocity vector makes an
angle with the horizontal, the motion
must be resolved into its components.
Use sine and cosine functions
Vx,i = Vi(cos) and Vy,i = Vi(sin)
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Formulas for Projectiles Launched
at an angle
 Vx=
Vi(cos) = constant
 x = Vi(cos) t
 Vy,f= Vi(sin) - g t
 Vy,f2= Vi2(sin)2 – 2g y
 y = Vi(sin) t – 1/2g (t)2
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Example one
Let’s try a complete projectile motion
problem.
 A golfer hits the golf ball, giving it an
initial velocity of 28 m/s, at an angle of
52o above the horizontal. If we ignore
air resistance, can you find
 How long the ball will be in the air? (t)
 How high will it go?(y)
 How far will it go? (x)
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How long the ball will be in the air?
(T)

Vi = 28m/s
Viy
52o
Vix
Given
Vi = 28m/s
 = 52
g = 9.81 m/s2

We know that ½ way
through it’s parabolic
path, the ball will reach
a velocity of 0 m/s in
the vertical (y)
dimension.
In other words, we say
at max Y, the Vy,f = 0
m/s.
At max Y, the Vy,f = 0 m/s
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How long the ball will be in the air?
(T)

Vi = 28m/s
52o
Vix
Given
Vi = 28m/s
 = 52
g = 9.81 m/s2
Viy

t = Vi sin  = 28 m/s sin 52 = 2.2s

At max Y, the Vy,f = 0 m/s
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Vy,f = Vi sin - gt
Because Vy,f = 0 m/s,
we can cross it out.
Isolating t we get
g
9.81 m/s2
Of course, this really
represents 1/2 T, so the
ball is in the air 4.4 s.
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How high will it go?(Y)
Given
Vi = 28m/s
 = 52
g = 9.81 m/s2
Formula: Vy,f2 = Vi2 (sin )2 - 2g Y
Again, at the max. Y , Vy,f = 0 m/s
Isolating Y, we get
Vi2 (sin )2 487 m2/s2
Y = ------------ = ---------------- = 25 m
T = 4.4 s
2)
2g
2(9.81
m/s
At max Y, the Vfy = 0 m/s
Find Y
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How far will it go? ( X)
Given
Vi = 28m/s
 = 52
g = 9.81 m/s2
Formula: X = VixT
X = VicosT
= (17.24 m/s)(4.4 s) =76 m
T = 4.4 s
y = 25 m
At max y, the Vy,f = 0 m/s
80
Find X
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Example 3E is good too!!
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Practice 3E on page 104
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Skip 3F…
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Relative Velocity
 Observers
using different frames
of reference may measure
different displacements or
velocities for an object in motion.
– Figure 3-23, page 106
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Example
A
boat heading north crosses a
wide river with a velocity of 10.00
km/h relative to the water. The
river has a uniform velocity of
5.00 km/h due east. Determine
the boat’s velocity with respect to
an observer on shore.
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List givens and draw a picture
V
(boat/river) = 10.00 km/h
 V (river/shore) = 5.00 km/h
 V (boat/shore) = ?
Vr/s
Vb/r
86

Vb/s
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List givens and draw a picture
V
(boat/river) = 10.00 km/h
 V (river/shore) = 5.00 km/h
 V (boat/shore) = ?
5.00km/h
10.00km/h

87
Vb/s
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List givens and draw a picture
Find Vb/s
 Pythagorean Theorem
 5.002 + 10.002 = Vb/s2
5.00km/h
 Vb/s = 11.18km/h
 Find 
10.00km/h
 Tan-1 (5.00/10.00)
Vb/s

 =  = 26.6 East of North

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Problems 3F on page 109
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