PES 1110 Fall 2013, Spendier
Lecture 5/Page 1
Today:
- Announcements: Quiz moved to next Monday, Sept 9th due to website glitch!
- Finish chapter 3: Vectors
- Chapter 4: Motion in 2D and 3D (sections 4.1-4.4)
- Questions on HW 1?
- next week Monday: Projectile Motion 4.5-4.6 and Quiz 1!
Review last time: Vectors
- Vectors have a magnitude and a direction
- Draw vectors: use standard angle from the positive x-axis (counterclockwise, CCW)
- Vector addition: tip to tail (pictorially) and by components
Ax = Ax iˆ
Ay = Ay ˆj
A = Ax + Ay = Ax iˆ + Ay ˆj
Ax = A cos θ
Ay = A sin θ
magnitude: A =
( Ax )
2
 Ay
direction: θ = tan −1 
 Ax
We use the symbols iˆ ,
+ ( Ay )
2



ĵ , and k̂ for the unit vectors along the x, y, and z axes.
Adding vectors using components:
A + B = ( Ax + Bx ) iˆ + ( Ay + B y ) ˆj + ( Az + Bz ) kˆ
Examples: Using unit vectors
Given the displacements D = 6iˆ + 3 ˆj − kˆ m and E = 4iˆ − 5 ˆj + 8kˆ m . Find the
magnitude of the displacement 2D − E
(
)
(
)
We are first multiplying the vector D by 2 (a scalar) and then subtract the vector E from
the result.
F = 2 D − E = 2 6iˆ + 3 ˆj − kˆ m − 4iˆ − 5 ˆj + 8kˆ m
(
)
(
)
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PES 1110 Fall 2013, Spendier
Lecture 5/Page 2
F = 12iˆ + 6 ˆj − 2kˆ m − 4iˆ − 5 ˆj + 8kˆ m
F =  (12 − 4)iˆ + (6 + 5) ˆj + (−2 − 8)kˆ m




ˆ
ˆ
ˆ
F = 8i + 11 j −10k  m


(
)
(
)
magnitude: F = Fx2 + Fy2 + Fz2 = 82 + 112 + (−10) 2 =17 m
Two-Dimensional Motion
Moving objects in two dimensions, everything is vectors; everything is components –
position, velocity, and acceleration
In 2D - It boils down to taking track of twice as many equations and twice as many
variables to keep track of, because we must keep track of all the components we just
talked about.
To describe motion, we still need to know position, velocity, and acceleration at all times.
In 2D, this means we have to know the components of the position, velocity, and
acceleration vectors. To locate an object, we have to give two numbers: (x, y). They are
the Cartesian coordinates AND they are the components of the position vector.
Position Vector:
In Physics components have real and useful meaning. In physics it is all about
components: how do you locate something in two dimensions? You give two numbers of
x and y.
Example: What are the components of the position vector in 2d? The components are the
a and y Cartesian coordinates:
x = how far to the right or left of the origin
y= how far up or down of the origin is the position?
We have the same procedure for finding components:
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PES 1110 Fall 2013, Spendier
Lecture 5/Page 3
x = position vector’s x-component (horizontal motion)
y = position vector’s y-component (vertical motion)
r = position vector – from the origin (tip to tail addition)
In unit-vector notation:
2D: r = x + y = x iˆ + y ĵ
3D: r = x + y = x iˆ + y ĵ + z k̂
Example: Draw a particle with position vector r = (-3m) iˆ + (5m) ĵ
Along the x-axis the particle is 3 m from the origin in the minus iˆ direction
Along the y-axis the particle is 5 m from the origin in the plus ĵ direction
Displacement Vector
A real application of all this we just learned – adding and subtracting vectors – is the
displacement vector.
→
The displacement vector ∆r = r2 − r1 is a vector subtraction:
the final position vector r2 minus the initial position vector r1
→
∆r points from to r1 to r2
( what the motion does in between – we do not know in this case)
Position vectors are from the origin.
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PES 1110 Fall 2013, Spendier
Lecture 5/Page 4
Example: A hiker walks 10.0 km west and then 7.50 km south. What is the hikers
displacement (magnitude and direction)?
N (+y)
10
E (+x)
W
7.5
+
=
θ
S
Magnitude:
x = −10
y = −7.5
magnitude : A =
( x)
2
+ ( y) =
2
( −10 )
2
+ ( −7.5 ) = 156.25 = 12.5km
2
Direction:
o 7.5
tan θ = =
a 10
 7.5 
θ = tan −1 
 = 36.9
 10 
use relative bearing: 90º - θ = 90º-36.9º = 53.1 º
θ
In α
Answer: The hiker’s displacement is 12.5 km at 53.1 º west of south.
Example:
An old man walks on three different bearings:
72.4 m, 32.0º east of north
57.3 m, 36.0 º south of west
17.8 m straight south
What is the man's displacement at the end of his walk? Find the sum (resultant) of the
three displacements.
a) draw a picture
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PES 1110 Fall 2013, Spendier
Lecture 5/Page 5
b) find vector components:
The angles of the vectors, measured from the +x-axis toward the +y-axis are
90.0 º – 32.0 º = 58.0 º
180 º + 36.0 º = 216 º
270 º
then
Ax = AcosθA = (74.4 m)(cos 58.0 º) = 38.37 m
Ay = AsinθA = (74.4 m)(sin 58.0 º) = 61.40 m
Bx = BcosθB = (57.3 m)(cos 216 º) = -46.36 m
By = BsinθB = (57.3 m)(sin 216 º) = -33.68 m
Cx = CcosθC = (17.8 m)(cos 270 º) = 0.00 m
Cy = CsinθC = (17.8 m)(sin 270 º) = 61.40 m
Rx = Ax + Bx + Cx = 38.37 m + (-46.36 m) + 0.00 m = -7.99 m
Ry = Ay + Ay + Cy = 61.40 m + (-33.68 m) + 61.40 m = 9.92 m
R = Rx2 + R y2 = (−7.99) 2 + 9.922 = 12.7 m
 Ry 
 9.92 
θ = tan −1   = tan−1 
 = 129
 Rx 
 −7.99 
Or 129º – 90º = 39º west of north
Average velocity vector
Let’s think about the average velocity vector. It is in the same direction as displacement
vector. If it is actually longer or shorter we don’t know at the moment.
→
The average velocity is parallel to ∆r :
→
→
∆r x2 − x1 ∆xiˆ + ∆yjˆ + ∆zkˆ ∆xiˆ ∆yjˆ ∆zkˆ
vavg =
=
=
=
+
+
∆t
∆t
∆t
∆t
∆t
∆t
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PES 1110 Fall 2013, Spendier
Lecture 5/Page 6
Example:
What is the resultant velocity of a plane that is heading north at 290 m/s when also aided
by a wind of 46.0 m/s NW?
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Lecture 5/Page 7
Trajectory:
Picture of the actual path taken by an object. A trajectory tells you where the object was
located at all the different points of motion.
There are two separate position plots which give the velocity vector’s components
At every point we measure its value of x and y
Both of these graphs of course have slopes. Slope of blue graph gives me the x
component of the velocity, vx. The green graph gives me the y component of the velocity
vector, vy. Two different positions – two different slopes – neither gives me the full story
of the component.
When we do make this trajectory graph, the velocity components vx and vy they will
make a vector which is tangent to the trajectory - this is the velocity vector. The
velocity vector is at the same angle as the slope of the trajectory graph.
Instantaneous Velocity vector
dr d ˆ ˆ ˆ
dx
dy ˆ dz ˆ
v= =
xi + yj + zk = iˆ +
j+ k
dt dt
dt
dt
dt
v=v ˆi+v ˆj+v kˆ
(
x
vx =
y
)
z
dx
dy
dz
; vy = ; vz = ;
dt
dt
dt
For example with differentiation look at book: page 62
The velocity vector will go in one direction – how do we determine which way it should
go? The orange line has as arrow on it. We use arrows to indicate the direction – the
arrow on the trajectory line will indicate which direction the velocity vector should point.
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PES 1110 Fall 2013, Spendier
Lecture 5/Page 8
Technically speaking I should not mix velocities with positions. Because they are
completely different quantities, completely different units. We really should not be
drawing them on the same graph. The directions and length of the components are good.
Just don’t compare the length of vx to x
Speed:
Speed is just how fast – the speed at this point – how fast is it going, how big is the
velocity vector, speed is the magnitude of the velocity vector. From the x and y
components of the velocity vector use the Pythagorean Theorem (speed is v with no
arrow)
speed = v = v 2x + v 2y
Average acceleration vectors:
→
The average velocity is parallel to ∆v :
→
→
∆v v2 − v1 ∆vx iˆ + ∆v y ˆj + ∆vz kˆ ∆vx iˆ ∆v y ˆj ∆vz kˆ
aavg =
=
=
=
+
+
∆t
∆t
∆t
∆t
∆t
∆t
Instantaneous acceleration vector:
In 2D I can make two separate velocity plots, give the components of the acceleration,
two plots vx and vy both functions of time, x and y components versus time.
We can find the acceleration components in the same way as velocity
There are two separate VELOCITY plots which give the acceleration vector’s
components
Slopes – will give me the x or y component of the acceleration vector:
Instantaneous acceleration
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PES 1110 Fall 2013, Spendier
Lecture 5/Page 9
Find tangent line at each x-y point on trajectory, find slope of tangent – get vx and vy at a
given time point - use this to make a velocity vs time graph – where the slope will give
me the x and y components of the acceleration.
vy…. how fast is it traveling vertically
vx…. how fast is it traveling horizontally
vertical component of position gives me vertical component of velocity which gives me
vertical component of acceleration!
dv y
dv
dv d
ˆj + dvz kˆ
a=
=
v x iˆ + v y ˆj + vz kˆ = x iˆ +
dt
dt
dt
dt dt
ˆ ˆ
ˆ
a=a x i+a y j+a z k
(
vx =
)
dv y
dvx
dv
; vy =
; vz = z ;
dt
dt
dt
For example with differentiation look at book: page 63
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PES 1110 Fall 2013, Spendier
Lecture 5/Page 10
Example:
What is the average acceleration of a plane that changes its velocity from 350m/s south to
350m/s NW in a time of 14.00s?
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