Vectors
hypotenuse
opposite
q
adjacent
Knowing the length of any one side as well as an angle
(other then the 90) will uniquely determine the other sides
of the right triangle
Knowing the length of any two sides will uniquely
determine the third side of the right triangle as well as
the angles
Knowing only the angles is indeterminate
opp
sin q 
hyp
opp  hyp * sin q
opp  50sin 30  25
adj
cosq 
hyp
adj  hyp * cosq
adj  50cos 30  43.3
Other vertex angle must be 60o, since θ1 + θ2 + θ3 = 180° always.
hyp
opp
sin 27 
hyp
adj
cos 27 
hyp
opp
tan 27 
adj
All unsolvable.
At least one side must be determined in
order to determine the other two
Vectors
A vector quantity has both magnitude and direction
y
magnitude
40o
x
Adding vectors

A
 
AB

B
 
AB
 
AB
6

B

A

R

R

R

B

A

B

A
Expandable
This graphical technique can be expanded to add as many vectors
are necessary
For example: a person walks 50 m east followed by 30 m north
followed by 60 m west and then 50 m south.
d2
N
The resultant is drawn from the tail of the
first vector to the head of the last vector.
W
As an exercise determine what the resultant
would be if the vectors were added in a
different order.
E
S
d1
Vector components
• Suppose you drive from State College to Philadelphia.
 How far east did you
end up from where
you began?
 How far south did you
end up from where
you began?
http://maps.google.com
Decomposing a vector into the two mutually perpendicular vectors that
add together to be equivalent to the original vector is called finding the
components
Distance versus Displacement
N
W
dN
30o
dw
dW
dW  300 cos 30  260 meters
300
dN
sin 30 
d N  300 sin 30  150 meters
300
total distance walked  dW  d N  410 meters
cos 30 
E
S
A student walks from the
HUB to Pattee Library
which gives her a
displacement of about
300 m at an angle of 30
degrees north of the west
axis from her starting
point at the HUB .
However, to get to Pattee,
she must walk West along
Pollock road and then turn
North on the Pattee Mall.
What is the distance that
she walked?
Lost
A hiker is to meet his buddies at a camp that is located 7 km due north of his current
location. The hiker becomes disoriented in a driving rainstorm and finds that he has
actually traveled 11.3 km at an angle of 20o to the West of due North. What
displacement vector will get him to his buddies camp?
dE N
Find components of the hike that he did
d
sin 20  w d w  11.3 sin 20  3.86 km
11.3
d
cos20  N d N  11.3 cos 20  10.6 km
11.3
dS
20o
dN
W
dW
E
S
d 2  d S2  d E2  3.6 2  3.86 2  27.9
d  27.9  5.28 km
3.6
3.6
tan q 
q  tan 1
 43o S of E
3.86
3.86
The components indicate that he is
3.86 km W of where he wants to
be so he will need to walk 3.86 km
E. He is also 3.6 km further N than
anticipated so he will need to walk
3.6 km S. These are the
components of his displacement
back to his buddies.
Two ways to multiply vectors
 
A  B  AB cos q  C where q is the angle between th eir tails
(the tendency of two vectors to align with each other)
  
A  B  R where the magnitude of vector R  AB sin q and whose direction
is given by the right hand rule
In one case the product produces a scalar (dot product) and in the other it
produces another vector (the cross product)
We’ll make use of the dot product in the near future and come back to
the cross product later in the course.
Tendency to Align

B
q

B cosq

A
Notice from the definition of the dot product that Bcosq is
simply the component of B in the same direction of vector
A
Dot Product

B
y
30o
40o
or,
 

A  Ax  Ay  5 x  4.2 y (mag  6.53)


B  Bx  B y  3 x  8.2 y (mag  8.73)

A
Two methods:
x
 
A  B  A B cos q where q is angle between ta ils
 
A  B  6.53(8.73) cos 30  49.4
 
A  B  Ax Bx  Ay By  5(3)  4.2(8.2)
 
A  B  15  34.4  49.4
2D Kinematics
Position and Displacement Vectors
y
A
r
rA

r
rAB
B
r r
r
rB  rA  rAB
x


Object follows dashed line.
Position vectors start at origin,
extend to location of object.
Displacement vectors connect
two position vectors.
r
r r
rAB  rB  rA
r r r
r  rf  ri
Extend the 1D Definitions
r
r
v avg  r t
r
r
aavg  v t
If I choose to express as an ordered pair, then

rX

r
r
Y

v avg  v avg, X , v avg, Y   t ,
t




Horizontal and Vertical Motion
Recall Uniform Motion
Equal distance covered in equal time intervals
vave 
Vertical Motion (Freely falling object)
Equal increments of speed gained in equal
increments of time
Distance increases in each time interval
a
v f  vi
t
Falling from rest:
v yf2  v yi2  2ay
y f  yi  viyt  1 2 at 2
d f  di
t
Resolution into Components
Vertical
Motion
Is
Uniform
Acceleration
Horizontal Motion is Uniform Motion
Notice that the Horizontal motion is in no way affected by the Vertical motion.
Cart through tunnel
Independence of horizontal motion with vertical motion
(run at different speeds)
Plan of Attack
When we analyze projectile motion we can greatly simplify the problem by separating the
problem into its horizontal and vertical components.
Once separated we can apply the equations for straight line motion as we did in the
previous problems.
Horizontally (uniform Motion)
vave 
d f  di
t
Vertically (uniform Acceleration)
a
v f  vi
t
And the derived equations..
v yf2  v yi2  2ay
y f  yi  viyt  1 2 at 2
Example
A bus load of terrorists traveling at 25 m/s drive off
of the edge of a cliff 50 m high. Where do they
land?
25 m/s
Initial Conditions
vxi = 25 m/s
Horizontally
vavg,X = (xf-xi)/(t-ti)… xf = 25*t
xf = 25 *3.19
= 79.8 m
Vertically
vyi = 0 m/s
a = (vyf-vyi)/(t)…. 9.8 = vyf/t
a x = 0 m / s2
vyf2 = vyi2 + 2ayf ….
Vyf2 = + 2(9.8)50 … Vyf = 31.3 m/s
ay = 9.8 m/s2
yf = yi +vyit + 1/2at2 ….
50 = 1/2(9.8)t2 … t = 3.19 s
ti = 0 s
xi = 0 m
yi = 0 m
yf = 50 m
79.8 m
Sample Problem Solving Technique
60°
Example 2 - An object is projected at an angle of 60o to
the horizontal at a speed of 40 m/sec. How much time
does it spend in the air, where does it hit, and what is the
maximum height that it obtains?
Resolve the motion into components
The components of the velocity are
found by drawing (or Trig)
vyi = 34.6 m/sec
Look to find the time in the air
Vertical Motion
The object is thrown upward with a speed of 34.6
m/sec.
The time that it takes for the object’s speed to
become zero is given by:
a = (vf – vi)/ t
where t = (0 - 34.6)/-9.8 = 3.53 sec
It takes the same amount of time to fall back
down, so the total time in the air is 7.06 sec.
vx = 20 m/sec
Horizontal Motion
Horizontally the object moves with
Uniform motion and is governed by the
equation:
vavg = x / t
x = 20 (7.06) = 141.2 m
Survivors?
An aircraft is to drop a raft
and supplies to survivors
of a ship wreck.
Vertically (uniform Acceleration)
a = (vf-vi)/(tf-ti)
And the derived equations..
vyf2 = vyi2 + 2ay
yf = yi +vyit + 1/2at2
100 m/s
Horizontally (uniform Motion)
vavg = (xf-xi)/(tf-ti)
500 m
1000 m
Solution
yf = yi +vyit + 1/2at2
During the time of the fall the
supplies continue to move
forward at 100 m/s, so
0 = 500 + 0 +1/2(-9.8)t2
vavg = (xf-xi)/t
t = 10.10 s
100 = (xf – 0)/10.10
Time in the air:
xf = 1010 m
100 m/s
500 m
1000 m
(they miss being crushed by the
supplies by only 10m or ~ 30ft)
Projectile Motion
An object is thrown with an initial velocity of 25 m/s at an angle of
60o to the horizontal. Where does it hit?
Resolve the vector:
Time in the air:
Distance Traveled:
vx = 12.5 m/s
a = (vyf – vyi)/t
vx = (xf – xi)/t
vyi = 21.7
-9.8 = (-21.7 – 21.7)/t
12.5 = xf/4.43
t = 4.43 s
xf = 55.4 m
Projectile Motion
An object is thrown with an initial velocity of 25 m/s at an angle of
30o to the horizontal. Where does it hit?
Resolve the vector:
Time in the air:
Distance Traveled:
vx = 21.7 m/s
a = (vyf – vyi)/t
vx = (xf – xi)/t
vyi = 12.5
-9.8 = (-12.5 – 12.5)/t
21.7 = xf/2.55
t = 2.55 s
xf = 55.4 m
In the absence of air resistance projectiles launched at
complimentary angles have the same range.
Castle Siege
10 m
A Trebuchet located 60 m from
the walls of a castle launches
projectiles at 30 m/s at an angle
of 30o.
If the walls of the castle are 10 tall, do the
projectiles make it over the walls?
30 m
Resolve the vector:
vx = 26 m/s
vyi = 15 m/s
What height does the
How much time until
projectile have at that
it gets to the wall’s
time?
position?
vx = (xf – xi)/t
yf = yi +vyit + 1/2at2
26 = 60/t
yf = 0 +15(2.3) + 1/2 (-9.8)(2.3)2
t = 2.3sec
yf = 8.6 m Doesn’t make it!
CAN it make it?
Will the cannonball ever reach a height of 10 m?
We can determine this in two different
ways:
•Find out the time when the height of the
projectile is 10m, or
•Determine if the projectile still has a
vertical velocity when it reaches 10m
yf = yi +vyit + 1/2at2
10 = 0 + 15t +1/2(-9.8)t2
Quadratic Equation
 b  b 2  4ac
x
2a
Or….
vf2 = vi2 + 2ad
vf2 = 152 + 2(-9.8)10
v = +/- 5.39
vf = vi +at
5.39 = 15 + (-9.8)t
t1 = 0.981 s
-5.39 = 15 + (-9.8)t
t2 = 2.08 s
gives two roots:
t1 = 0.981 sec
x1 = vxt1 = 26(0.981)= 25.5m
t2 = 2.08 sec
x2 = vxt2 = 26(2.08)= 54.1m
A Demo
Magnet of Life
We’ve Shot Kenney
Magnet Releases when
ball is fired
(Kenney’s a goner anyway)
Where do we aim?
Shoot Kenney
Easy if there is no gravity
t=0.25 sec
V = 8 m/sec
If Kenny is 8 m away.
The ball takes 1 sec
to hit him.
Shoot Kenney
With Gravity?
t=0.25 sec
V = 8 m/sec
The Projectile arrives
at the same spot as
Kenny.
River
Problem
A boat travels directly across a river at a speed of 8 m/s.
The river flows at 4 m/s and is 40 m wide.
How long does it take the boat to travel across the river and
where does it land relative to its’ starting point?
8 m/s
40 m
4 m/s
River Problem Solution
@26.6o
8 m/s
8.9 m/s
4 m/s
One method (hard):
Join the velocity vectors to find the
velocity of the boat through the
water. (8.9 m/s @26.6o
downstream).
40 m
Much, much easier method:
The boat covers 8 horizontal meters
every second so it takes 5 sec to travel
across a 40 m wide river.
During that 5 sec, the boat goes
downstream 4 m each second so it
lands 20 m downstream
Determine the length of
the path that the boat
takes through the water.
(44.7 m @ 26.6o)
Time : v = d/t so
t = d/v = 44.7/8.9 = 5.0 s
The
Reverse
Problem
At what direction upstream must the boat point in order to
arrive directly across the river from its’ starting point?
4 m/s
40 m
If you know the trig….use it. If not,
Draw a horizontal line with the river
vector pointing to it.
Really a vector subtraction
problem.
But perhaps you recognize
this triangle (30/60/90)
We must find the direction
of the 8m/s so that when it
is added to the 4m/s the
resultant vector points in a
particular direction.
Draw the 8 m/s vector with its arrow
on the river vector so that 8 units falls
on the horizontal line
4 m/s
6.9 m/s
Speed of the boat
through the water