Vectors and Two Dimensional
Motion
Unit 2
Lesson 1 : Some Properties of Vectors
Adding Vectors
B
A
Resultant (R) is drawn from the tail of
the first vector to the tip of the last
vector
Commutative Law of Addition
When two vectors are added, the sum is
independent of the order of the addition.
A+B=B+A
A
B
R
B
A
Example 1
A car travels 20.0 km due north and then
35.0 km in a direction 60.0o west of north.
Find the magnitude and direction of the
car’s resultant displacement.
R
20.0 km
Negative of a Vector
The vector that when added to A
gives zero for the vector sum.
A + (-A) = 0
A
-A
A and –A have the same magnitude but
point in opposite directions
Subtracting Vectors
We define the operation A – B as
vector –B added to vector A.
A – B = A + (-B)
A
C=A-B
Multiplying a Vector by a Scalar
When vector A is multiplied by a positive
scalar quantity m, then the product mA is a
vector with the same direction of A and
magnitude mA.
When vector A is multiplied by a negative
scalar quantity -m, then the product -mA is
a vector directed opposite A and
magnitude mA.
Lesson 2 : Components of a Vector and
Unit Vectors
Ay
A
q
Ax
A = Ax + Ay
Ax = A cosq
Ay = A sinq
Signs of the Components Ax and Ay
Ax negative
Ax positive
Ay positive
Ay positive
Ax negative
Ax positive
Ay negative
Ay negative
Unit Vectors
A unit vector is a dimensionless vector
having a magnitude of exactly 1.
Units vectors specify a given
direction in space.
^i (x direction)
^j
^i
^j (y direction)
^
k (z direction)
^
k
Ax
Ay
x
x
^
^
^
^
i = Axi
j = Ay j
^
^
A = Ax i + Ay j
Position Vector (r)
y
(x,y)
r
^
^
r=xi+yj
x
Vector Addition Using Unit Vectors
Given :
A
^
^
^
^
^
^
A = Ax i + Ay j
B
B = Bx i + B y j
^
^
R = A + B = (Ax i + Ay j ) + ( Bx i + By j )
^
^
R = (Ax + Bx ) i + ( Ay + By ) j
R x = A x + Bx
Ry = Ay + By
Since R =
R=
Rx2 + Ry2
(Ax + Bx)2 + (Ay + By)2
(magnitude)
tan q =
Ry
Rx
tan q =
Ay + By
Ax + B x
(direction)
Example 1
Find the magnitude and direction of
the position vector below.
y
x
^
^
r = 10 i – 6 j
Example 2
Given the vectors
^
^
A = -7 i + 4 j
^
^
B=5i+9j
a) find an expression for the resultant
A + B in terms of unit vectors.
b) find the magnitude and direction
of the resultant.
Example 3
A hiker begins a trip by first walking 25.0 km
southeast from her car. She stops and sets up
her tent for the night. On the second day, she
walks 40.0 km in a direction 60.0o north of east.
a) Determine the components of the hiker’s
displacement for each day.
b) Determine the components of the hiker’s
resultant displacement (R) for the trip.
c) Find an expression for R in terms of unit
vectors.
Lesson 3 : Vector Multiplication
Vector x Vector
Dot Product
(scalar product)
Cross Product
(vector product)
X
Dot Product
To what extent are these two
vectors in the same direction ?
A
q
B
Answer : Dot Product
A . B = AB cosq
A
q
B
When vectors are parallel, dot product is a maximum.
When vectors are perpendicular, dot product is a
minimum.
A . B = (AxBx + AyBy)
A . A = (Ax2 + Ay2) = A2
Example 1
Find the angle between the two vectors
^
^
^
^
A = -7 i + 4 j
B = -2 i + 9 j
Example 2
Two vectors r and s lie in the x-y plane.
Their magnitudes are 4.50 and 7.30,
respectively, and their directions are 320o
and 85.0o, respectively, as measured
counterclockwise from the +x axis. What
is the value of r . s ?
Example 3
^
^
Find the component of A = 5 i + 6 j
^
^
that lies along the vector B = 4 i – 8 j.
Cross Product
The vector product a x b produces a
third vector c whose magnitude is
C = AB sinq
The cross product is maximum when vectors are perpendicular.
The cross product is minimum (0) when vectors are parallel.
Direction of the Cross Product
The direction of c is perpendicular to
the plane that contains a and b.
Right-Hand Rule
1. Place vectors a and b tail-to-tail.
2. Imagine a perpendicular line to their
plane where they meet.
3. Pretend to place your right hand around
that line so that your fingers sweep a
into b through the smaller angle
between them.
4. Your outstretched thumb points in the
direction of c.
Order of Cross Product is Important
Commutative law does not apply to a
vector product.
A x B = -B x A
In unit-vector notation :
^
^
^
^
^
^
A x B = (Axi + Ayj + Azk) x (Bxi + Byj + Bzk)
Example 4
Vector A lies in the x-y plane, has a
magnitude of 18 units, and points in a
direction 250o from the + x axis. Vector B
has a magnitude of 12 units and points
along the +z axis. What is the vector
product c = a x b ?
Example 5
^
^
^
^
If A = 3 i – 4 j and B = -2 i + 3 k, what is
c=axb?
Lesson 4 : Projectile Motion
To describe motion in two dimensions
precisely, we use the position vector, r.
Dr
r(t1)
Dr = r(t2) – r(t1)
r(t2)
Dr
Dt
vav =
v = lim
Dt  0
v=
dr
dt
a=
=
Dr
Dt
=
^
i +
dx
dt
dv
dt
dr
dt
=
d2r
dt2
dy
dt
^
j
Example 1
An object is described by the position vector
^
^
r(t) = (3t3 - 4t) i + (1 – ½ t2) j
Find its velocity and acceleration for
arbitrary times.
Example 2
A rabbit runs across a parking lot. The
coordinates of the rabbit’s position as
functions of time t are given by
x = -0.31t2 + 7.2t + 28
y = 0.22t2 – 9.1t + 30
a) Find its velocity v at time t = 15s in unitvector notation and magnitude-angle
notation.
b) Find its acceleration a at time t = 15s in
unit-vector notation and magnitudeangle notation.
Analyzing Projectile Motion
vy
v
vx
In projectile motion, the horizontal motion
and the vertical motion are independent of
each other. Neither motion affects the other.
X-Direction
Constant
Velocity
Y-Direction
Constant
Acceleration
Initial x and y Components
q
viy
vix
vix = vi cosq
viy = vi sinq
Horizontal Motion
Equations
Vertical Motion
Equations
vy = viy - gt
vx = vix
Dy = ½ (vy + viy) t
Dx = vix t
Dy = viy t – ½ gt2
vy2 = viy2 – 2 gDy
Upward and toward right is +
ay = -g
Proof that Trajectory is a Parabola
Dx = vix t
t=
Dx
vix
Dy = viy t – ½ gt2
Dy = viy ( Dx ) – ½ g ( Dx )2
vix
vix
viy
g
y= (
x + (x2
)
)
vix
2vix
(equation of a parabola)
Maximum Height of a Projectile
vy = viy - gt
0 = vi sinq - gt
t=
(at peak)
vi sinq
g
(at peak)
Dy = viy t – ½ gt2
vi sinq
-½g
h = (vi sinq)
g
h=
(
vi2 sin2q
2g
vi sinq
g
2
)
Horizontal Range of a Projectile
Dx = R = vix t
R = vi cosq 2t (twice peak time)
t=
vi sinq
g
(at peak)
2vi sinq
g
R = vi cosq
sin 2q = 2sinqcosq (trig identity)
R=
vi2 sin 2q
g
Example 3
A ball rolls off a table 1.0 m high with a
speed of 4 m/s. How far from the base
of the table does it land ?
Example 4
An arrow is shot from a castle wall 10. m high.
It leaves the bow with a speed of 40. m/s
directed 37o above the horizontal.
a) Find the initial velocity components.
b) Find the maximum height of the arrow.
c) Where does the arrow land ?
d) How fast is the arrow moving just
before impact ?
Example 5
A stone is thrown from the top of a building
upward at an angle of 30o to the horizontal
with an initial speed of 20.0 m/s.
a) If the building is 45.0 m high, how long
does it take the stone to reach the
ground ?
b) What is the speed of the stone just
before it strikes the ground ?
Example 6 : 1985 #1
A projectile is launched from the top of a cliff above level ground. At launch
the projectile is 35 m above the base of the cliff and has a velocity of 50 m/s at
an angle of 37o with the horizontal. Air resistance is negligible. Consider the
following two cases and use g = 10 m/s2, sin 37o = 0.60, and cos 37o = 0.80.
Case I : The projectile follows the path shown by the curved line in the following
diagram.
a) Calculate the total time from launch until the projectile hits the ground at point C.
b) Calculate the horizontal distance R that the projectile travels before it hits the ground.
c) Calculate the speed of the projectile at points A, B, and C.
Example 7 : The Monkey Gun
Prove that the monkey will hit the dart if the
monkey lets go of the branch (free-fall starting
from rest) at the instant the dart leaves the gun.
xmonkey = Dx
xdart = (vi cosq)(t)
ydart = (vi sinq)(t) – ½ gt2
ymonkey = Dy – ½ gt2
When the two objects collide :
xmonkey = xdart and ymonkey = ydart
(vi cosq)(t) = Dx
Dx
t=
(vi cosq)
ydart = (vi sinq)(t) – ½ gt2 = ymonkey = Dy – ½ gt2
Dy = (vi sinq)(t)
Dy = (vi sinq)
Dx
= Dx (tanq)
(vi cosq)
Lesson 5 : Uniform Circular Motion
v
v
Object moves in a circular path with constant speed
Object is accelerating because velocity vector changes
Centripetal Acceleration
The direction of Dv is toward the center
of the circle
vi
Dv = vf - vi
vf
-vi
Dv
vf
v
a
Since the magnitude of the velocity is
constant, the acceleration vector can only
have a component perpendicular to the path.
Dr
r
Dv
=
v
(similar triangles)
Dv =
v Dr
r
a=
Dv
Dt
a=
ac =
vDr
r
Dt
v2
r
centripetal
acceleration
(center-seeking)
Speed in Uniform Circular Motion
v
Period (T)
time required for one
complete revolution
T=
2pr
v
v=
2pr
T
Lesson 6 : Tangential and Radial Acceleration
Velocity changing in direction and magnitude
at = tangential acceleration (changes speed)
ar = radial acceleration (changes direction)
at =
perpendicular
components
of a
ar =
dv
dt
v2
r
(ar)2 + (at )2
a=
a
ar
(magnitude of a)
a = at + ar
a=
dv
dt
^
q -
at
v2
r
(total acceleration)
^
r
Example 1
The diagram below represents the total acceleration of a
particle moving clockwise in a circle of radius 2.50 m at a
certain instant of time. At this instant, find
a) the radial acceleration
b) the speed of the particle
c) the tangential acceleration
Example 2
A ball swings in a vertical circle at the end
of a rope 1.5 m long. When the ball is 36.9o
past the lowest point on its way up, its
^
^
total acceleration is (-22.5 i + 20.2 j) m/s2.
At that instant,
a) sketch a vector diagram showing the
components of its acceleration
b) determine the magnitude of its radial
acceleration
c) determine the speed and velocity of the
ball
Example 3
A boy whirls a stone on a horizontal circle
of radius 1.5 m and at height 2.0 m above
level ground. The string breaks, and the
stone flies off horizontally and strikes the
ground after traveling a horizontal
distance of 10. m. What is the magnitude
of the centripetal acceleration of the stone
during the circular motion ?