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Vectors and Two Dimensional Motion
Unit 2
Lesson 1 : Some Properties of Vectors
Adding Vectors
A
BR = A + B
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
RB
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.
20.0 km
35.0 km
R
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)
AB
-B
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
A
Ax
Ay
A = Ax + Ay
Ax = A cos
Ay = A sin
Signs of the Components Ax and Ay
Ax positive
Ay positive
Ax positive
Ay negative
Ax negative
Ay positive
Ax 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
i (x direction)j
j (y direction)
kk (z direction)
Ax i = Axi x^ ^
Ay j = Ay j x^ ^
A = Ax i + Ay j^ ^
y
x
(x,y)
r
Position Vector (r)
r = x i + y j^ ^
Vector Addition Using Unit Vectors
R = A + B = (Ax i + Ay j ) + ( Bx i + By j )^ ^ ^ ^
R = (Ax + Bx ) i + ( Ay + By ) j^ ^
Rx = Ax + Bx
Ry = Ay + By
Given : A = Ax i + Ay j^ ^
B = Bx i + By j^ ^
AB
Since R = Rx2 + Ry
2
R = (Ax + Bx)2 + (Ay + By)2
(magnitude)
tan = Ry
Rx
tan = Ay + By
Ax + Bx
(direction)
Example 1
Find the magnitude and direction of the position vector below.
r = 10 i – 6 j^ ^x
y
Example 2
b) find the magnitude and direction of the resultant.
Given the vectors
A = -7 i + 4 j
B = 5 i + 9 j
^ ^
^ ^
a) find an expression for the resultant A + B in terms of unit vectors.
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
A
B
To what extent are these two vectors in the same direction ?
Answer : Dot Product
A
B
A cos
When vectors are parallel, dot product is a maximum.
When vectors are perpendicular, dot product is a minimum.
A . B = AB cos
A . B = (AxBx + AyBy)
A . A = (Ax2 + Ay
2) = 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 sin
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.
3. Pretend to place your right hand around that line so that your fingers sweep a
into b through the smaller angle between them.
2. Imagine a perpendicular line to their plane where they meet.
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 = a x b ?
^ ^ ^ ^
Lesson 4 : Projectile Motion
To describe motion in two dimensions precisely, we use the position vector, r.
r(t1)r(t2)
r
r = r(t2) – r(t1)
vav =r
t
v =r
tlim =
dr
dtt 0
v =dr
dt=
dx
dt+i
^ dy
dtj^
a =dv
dt=
d2r
dt2
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 unit-vector notation and magnitude-angle notation.b) Find its acceleration a at time t = 15s in unit-vector notation and magnitude- angle notation.
Analyzing Projectile Motion
In projectile motion, the horizontal motion and the vertical motion are independent of
each other. Neither motion affects the other.
vvy
vx
X-Direction Constant Velocity
Y-Direction Constant
Acceleration
Initial x and y Components
v i
vix
viy
vix = vi cos
viy = vi sin
Horizontal Motion Equations
Vertical Motion Equations
x = vix t
vx = vix
vy = viy - gt
y = ½ (vy + viy) t
y = viy t – ½ gt2
vy2 = viy
2 – 2 gy
Upward and toward right is +
ay = -g
Proof that Trajectory is a Parabola
x = vix t
t = xvix
y = viy t – ½ gt2
y = viy ( ) – ½ g ( )2xvix
xvix
y = viy
vix( )x + (-
g
2vix
)x2
(equation of a parabola)
Maximum Height of a Projectile
vy = viy - gt
0 = vi sin - gt (at peak)
t =vi sin
g(at peak)
y = viy t – ½ gt2
( )h = (vi sin)vi sin
g
vi sing
- ½ g
2
h =vi
2 sin2
2g
Horizontal Range of a Projectile
x = R = vix t
R = vi cos 2t (twice peak time)
t =vi sin
g(at peak)
R = vi cos 2vi sing
sin 2 = 2sincos(trig identity)
R =vi
2 sin 2
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 #1A 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.
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.
a) Calculate the total time from launch until the projectile hits the ground at point C.
Case I : The projectile follows the path shown by the curved line in the following diagram.
Example 7 : The Monkey GunProve 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.
xdart = (vi cos)(t)
ydart = (vi sin)(t) – ½ gt2
xmonkey = x
ymonkey = y – ½ gt2
When the two objects collide :
xmonkey = xdart and ymonkey = ydart
(vi cos)(t) = x
t = x
(vi cos)
ydart = (vi sin)(t) – ½ gt2 = ymonkey = y – ½ gt2
y = (vi sin)(t)
y = (vi sin) x
(vi cos)= x (tan)
Lesson 5 : Uniform Circular Motion
Object moves in a circular path with constant speed
Object is accelerating because velocity vector changes
v
v
Centripetal Acceleration
The direction of v is toward the center of the circle
vf
-vi
v = vf - vi
v
vivf
v
Since the magnitude of the velocity is constant, the acceleration vector can only
have a component perpendicular to the path.
a
v
vr
r=
(similar triangles)
v =v r
r
a = v
t
a =
vr
t
r
ac = v2
r
centripetal acceleration
(center-seeking)
Speed in Uniform Circular Motion
vPeriod (T)
time required for one complete revolution
T =2r
v
v =2r
T
Lesson 6 : Tangential and Radial Acceleration
Velocity changing in direction and magnitude
at = tangential acceleration (changes speed)
ar = radial acceleration (changes direction)
at =dv
dt
ar =v2
r
perpendicular components
of a ar
ata
a = (ar)2 + (at )2 (magnitude of a)
a =dv
dt-
^ v2
rr
(total acceleration)
a = at + ar
Example 1The 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 ?