Vectors in Physics
Transcript of Vectors in Physics
Vectors in Physics
Scalars vs. Vectors
Some quantities in Physics are uniquely determined by one number
mass, temperature, speed, distance, etc.
These are called SCALARS
Scalars vs. Vectors
There are other quantities that require more than one number
velocity, acceleration, displacement, etc.
These are called VECTORS
Characteristics of VectorsA vector is represented as an arrow.
An arrow has length and direction.
So does a vector.
Visual representation:
Out of page:
Into page:
Vector Addition
A table in the room...
Vector Addition
O
A table in the room...A bug moves from point O...
Vector Addition
O
PA table in the room...A bug moves from point O...to point P on the table.
Vector Addition
O
PA table in the room...A bug moves from point O...to point P on the table.But, the table also moves
Vector Addition
O
PA table in the room...A bug moves from point O...to point P on the table.But, the table also moves
Vector Addition
O
P
S
A table in the room...A bug moves from point O...to point P on the table.But, the table also moves
The bug’s position in the room is then point S.
Vector Addition
O
P
S
A table in the room...A bug moves from point O...to point P on the table.But, the table also moves
The bug’s position in the room is then point S.What you observe if these two motions happen simultaneously
Vector Addition
O
P
S
�⇥OS =
�⇥OP +
�⇥PS
The vector OS is the vector OP plus the vector PS
Vector Addition Head-to-Tail Method
�⇥A
�⇥B
To add vectors A & B...
Vector Addition Head-to-Tail Method
�⇥A
�⇥B
To add vectors A & B...➡Translate one vector to the other, head-to-tail
➡Magnitude & direction do not change➡So, same vector
Vector Addition Head-to-Tail Method
�⇥A
�⇥B
�⇥C =
�⇥A +
�⇥B
To add vectors A & B...➡Translate one vector to the other, head-to-tail
➡Magnitude & direction do not change➡So, same vector
➡Draw resultant vector from tail of first to head of last
Vector Addition Head-to-Tail Method
�⇥A
�⇥B
�⇥C =
�⇥A +
�⇥B
To add vectors A & B...➡Translate one vector to the other, head-to-tail
➡Magnitude & direction do not change➡So, same vector
➡Draw resultant vector from tail of first to head of last
➡Result does not depend on order, so�⇥C =
�⇥B +
�⇥A
Vector Addition Parallelogram Method
�⇥A
�⇥B
To add vectors A & B...
Vector Addition Parallelogram Method
�⇥A
�⇥B
To add vectors A & B...➡Translate one vector to the other, tail-to-tail
➡Magnitude & direction do not change➡So, same vector
Vector Addition Parallelogram Method
�⇥A
�⇥B
To add vectors A & B...➡Translate one vector to the other, tail-to-tail
➡Magnitude & direction do not change➡So, same vector
➡Complete the parallelogram
Vector Addition Parallelogram Method
�⇥A
�⇥B
�⇥C =
�⇥A +
�⇥B
To add vectors A & B...➡Translate one vector to the other, tail-to-tail
➡Magnitude & direction do not change➡So, same vector
➡Complete the parallelogram➡Draw resultant vector from tails of vectors to
opposite corner of parallelogram
Vector Addition Parallelogram Method
�⇥A
�⇥B
�⇥C =
�⇥A +
�⇥B
To add vectors A & B...➡Translate one vector to the other, tail-to-tail
➡Magnitude & direction do not change➡So, same vector
➡Complete the parallelogram➡Draw resultant vector from tails of vectors to
opposite corner of parallelogram➡Easy to see that...
�⇥A +
�⇥B =
�⇥B +
�⇥A
Vector Addition Subtraction
�⇥A
�⇥A + (��⇥
A ) = 0
Vector Addition Subtraction
�⇥A ➡What is the meaning of ?
�⇥A + (��⇥
A ) = 0
��⇥A
Vector Addition Subtraction
�⇥A ➡What is the meaning of ?
➡If this is displacement, it is the vector required to return to the start.
�⇥A + (��⇥
A ) = 0
��⇥A
Vector Addition Subtraction
�⇥A ➡What is the meaning of ?
➡If this is displacement, it is the vector required to return to the start.
➡That would make the displacement zero.
�⇥A + (��⇥
A ) = 0
��⇥A
Vector Addition Subtraction
�⇥A ➡What is the meaning of ?
➡If this is displacement, it is the vector required to return to the start.
➡That would make the displacement zero.➡It is simply the same vector in the opposite
direction.
�⇥A + (��⇥
A ) = 0
��⇥A
Vector Addition Subtraction
�⇥A ➡What is the meaning of ?
➡If this is displacement, it is the vector required to return to the start.
➡That would make the displacement zero.➡It is simply the same vector in the opposite
direction.
�⇥A + (��⇥
A ) = 0
��⇥A
��⇥A
Vector Addition Subtraction
�⇥A
�⇥B
To determine
�⇥B
�⇥A ��⇥
B =�⇥C
Vector Addition Subtraction
�⇥A
�⇥B
To determine ➡Rewrite as
�⇥B
�⇥A + (��⇥
B ) =�⇥C
�⇥A ��⇥
B =�⇥C
Vector Addition Subtraction
�⇥A
�⇥B
To determine ➡Rewrite as➡Use head-to-tail method with &
��⇥B
��⇥B
�⇥A
�⇥A + (��⇥
B ) =�⇥C
�⇥A ��⇥
B =�⇥C
Vector Addition Subtraction
�⇥A
�⇥B
To determine ➡Rewrite as➡Use head-to-tail method with &➡Draw resultant vector from tail of first
to head of last
��⇥B
�⇥C
��⇥B
�⇥A
�⇥A + (��⇥
B ) =�⇥C
�⇥A ��⇥
B =�⇥C
Vector Addition Subtraction
�⇥A
�⇥B
To determine ➡Rewrite as➡Use head-to-tail method with &➡Draw resultant vector from tail of first
to head of last➡Order does matter
��⇥B
�⇥C
��⇥B
�⇥A
�⇥A + (��⇥
B ) =�⇥C
�⇥A ��⇥
B =�⇥C
�⇥A ��⇥
B ⇥= �⇥B ��⇥
A
Right-Triangle Math
a
bc Pythagorean Theorem
c2 = a2 + b2✓
SOH CAH TOA
sin ✓ = opposite
hypotenuse
= bc
cos ✓ =
adjacent
hypotenuse
=
ac
tan ✓ = opposite
adjacent
= ba
Right-Triangle Math
a
bc Pythagorean Theorem
c2 = a2 + b2✓
SOH CAH TOA
sin ✓ = opposite
hypotenuse
= bc
cos ✓ =
adjacent
hypotenuse
=
ac
tan ✓ = opposite
adjacent
= ba
a = c cos ✓
b = c sin ✓
b = a tan ✓a = b
tan ✓
c = acos ✓
c = bsin ✓
Right-Triangle Math
a
bc Pythagorean Theorem
c2 = a2 + b2✓
SOH CAH TOA
sin ✓ = opposite
hypotenuse
= bc
cos ✓ =
adjacent
hypotenuse
=
ac
tan ✓ = opposite
adjacent
= ba
a = c cos ✓
b = c sin ✓
b = a tan ✓a = b
tan ✓
c = acos ✓
c = bsin ✓
✓ = sin�1�bc
�
✓ = cos
�1�ac
�
✓ = tan�1�ba
�
Right-Triangle MathPythagorean Theorem
c2 = a2 + b2✓
SOH CAH TOA
sin ✓ = opposite
hypotenuse
= bc
cos ✓ =
adjacent
hypotenuse
=
ac
tan ✓ = opposite
adjacent
= ba
~A
x
y
Right-Triangle MathPythagorean Theorem
c2 = a2 + b2✓
SOH CAH TOA
sin ✓ = opposite
hypotenuse
= bc
cos ✓ =
adjacent
hypotenuse
=
ac
tan ✓ = opposite
adjacent
= ba
~A
~Ax
x
y
Right-Triangle MathPythagorean Theorem
c2 = a2 + b2✓
SOH CAH TOA
sin ✓ = opposite
hypotenuse
= bc
cos ✓ =
adjacent
hypotenuse
=
ac
tan ✓ = opposite
adjacent
= ba
~A
~Ax
~Ay
x
y
Right-Triangle MathPythagorean Theorem
✓SOH CAH TOA
sin ✓ = opposite
hypotenuse
= bc
cos ✓ =
adjacent
hypotenuse
=
ac
tan ✓ = opposite
adjacent
= ba
~A
~Ax
~Ay | ~A|2 = | ~Ax
|2 + | ~Ay
|2
x
y
Right-Triangle MathPythagorean Theorem
✓SOH CAH TOA
cos ✓ =
adjacent
hypotenuse
=
ac
tan ✓ = opposite
adjacent
= ba
~A
~Ax
~Ay
sin ✓ = opposite
hypotenuse
= | ~Ay|| ~A|
| ~A|2 = | ~Ax
|2 + | ~Ay
|2
x
y
Right-Triangle MathPythagorean Theorem
✓SOH CAH TOA
tan ✓ = opposite
adjacent
= ba
~A
~Ax
~Ay
sin ✓ = opposite
hypotenuse
= | ~Ay|| ~A|
cos ✓ =
adjacent
hypotenuse
=
| ~Ax
|| ~A|
| ~A|2 = | ~Ax
|2 + | ~Ay
|2
x
y
Right-Triangle MathPythagorean Theorem
✓SOH CAH TOA
~A
~Ax
~Ay
sin ✓ = opposite
hypotenuse
= | ~Ay|| ~A|
cos ✓ =
adjacent
hypotenuse
=
| ~Ax
|| ~A|
tan ✓ = opposite
adjacent
= | ~Ay
|| ~A
x
|
| ~A|2 = | ~Ax
|2 + | ~Ay
|2
x
y
Vector Components In Two Dimensions
Given magnitude & directionAx = |A| cos �
Ay = |A| sin �
Given components|A| =
�A2
x + A2y
� = tan�1
�Ay
Ax
⇥
Vector Components In Two Dimensions
Given magnitude & directionAx = |A| cos �
Ay = |A| sin �
Given components|A| =
�A2
x + A2y
� = tan�1
�Ay
Ax
⇥Practice1. ~A = 328 km @ 13.2� N of E
2. ~v = 142 m/s @ 27
�N of W
3. ~r1 = 7 m @ 6�
4. ~r2 = 7 m @ � 6�
Vector Components In Two Dimensions
Given magnitude & directionAx = |A| cos �
Ay = |A| sin �
Given components|A| =
�A2
x + A2y
� = tan�1
�Ay
Ax
⇥Practice5. ~�r1 = 45.0 m N, ~�r2 = 7.5 m E6. ~�r1 = 6.0 m @ 0�,
~�r2 = 14.5 m @ 90�
Vector Components In Two Dimensions
Angle may be measured relative to the x or y axis. Be careful.
�
Find total displacement...Each pace = 0.75 m
Vector x Components y Components
A 0 3.75 mB 2.25 m 0C 2.12 m –2.12 mD 4.37 m 1.63 m
Vector Components In Two Dimensions
Find total displacement...Each pace = 0.75 m
Vector x Components y Components
A 0 3.75 mB 2.25 m 0C 2.12 m –2.12 mD 4.37 m 1.63 m
Vector Components In Two Dimensions
Find total displacement...Each pace = 0.75 m
Vector x Components y Components
A 0 3.75 mB 2.25 m 0C 2.12 m –2.12 mD 4.37 m 1.63 m
Vector Components In Two Dimensions
Find total displacement...Each pace = 0.75 m
Vector x Components y Components
A 0 3.75 mB 2.25 m 0C 2.12 m –2.12 mD 4.37 m 1.63 m
Vector Components In Two Dimensions
Find total displacement...Each pace = 0.75 m
Vector x Components y Components
A 0 3.75 mB 2.25 m 0C 2.12 m –2.12 mD 4.37 m 1.63 m
Vector Components In Two Dimensions
Find total displacement...Each pace = 0.75 m
Vector x Components y Components
A 0 3.75 mB 2.25 m 0C 2.12 m –2.12 mD 4.37 m 1.63 m
Vector Components In Two Dimensions
Find total displacement...Each pace = 0.75 m
Vector x Components y Components
A 0 3.75 mB 2.25 m 0C 2.12 m –2.12 mD 4.37 m 1.63 m
Vector Components In Two Dimensions
Find total displacement...Each pace = 0.75 m
Vector x Components y Components
A 0 3.75 mB 2.25 m 0C 2.12 m –2.12 mD 4.37 m 1.63 m
Vector Components In Two Dimensions
Find total displacement...Each pace = 0.75 m
Vector x Components y Components
A 0 3.75 mB 2.25 m 0C 2.12 m –2.12 mD 4.37 m 1.63 m
Vector Components In Two Dimensions
Find total displacement...Each pace = 0.75 m
Vector x Components y Components
A 0 3.75 mB 2.25 m 0C 2.12 m –2.12 mD 4.37 m 1.63 m
Vector Components In Two Dimensions
Find total displacement...Each pace = 0.75 m
Vector x Components y Components
A 0 3.75 mB 2.25 m 0C 2.12 m –2.12 mD 4.37 m 1.63 m
Vector Components In Two Dimensions
Find total displacement...Each pace = 0.75 m
Vector x Components y Components
A 0 3.75 mB 2.25 m 0C 2.12 m –2.12 mD 4.37 m 1.63 m
Vector Components In Two Dimensions
Find total displacement...Each pace = 0.75 m
Vector x Components y Components
A 0 3.75 mB 2.25 m 0C 2.12 m –2.12 mD 4.37 m 1.63 m
~Dx
= 4.37 m
Vector Components In Two Dimensions
Find total displacement...Each pace = 0.75 m
Vector x Components y Components
A 0 3.75 mB 2.25 m 0C 2.12 m –2.12 mD 4.37 m 1.63 m
~Dx
= 4.37 m
~Dx
Vector Components In Two Dimensions
Find total displacement...Each pace = 0.75 m
Vector x Components y Components
A 0 3.75 mB 2.25 m 0C 2.12 m –2.12 mD 4.37 m 1.63 m
~Dx
= 4.37 m ~Dy = 1.63 m
~Dx
Vector Components In Two Dimensions
Find total displacement...Each pace = 0.75 m
Vector x Components y Components
A 0 3.75 mB 2.25 m 0C 2.12 m –2.12 mD 4.37 m 1.63 m
~Dx
= 4.37 m ~Dy = 1.63 m
~Dy
~Dx
Vector Components In Two Dimensions
Find total displacement...Each pace = 0.75 m
|D| =�
(4.37)2 + (1.63)2 = 4.66 m
Vector x Components y Components
A 0 3.75 mB 2.25 m 0C 2.12 m –2.12 mD 4.37 m 1.63 m
~Dx
= 4.37 m ~Dy = 1.63 m
~Dy
~Dx
Vector Components In Two Dimensions
Find total displacement...Each pace = 0.75 m
|D| =�
(4.37)2 + (1.63)2 = 4.66 m
� = tan�1
�1.634.37
⇥= 20.5o
Vector x Components y Components
A 0 3.75 mB 2.25 m 0C 2.12 m –2.12 mD 4.37 m 1.63 m
~Dx
= 4.37 m ~Dy = 1.63 m
~Dy
~Dx
Vector Components In Two Dimensions
Vector Components In Two Dimensions
PracticeIn 1976, the Lockheed SR-71 Blackbird set the record speed for any airplane: 3.53 × 103 km/h. Suppose you observe this plane ascending at this speed. For 20.0 s, it flies at an angle of 15.0° above the horizontal (displacement, ∆r1, is 19.6 km), then for another 10.0 s, its angle of ascent is increased to 35.0° (∆r2 = 9.81 km). Calculate the plane’s total gain in altitude, its total horizontal displacement, and its resultant displacement.
Homework: Workbook Problem 3C, 1 & 5
Vector Components In Two Dimensions
PracticeIn 1976, the Lockheed SR-71 Blackbird set the record speed for any airplane: 3.53 × 103 km/h. Suppose you observe this plane ascending at this speed. For 20.0 s, it flies at an angle of 15.0° above the horizontal (displacement, ∆r1, is 19.6 km), then for another 10.0 s, its angle of ascent is increased to 35.0° (∆r2 = 9.81 km). Calculate the plane’s total gain in altitude, its total horizontal displacement, and its resultant displacement.
Vector Components In Two Dimensions
PracticeIn 1976, the Lockheed SR-71 Blackbird set the record speed for any airplane: 3.53 × 103 km/h. Suppose you observe this plane ascending at this speed. For 20.0 s, it flies at an angle of 15.0° above the horizontal (displacement, ∆r1, is 19.6 km), then for another 10.0 s, its angle of ascent is increased to 35.0° (∆r2 = 9.81 km). Calculate the plane’s total gain in altitude, its total horizontal displacement, and its resultant displacement.
~�r1
= 19.6 km @ 15.0�~�r
2
= 9.81 km @ 35.0�~�r
y
= ?, ~�rx
= ?
Vector Components In Two Dimensions
Practice~�r
1
= 19.6 km @ 15.0�~�r
2
= 9.81 km @ 35.0�~�r
y
= ?, ~�rx
= ?
x
y
~�r1
~�r2
Vector Components In Two Dimensions
Practice~�r
1
= 19.6 km @ 15.0�~�r
2
= 9.81 km @ 35.0�~�r
y
= ?, ~�rx
= ?
x
y
~�r1
~�r2
~�r1x
~�r1y
Vector Components In Two Dimensions
Practice~�r
1
= 19.6 km @ 15.0�~�r
2
= 9.81 km @ 35.0�~�r
y
= ?, ~�rx
= ?
x
y
~�r1
~�r2
~�r1x
~�r1y
~�r2y~�r2x
Vector Components In Two Dimensions
Practice~�r
1
= 19.6 km @ 15.0�~�r
2
= 9.81 km @ 35.0�~�r
y
= ?, ~�rx
= ?
x
y
~�r1
~�r2
~�r1x
~�r1y
~�r2y
~�r2x
Vector Components In Two Dimensions
Practice~�r
1
= 19.6 km @ 15.0�~�r
2
= 9.81 km @ 35.0�~�r
y
= ?, ~�rx
= ?
x
y
~�r1
~�r2
~�r1x
~�r1y
~�r2y
~�r2x
~�rx
Vector Components In Two Dimensions
Practice~�r
1
= 19.6 km @ 15.0�~�r
2
= 9.81 km @ 35.0�~�r
y
= ?, ~�rx
= ?
x
y
~�r1
~�r2
~�r1x
~�r1y
~�r2y
~�r2x
~�rx
~�ry
Vector Components In Two Dimensions
Practice~�r
1
= 19.6 km @ 15.0�~�r
2
= 9.81 km @ 35.0�~�r
y
= ?, ~�rx
= ?
x
y
~�r1
~�r2
~�r1x
~�r1y
~�r2y
~�r2x
~�rx
~�ry~�r
Vector Components In Two Dimensions
Practice~�r
1
= 19.6 km @ 15.0�~�r
2
= 9.81 km @ 35.0�~�r
y
= ?, ~�rx
= ?
x
y
~�r1
~�r2
~�r1x
~�r1y
~�r2y
~�r2x
~�rx
~�ry~�r
~�r1x =
~�r1 cos 15.0�
~�r1x = 19.6 km (cos 15.0�)~�r1x = 18.9 km
Vector Components In Two Dimensions
Practice~�r
1
= 19.6 km @ 15.0�~�r
2
= 9.81 km @ 35.0�~�r
y
= ?, ~�rx
= ?
x
y
~�r1
~�r2
~�r1x
~�r1y
~�r2y
~�r2x
~�rx
~�ry~�r
~�r1x = 18.9 km
Vector Components In Two Dimensions
Practice~�r
1
= 19.6 km @ 15.0�~�r
2
= 9.81 km @ 35.0�~�r
y
= ?, ~�rx
= ?
x
y
~�r1
~�r2
~�r1x
~�r1y
~�r2y
~�r2x
~�rx
~�ry~�r
~�r1x = 18.9 km~�r1y = ~�r1 sin 15.0�
~�r1y = 19.6 km (sin 15.0�)~�r1y = 5.07 km
Vector Components In Two Dimensions
Practice~�r
1
= 19.6 km @ 15.0�~�r
2
= 9.81 km @ 35.0�~�r
y
= ?, ~�rx
= ?
x
y
~�r1
~�r2
~�r1x
~�r1y
~�r2y
~�r2x
~�rx
~�ry~�r
~�r1x = 18.9 km~�r1y = 5.07 km
Vector Components In Two Dimensions
Practice~�r
1
= 19.6 km @ 15.0�~�r
2
= 9.81 km @ 35.0�~�r
y
= ?, ~�rx
= ?
x
y
~�r1
~�r2
~�r1x
~�r1y
~�r2y
~�r2x
~�rx
~�ry~�r
~�r1x = 18.9 km~�r1y = 5.07 km
~�r2x = 8.04 km
~�r2x =
~�r2 cos 35.0�
~�r2x = 9.81 km (cos 35.0�)
Vector Components In Two Dimensions
Practice~�r
1
= 19.6 km @ 15.0�~�r
2
= 9.81 km @ 35.0�~�r
y
= ?, ~�rx
= ?
x
y
~�r1
~�r2
~�r1x
~�r1y
~�r2y
~�r2x
~�rx
~�ry~�r
~�r1x = 18.9 km~�r1y = 5.07 km~�r2x = 8.04 km
Vector Components In Two Dimensions
Practice~�r
1
= 19.6 km @ 15.0�~�r
2
= 9.81 km @ 35.0�~�r
y
= ?, ~�rx
= ?
x
y
~�r1
~�r2
~�r1x
~�r1y
~�r2y
~�r2x
~�rx
~�ry~�r
~�r1x = 18.9 km~�r1y = 5.07 km~�r2x = 8.04 km~�r2y = ~�r2 sin 35.0�
~�r2y = 9.81 km (sin 35.0�)
~�r2y = 5.63 km
Vector Components In Two Dimensions
Practice~�r
1
= 19.6 km @ 15.0�~�r
2
= 9.81 km @ 35.0�~�r
y
= ?, ~�rx
= ?
x
y
~�r1
~�r2
~�r1x
~�r1y
~�r2y
~�r2x
~�rx
~�ry~�r
~�r1x = 18.9 km~�r1y = 5.07 km~�r2x = 8.04 km~�r2y = 5.63 km
Vector Components In Two Dimensions
Practice~�r
1
= 19.6 km @ 15.0�~�r
2
= 9.81 km @ 35.0�~�r
y
= ?, ~�rx
= ?
x
y
~�r1
~�r2
~�r1x
~�r1y
~�r2y
~�r2x
~�rx
~�ry~�r
~�r1x = 18.9 km~�r1y = 5.07 km~�r2x = 8.04 km~�r2y = 5.63 km
~�r = ~�rx
+ ~�ry
| ~�r| =q
(| ~�r1x|+ | ~�r2x|)2 + (| ~�r1y|+ | ~�r2y|)2
| ~�r| =p
(18.9 + 8.04)2 + (5.07 + 5.63)2
| ~�r| =p27.02 + 10.72
Vector Components In Two Dimensions
Practice~�r
1
= 19.6 km @ 15.0�~�r
2
= 9.81 km @ 35.0�~�r
y
= ?, ~�rx
= ?
x
y
~�r1
~�r2
~�r1x
~�r1y
~�r2y
~�r2x
~�rx
~�ry~�r
~�r1x = 18.9 km~�r1y = 5.07 km~�r2x = 8.04 km~�r2y = 5.63 km
~�r = ~�rx
+ ~�ry
| ~�r| =q
(| ~�r1x|+ | ~�r2x|)2 + (| ~�r1y|+ | ~�r2y|)2
| ~�r| =p
(18.9 + 8.04)2 + (5.07 + 5.63)2
| ~�r| =p27.02 + 10.72
| ~�r| = 29.0 km
Vector Components In Two Dimensions
Practice~�r
1
= 19.6 km @ 15.0�~�r
2
= 9.81 km @ 35.0�~�r
y
= ?, ~�rx
= ?
x
y
~�r1
~�r2
~�r1x
~�r1y
~�r2y
~�r2x
~�rx
~�ry~�r
~�r1x = 18.9 km~�r1y = 5.07 km~�r2x = 8.04 km~�r2y = 5.63 km
~�r = ~�rx
+ ~�ry
| ~�r| =q
(| ~�r1x|+ | ~�r2x|)2 + (| ~�r1y|+ | ~�r2y|)2
| ~�r| =p
(18.9 + 8.04)2 + (5.07 + 5.63)2
| ~�r| =p27.02 + 10.72
| ~�r| = 29.0 km
✓ = tan�1⇣
| ~�ry
|| ~�r
x
|
⌘
✓ = tan�1�10.727.0
�
Vector Components In Two Dimensions
Practice~�r
1
= 19.6 km @ 15.0�~�r
2
= 9.81 km @ 35.0�~�r
y
= ?, ~�rx
= ?
x
y
~�r1
~�r2
~�r1x
~�r1y
~�r2y
~�r2x
~�rx
~�ry~�r
~�r1x = 18.9 km~�r1y = 5.07 km~�r2x = 8.04 km~�r2y = 5.63 km
~�r = ~�rx
+ ~�ry
| ~�r| =q
(| ~�r1x|+ | ~�r2x|)2 + (| ~�r1y|+ | ~�r2y|)2
| ~�r| =p
(18.9 + 8.04)2 + (5.07 + 5.63)2
| ~�r| =p27.02 + 10.72
| ~�r| = 29.0 km
✓ = tan�1⇣
| ~�ry
|| ~�r
x
|
⌘
✓ = tan�1�10.727.0
�
✓ = 21.6�
Vector Components In Two Dimensions
Practice~�r
1
= 19.6 km @ 15.0�~�r
2
= 9.81 km @ 35.0�~�r
y
= ?, ~�rx
= ?
x
y
~�r1
~�r2
~�r1x
~�r1y
~�r2y
~�r2x
~�rx
~�ry~�r
~�r1x = 18.9 km~�r1y = 5.07 km~�r2x = 8.04 km~�r2y = 5.63 km
~�r = ~�rx
+ ~�ry
| ~�r| =q
(| ~�r1x|+ | ~�r2x|)2 + (| ~�r1y|+ | ~�r2y|)2
| ~�r| =p
(18.9 + 8.04)2 + (5.07 + 5.63)2
| ~�r| =p27.02 + 10.72
| ~�r| = 29.0 km
✓ = tan�1⇣
| ~�ry
|| ~�r
x
|
⌘
✓ = tan�1�10.727.0
�
✓ = 21.6�
~�r = 29.0 km @ 21.6�
Position Vector
position vector = �⇥rSI unit: meter, m
Position:
~ry
~rx
~r = ~rx
+ ~ry
Displacement Vector
SI unit: meter, m��⇥r = �⇥r ��⇥r 0
�⇥r = �⇥r 0 + ��⇥r
Displacement:
Written as a sum:
Velocity Vectors
�⇥v av =��⇥r�t
SI unit: meter per second, m/s
Average velocity:
SI unit: meter per second, m/s
Instantaneous velocity:�⇥v = lim
�t�0
��⇥r�t
Acceleration VectorsAverage acceleration:
Instantaneous acceleration:
�⇥a av =��⇥v�t
�⇥a = lim�t�0
��⇥v�t
SI unit: meter per second, m/s2
SI unit: meter per second, m/s2
Acceleration Vectors
Average acceleration:➡Direction is the same as change in velocity
Relative Motion
Two objects in motion relative to each other, and to a third
Object one moves relative to object two
Object two moves relative to object three
These two relationships can be used to determine object one’s motion relative to object three
Relative MotionExample:
O
P ➡Bug’s motion relative to table
Relative MotionExample:
O
P
S
➡Bug’s motion relative to table➡Table’s motion relative to room
Relative MotionExample:
O
P
S
➡Bug’s motion relative to table➡Table’s motion relative to room➡Bug’s motion relative to room
Relative MotionExample:
O
P
S
➡Bug’s motion relative to table➡Table’s motion relative to room➡Bug’s motion relative to room
For displacement, we would write:
Relative MotionExample:
O
P
S
➡Bug’s motion relative to table➡Table’s motion relative to room➡Bug’s motion relative to room
For displacement, we would write:�⇥rbt � Bug’s displacement relative to Table
�⇥rbt
Relative MotionExample:
O
P
S
➡Bug’s motion relative to table➡Table’s motion relative to room➡Bug’s motion relative to room
For displacement, we would write:�⇥rbt � Bug’s displacement relative to Table�⇥rtr � Table’s displacement relative to Room
�⇥rbt
�⇥rtr
Relative MotionExample:
O
P
S
➡Bug’s motion relative to table➡Table’s motion relative to room➡Bug’s motion relative to room
For displacement, we would write:�⇥rbt � Bug’s displacement relative to Table�⇥rtr � Table’s displacement relative to Room�⇥rbr � Bug’s displacement relative to Room
�⇥rbt
�⇥rtr
�⇥rbr
Relative MotionExample:
O
P
S
➡Bug’s motion relative to table➡Table’s motion relative to room➡Bug’s motion relative to room
For displacement, we would write:�⇥rbt � Bug’s displacement relative to Table�⇥rtr � Table’s displacement relative to Room�⇥rbr � Bug’s displacement relative to Room
�⇥rbr = �⇥rbt + �⇥rtr
�⇥rbt
�⇥rtr
�⇥rbr
Relative MotionAnother ExampleSuppose that you are standing on the ground as a train goes by at 15.0 m/s, as shown in the figure. Inside the train, a free-riding passenger is walking in the forward direction at 1.2 m/s relative to the train. How fast is the passenger moving relative to you?
~vpg = ~vpt + ~vtg~vpg = +1.2 m/s + 15.0 m/s
~vpt = +1.2 m/s~vtg = 15.0 m/s~vpg = ? ~vpg = +16.2 m/s
Relative MotionAnd AnotherWhat if the passenger had been walking with the same speed, but toward the back of the train?
~vpg = ~vpt + ~vtg
~vpt = �1.2 m/s~vtg = 15.0 m/s~vpg = ?
~vpg = �1.2 m/s + 15.0 m/s
~vpg = +13.8 m/s
Relative MotionPracticeSuppose the passenger in the figure is climbing a vertical ladder with a speed of 0.20 m/s, and the train is slowly coasting forward at 0.70 m/s. Find the speed and direction of the passenger relative to the ground.
Steps1. Draw diagram2. Find resultant (both magnitude & direction)