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...


�⇥A

�⇥B

To add vectors A & B...➡Translate one vector to the other, head-to-tail

➡Magnitude & direction do not change➡So, same vector


�⇥A

�⇥B

�⇥C =

�⇥A +

�⇥B



➡Draw resultant vector from tail of first to head of last


�⇥A

�⇥B

�⇥C =

�⇥A +

�⇥B



➡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...


�⇥A

�⇥B

To add vectors A & B...➡Translate one vector to the other, tail-to-tail



�⇥A

�⇥B



➡Complete the parallelogram


�⇥A

�⇥B

�⇥C =

�⇥A +

�⇥B



➡Complete the parallelogram➡Draw resultant vector from tails of vectors to

opposite corner of parallelogram


�⇥A

�⇥B

�⇥C =

�⇥A +

�⇥B



➡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


�⇥A ➡What is the meaning of ?

�⇥A + (��⇥

A ) = 0

��⇥A



➡If this is displacement, it is the vector required to return to the start.

�⇥A + (��⇥

A ) = 0

��⇥A




➡That would make the displacement zero.

�⇥A + (��⇥

A ) = 0

��⇥A




➡That would make the displacement zero.➡It is simply the same vector in the opposite

direction.

�⇥A + (��⇥

A ) = 0

��⇥A




➡That would make the displacement zero.➡It is simply the same vector in the opposite

direction.

�⇥A + (��⇥

A ) = 0

��⇥A

��⇥A


�⇥A

�⇥B

To determine

�⇥B

�⇥A ��⇥

B =�⇥C


�⇥A

�⇥B

To determine ➡Rewrite as

�⇥B

�⇥A + (��⇥

B ) =�⇥C

�⇥A ��⇥

B =�⇥C


�⇥A

�⇥B

To determine ➡Rewrite as➡Use head-to-tail method with &

��⇥B

��⇥B

�⇥A

�⇥A + (��⇥

B ) =�⇥C

�⇥A ��⇥

B =�⇥C


�⇥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


�⇥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


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


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


c2 = a2 + b2✓

SOH CAH TOA

sin ✓ = opposite

hypotenuse

= bc

cos ✓ =

adjacent

hypotenuse

=

ac

tan ✓ = opposite

adjacent

= ba

~A

~Ax

x

y


c2 = a2 + b2✓

SOH CAH TOA

sin ✓ = opposite

hypotenuse

= bc

cos ✓ =

adjacent

hypotenuse

=

ac

tan ✓ = opposite

adjacent

= ba

~A

~Ax

~Ay

x

y


✓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


✓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


✓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


✓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

⇥



Ay = |A| sin �


�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�



Ay = |A| sin �


�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�


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




A 0 3.75 mB 2.25 m 0C 2.12 m –2.12 mD 4.37 m 1.63 m




A 0 3.75 mB 2.25 m 0C 2.12 m –2.12 mD 4.37 m 1.63 m




A 0 3.75 mB 2.25 m 0C 2.12 m –2.12 mD 4.37 m 1.63 m




A 0 3.75 mB 2.25 m 0C 2.12 m –2.12 mD 4.37 m 1.63 m




A 0 3.75 mB 2.25 m 0C 2.12 m –2.12 mD 4.37 m 1.63 m




A 0 3.75 mB 2.25 m 0C 2.12 m –2.12 mD 4.37 m 1.63 m




A 0 3.75 mB 2.25 m 0C 2.12 m –2.12 mD 4.37 m 1.63 m




A 0 3.75 mB 2.25 m 0C 2.12 m –2.12 mD 4.37 m 1.63 m




A 0 3.75 mB 2.25 m 0C 2.12 m –2.12 mD 4.37 m 1.63 m




A 0 3.75 mB 2.25 m 0C 2.12 m –2.12 mD 4.37 m 1.63 m




A 0 3.75 mB 2.25 m 0C 2.12 m –2.12 mD 4.37 m 1.63 m




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




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




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




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



|D| =�

(4.37)2 + (1.63)2 = 4.66 m


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



|D| =�

(4.37)2 + (1.63)2 = 4.66 m

� = tan�1

�1.634.37

⇥= 20.5o


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



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





~�r1

= 19.6 km @ 15.0�~�r

2

= 9.81 km @ 35.0�~�r

y

= ?, ~�rx

= ?


Practice~�r

1

= 19.6 km @ 15.0�~�r

2

= 9.81 km @ 35.0�~�r

y

= ?, ~�rx

= ?

x

y

~�r1

~�r2


Practice~�r

1

= 19.6 km @ 15.0�~�r

2

= 9.81 km @ 35.0�~�r

y

= ?, ~�rx

= ?

x

y

~�r1

~�r2

~�r1x

~�r1y


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


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


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


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


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


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


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


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


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


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�)


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


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


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


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


~�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


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


~�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


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


~�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

�


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


~�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�


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


~�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


O

P

S

➡Bug’s motion relative to table➡Table’s motion relative to room


O

P

S

➡Bug’s motion relative to table➡Table’s motion relative to room➡Bug’s motion relative to room


O

P

S


For displacement, we would write:


O

P

S


For displacement, we would write:�⇥rbt � Bug’s displacement relative to Table

�⇥rbt


O

P

S


For displacement, we would write:�⇥rbt � Bug’s displacement relative to Table�⇥rtr � Table’s displacement relative to Room

�⇥rbt

�⇥rtr


O

P

S


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


O

P

S


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)

Date post:	09-Jan-2022
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Vectors in Physics

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