**Kinematics in Two Dimensions:**

Vectors!!!

Vectors!!!

This is big, the next ~7 chapters use this

Y

X

V

V

V

y

x

The idea is that every vector has

components (parts) that are in the

direction of the x and y axis.

O

V

y

V

y

V

x

V

x

V

V

V

V

Sin =

Tan =

= Sin

= Cos

O

O

O

O

V

x

V

Cos =

O

y

Vectors: By components

**If I walk 30 m north, and then 20 m east**

what is my displacement vector from the origin?

Magnitude = ?

Angle = ?

what is my displacement vector from the origin?

Magnitude = ?

Angle = ?

**magnitude 1st**

**a. 45 degrees north of east**

b. 56 degrees north of east

c. 34 degrees north of east

b. 56 degrees north of east

c. 34 degrees north of east

**A ridiculousness large river has a current of 3 m/s north. On a nearby lake a wind is able to give a sailboat a velocity of 5 m/s at an angle 30 north of east.**

What would the total velocity of this boat be with the wind and the current together?

What would the total velocity of this boat be with the wind and the current together?

**a) 4 m/s**

b) 5 m/s

c) 6 m/s

d) 7 m/s

e) 8 m/s

b) 5 m/s

c) 6 m/s

d) 7 m/s

e) 8 m/s

**angle as north of east**

**52**

Let vector A have a magnitude of 3.5 m at 25 degrees above the negative x direction. Let vector B be defined by Bx = 2.4 m and By = -1.2 m. And finally let C be 1.8 m pointing straight south.

Find vector D such that A+B+C+D = 0 (as a vector equation)

Find Vector E defined as E = 2A+3B -C

**A submarine fires a torpedo with a speed of 25 m/s in a direction of 30 degrees south of east.**

An aircraft carrier is moving at a speed of 15 m/s in a direction 10 degrees west of north.

How fast is the torpedo seen as approaching by sailors on the aircraft carrier

An aircraft carrier is moving at a speed of 15 m/s in a direction 10 degrees west of north.

How fast is the torpedo seen as approaching by sailors on the aircraft carrier

**Draw a picture**

Break down the vectors

Do each direction separately.

Break down the vectors

Do each direction separately.

**A light plane is headed due south, with a speed relative to still air of 185 km/h. After 1.00 h, the pilot notices that they have covered only 165 km and their direction is not south but 15.0 degrees east of south.**

**How far south did he fly?**

**What is the velocity of the wind**

**What do the two velocity vectors look like ( plane and wind)**

**how far east did he fly?**

"why is the hypotenuse called the resultant displacement?"

"Vectors are making sense more after today's class! Can we do some problems like we will have on the test sometime this week?"

" I just like to get more practice in adding vectors "

"Doing the Homework I guess I am still confused about finding the Magnitude of two vectors?"

Demonstration

A person walks 25 m straight west, then turns 30 degrees north and walks another 15 m.

How far is the person from their starting point?

What is the direction of this displacement?

Ax = -25m

Ay = 0

Bx = 15 cos 30 = -13 m

By=15 sin 30 = 7.5

Cx=Ax+Bx =-25+-13=-38

Cy = 0+7.5 =7.5 m

C^2 = 38^2+7.5^2 = 39 m

angle= tan^-1 (7.5/38) = 11 NofW

D = 1.706 at 63 NofE

E = 1.44 at 53.5 NofE

**49.8 km/h at 31 NofE**

**A. 20 m**

B. 30 m

C. 36 m

D. 40 m

E. 50 M

B. 30 m

C. 36 m

D. 40 m

E. 50 M

**Angle 2nd**

**A. Got it!**

B. Getting there, more time

C. Need some help

B. Getting there, more time

C. Need some help

**42.7 km**

**159 km**