Unit 1, Lessons 2-5: Vectors in Two Dimensions

Textbook Sign-Out

Put your name in it and let’s go!

Check-In

• Any questions from last day’s homework?

1. Find the resultant displacement when someone walks 100m East, then… a) 200m East

Step 1: draw arrows “head” to “tail”

Step 2: connect tail of first to head of last

100m 200m

300m Resultant: 300m East

b) 200m West

100m 200m

Step 1:

Step 2: 100m

Resultant: 100m West

Vector Addition

c) 100m East then 200m North

100m

200m

Step 1: Step 2:

Resultant: 224m, 63 North of East

Resultant magnitude: R2 = 1002 + 2002

R = 224m

ө

R

Resultant direction: tanө =

ө = 63

2. Find the resultant displacement for 150km [N] + 200km [W]

150km

200km

Resultant magnitude: R2 = 2002 + 1502

R = 250km

ө

R

Resultant: 250km, 53 West of North (or 37 North of West)

Resultant direction: tanө =

ө = 53

Subtracting Vectors

• Simply remember this: the negative of any vector is a vector of the same magnitude but the opposite direction.

• For example, to subtract a vector pointing South, simply add the same vector pointing North.

1) Find the Northward and Eastward components of the displacement 300m, 30 North of East.

E

N

30

300m

2) Find the components of 25m/s, 37 West of South.

W

S 37 25m/s

cos 30 =

E = 260m

sin 30 =

N = 150m

sin 37 =

W = 15m/s S = 20m/s

cos 37 =

Homework

• Practice today’s lesson: pg. 10 #3 and 4a, and pg. 19 #2

• More review: “One Dimensional Kinematics Problems” Sheet, Part B #1-5

Unit 1, Lesson 3: Velocity Problems

Check-In

• Any questions from last day’s homework?

A) Find the resultant velocity (relative to the shore) when a boat with a “still-water” speed of 15m/s heads due North across a river with a current of 5m/s due East.

15m/s

5m/s

ө

R

15m/s 5m/s

R2 = 52 + 152

R = 15.8m/s

Resultant: 15.8m/s, 18 East of North

B) If the river is 300m wide, how long will it take to cross? How far down stream will the boat land?

Note: motion of water and motion of boat are independent of each other

Answer: 20s, 100m

d = (20)(5) = 100m

tanө =

ө = 18

t =

= 20s

A swimmer swims at 1.0m/s in still water. There is a current of 0.5m/s in a river running due East. a) Find the velocity of the swimmer relative to the shore, if she heads due North. b) Find the heading required for her to end up going due North. c) If it takes 3.0min to go straight across, how wide is the river.

1m/s

0.5m/s

ө

v

Resultant: 1.12m/s, 27 E of N

b) Find the heading required for her to end up going due North. c) If it takes 3.0min to go straight across, how wide is the river.

1m/s

0.5m/s

ө

v

Answer: 30 W of N

Ө =

Example 3 A boat has a top speed of 20m/s in still water. There is a current of 12m/s due South. Find: a) the heading needed to travel due West. b) the distance due West of the starting point 20min later. c) How long it would take to get that far West if you pointed the boat due West.

20m/s 12m/s

ө

v

Ө =

Homework

• Practice today’s lesson: “Vector Kinematics Problems” Sheet #7-9

• Keep up that review: “One Dimensional Kinematics Problems” Sheet, Part B #6-10

Unit 1, Lesson 4: Projectile Motion

Check-In

• Any questions from last day’s homework?

Thought experiment: If you shoot a bullet horizontally from your gun and drop another bullet from the same height at the same time, which bullet hits the ground first? a) A first b) B first c) Both at the same time

Because the vertical and horizontal components of the motion are independent!

A B

Example 1 One of those old VW bugs is driven off a 50m tall cliff at 108km/h a) How long is it in the air? b) How far from the cliff does it land? c) What is its velocity just before impact? (ignore air resistance)

Note: horizontal and vertical motion is independent and we can look at each separately. Another note: the path of a projectile is always parabolic.

?

50m

“Moving straight right at 108 km/h, or 30m/s” (constant speed)

“falling straight down from rest”

Find the time for something to fall (from rest) 50m

How far does something go at 108km/h for 3.19s? (constant speed)

c) What is its velocity just before impact? (ignore air resistance)

31.36m/s

30m/s ө

v

Example 2 A Honda Civic with a 5000 Watt stereo is driven off a cliff at 25m/s and lands 60m from the base of the cliff. How high is the cliff?

60m Answer: 28m tall

HW: p113 #20-28

Example 3 A rock is thrown from a high cliff horizontally at 25m/s. Find its velocity when it is 120m below the top of the cliff.

120m

v? vv?

25m/s

25m/s

Homework

• “Vector Kinematics Problems” Sheet #1

• Pg. 52 #1, 3, 5

Unit 1, Lesson 5: Projectiles Launched at an Angle

Check-In

• Any questions from last day’s homework?

A soccer ball is kicked from ground level with an initial velocity of 25m/s at an angle of 37 with horizontal. Find

a) the horizontal and vertical components of the initial velocity. b) how far away does the ball land? c) what is the maximum height reached by the ball? d) what is the velocity of the ball 2.0 seconds after it is kicked?

vH

vY0 v0=25m/s Stays

constant

Note: Horizontal motion: constant speed Vertical motion: straight up then down (like something thrown straight upward)

how far away does the ball land?

what is the maximum height reached by the ball?

Method 2

Method 1

20m/s

h

what is the velocity of the ball 2.0 seconds after it is kicked?

20/s

4.6m/s

ө

v

20.5m/s, 13 below the horizontal

Example 2 A mortar shell is launched with an initial velocity of 100m/s at an angle of 30 with horizontal. Find (ignore air resistance)

a) horizontal range b) maximum height c) velocity at 3 seconds

vH

v0y

100m/s V0y =100 sin 30 = 50m/s

VH =100 cos 30 = 86.6m/s

b) maximum height c) velocity at 3 seconds

86.6

20.6 ө

v

HW: WS #1-4

Θ = tan-1(20.6/89.6)

Example 3 A tennis ball is hit by a student. The time of flight is 3.7 seconds, and the horizontal range is 27.5 m. Find the initial velocity.

Horizontal Vertical

smt

dvx /43.7

7.3

5.27

2

02

1attvd

2

0 )7.3(9.4)7.3(0 yv

smv y /1.187.39.40

smv /6.194.71.18 22

0

upwards01 68)4.7/1.18(tan

Homework

• “Vector Kinematics Problems” Sheet #2-6