Vectors• Vector: quantity that has magnitude and direction• Scalar: quantity that has magnitude only

– Example: 20 mi north (vector) vs. 20 mi (scalar)

• Arrows represent vectors graphically

Vector Scalar

Position Distance

Velocity Speed

Weight Mass

A

Magnitude = A

(arrow length proportional to vector magnitude)

(or A) (or A)

Vectors• Two vectors A and B are equal if |A| = |B| AND A B:

• For vector C antiparallel to A and B, with same length as A and B A = B = – C

• Vector addition: Must take direction into account!– Use tail-to-tip rule: place tail of 2nd vector to tip of 1st

vector– Works for any number of vectors

A B

A B C

Vector Addition and Subtraction• For A + B = C:

– Vector addition is commutative (order doesn’t matter) – try it!

• For multiple vectors A + B + D = E:

• Rules for vector subtraction are the same – just consider adding a negative vector:

A B

A B DE

A

–B

A – BA – B = A + (– B)

A

B

C

Vector Addition and Subtraction Interactive

Component Vectors• Imagine the hassle of measuring a triangle

each time you want to add vectors!– Would need protractor, ruler, etc.

• Use of vector components provides simple yet accurate method for adding vectors

A

x

y

Ax and Ay are determined from trigonometry:

Ax = Acos and Ay = Asin

Ax (Ay) could be positive or negative depending on whether or not it points in

the +x or –x (+y or –y) direction A = (Ax2 + Ay

2)1/2

Ax

Ax

AyAy

A = Ax + Ay

Component Vectors• It may be helpful to envision components as

“shadows” that are projected along each axis:

– The length of the “shadow” depends on the orientation of A

A

x

y

x

Ax

yAy

Component Vectors• Any vector can be represented by its x and y

component vectors

• Sometimes it is convenient to determine the components of vectors in a tilted coordinate system:

CQ1: Consider vector shown below. Which of the following could not be a pair of component

vectors for vector ?

A)

B)

C)

D)

Q

Q

Q

Component Vectors• Using components to add vectors A + B = C:

x

y

A

BC

Ax

Ay

Bx

By

Cx

Cy

Ax + Bx = Cx

Ay + By = Cy

In terms of components:

jAiAA yxˆˆ

jBiBB yxˆˆ

jBAiBAC yyxxˆˆ

Unit vector Magnitude Direction

1 +x

1 +y

1 +z

i

j

k

CQ2: A man entered a cave and walked 100 m north. He then made a sharp turn 150° to the

west and walked 87 m straight ahead. How far is the man from where he entered the cave?

(Note: sin30° = 0.50; cos30° = 0.87.)

A) 25 m

B) 50 m

C) 100 m

D) 150 m

Example Problem #3.20

Solution (details given in class):

(a) 185 N, = 77.8° from x axis

(b) 185 N, = 258° from x axis

The helicopter view in the figure at right shows two people pulling on a stubborn mule. Find

(a) the single force that is equivalent to the two forces shown, and

(b) the force that a third person would have to exert on the mule to make the net force equal to zero.

The forces are measured in units of Newtons (N).

2–D and 3–D Motion• Strong similarity between 1–D motion and 2–D (or

3–D) motion• Same kinematic quantities (displacement, velocity,

acceleration) are used• Only difference is vector quantities can have up to

three non-zero components• First, we describe position:

• Displacement (change in position):x

y

z

1r

2r

kzjyixr ˆˆˆ1111

kzjyixr ˆˆˆ2222

kzzjyyixxrrr ˆˆˆ12121212

2–D and 3–D Velocity• Average velocity:

– Or:

• Instantaneous velocity:

• Instantaneous speed:– Think Pythagorean theorem!

• Just like in 1–D motion, average velocity only depends on beginning and ending position

• Instantaneous velocity is tangent to the curve of versus time

t

r

tt

rrv

12

12ave

kt

zj

t

yi

t

xv ˆˆˆ

ave

kvjvivt

rv zyx

t

ˆˆˆlim0

222zyx vvvv

r

2–D and 3–D Acceleration• Average acceleration defined as:

– are the instantaneous velocities at t2 and t1, respectively

– points in the same direction as

• Instantaneous acceleration:

– points toward the inside of any turn particle is making

• Remember that an object can have a non-zero acceleration even though its speed remains unchanged

• Example: uniform circular motion– Speed remains the same– Velocity keeps changing due to changing direction

t

v

tt

vva

12

12ave

12 ,vv

(same as 1–D)

avea

12 vv

t

va

t

0lim

(same as 1–D)

a

CQ3: A weather balloon travels upward for 6 km while the wind blows it 10 km north and 8 km

east. Approximately what is its final displacement from its initial position?

A) 7 km

B) 10 km

C) 14 km

D) 20 km

Projectile Motion• Have you ever wondered how to predict how far a

batted baseball will travel, or how far a skier will travel after jumping from a ramp?

• These are examples of projectile motion– Projectile (baseball, skier) follows a trajectory (path in space)

determined by its initial velocity and effects of gravity

• Start with simple model– Projectile represented by single particle– Neglect air resistance, curvature & rotation of the earth

y

x

a

(Usually put start of motion at origin of coordinates)

v0

Projectile Motion• Notice that acceleration is constant (= g) and is in

vertical direction (due to gravity) pointing toward earth’s surface

• This is motion with constant acceleration in 2–D• Particle is given an initial velocity v0:

• We can analyze projectile motion as:– Horizontal motion with constant velocity (ax = 0)– Vertical motion with constant acceleration (due to gravity)

x

y

v0x

v0y

v0x

v0y

0v

ax = 0 vx = v0x = v0cos (constant in time)

ay = –g vy varies with time

CQ4: Two skydivers are playing catch with a ball while they are falling through the air. Ignoring air resistance, in which direction should one skydiver throw the ball relative to the other if the one wants

the other to catch it?

A) above the other since the ball will fall faster

B) above the other since the ball will fall more slowly

C) below the other since the ball will fall more slowly

D) directly at the other since there is no air resistance

Projectile Motion

• The equations describing the motion follow directly from our discussion of 1–D motion with constant acceleration:

• Notice that we treat the x– and y–components separately

x–direction (ax = 0): y–direction (ay = –g):

xx vv 0constant yy vgtv 0

00 xtvx x 002

2

1ytvgty y

Velocity Components Interactive

Position and Time Interactive

Projectile Motion• You can answer lots of questions concerning the

path the projectile takes by using these equations• Many times you need to restate the question to

determine what the question is looking fory

x

y = ymax

What is the maximum height reached? (or At what value of y is vy = 0?)x = xmax

What is the maximum range? (or What is the value of x when y = 0 [other than at the origin]?)

Example Problem #3.23

Partial solution (details given in class):(e) 3.19 s(f) 36.1 m/s, = 60.1° below +x axis

A student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of 18.0 m/s. The cliff is 50.0 m above a flat, horizontal beach, as shown in the figure at right. (a) What are the coordinates of the initial position of the stone? (b) What are the components of the initial velocity? (c) Write the equations for the x- and y-components of the velocity of the stone with time. (d) Write the equations for the position of the stone with time. (e) How long after being released does the stone strike the beach below the cliff? (f) With what speed and angle of impact does the stone land?

Example Problem #3.31

Solution (details given in class):(b) 1.78 s(a) 32.5 m

A car is parked on a cliff overlooking the ocean on an incline that makes an angle of 24.0° below the horizontal. The negligent driver leaves the car in neutral, and the emergency brakes are defective. The car rolls from rest down the incline with a constant acceleration of 4.00 m/s2 for a distance of 50.0 m to the edge of the cliff, which is 30.0 m above the ocean. Find (a) the car’s position relative to the base of the cliff when the car lands in the ocean, and (b) the length of time the car is in the air.

CQ5: Interactive Problem: Target Practice

The airplane cargo needs to be dropped at a horizontal position of:

ActivPhysics Problem #3.7, Pearson/Addison Wesley (1995–2007)

A) 0 m

B) 0.775 m

C) 1.35 m

D) 4.65 m

E) 6.00 m