M i i t f t d ’ l t

Physic 231 Lecture 4

• Main points of today’s lecture:• Vectors and components• Trajectories of objects in 2

dimensions:dimensions:

2 downwards m/s 9.8 g =

( )tvvyyy

gtvv

yoy

yy

0

0

21 +=−≡Δ

−=

ygvv

gttvy

y

y

y2

02

20

221

Δ−=−

−=Δ

tvxvv

x

xx

0

0

=Δ=

Conceptual question

A t di t th d f liff th b ll t i ht• A person standing at the edge of a cliff throws one ball straight up and another ball straight down at the same initial speed. Neglecting air resistance, the ball to hit the ground below the cliff with the greater speed is the one initially throwncliff with the greater speed is the one initially thrown– a) upward.– b) downward.– c) neither—they both hit at the same speed.

= downwards m/s 9.8 g 2

( )tvvyyy

gtvv

o+=−≡Δ

−=

21

0

0

ygvv

gttvy

Δ−=−

−=Δ

221

20

2

20

Two dimensional coordinate systems

y3 (xf,yf)

2

(x0,y0)

1

• Any position can be unique expressed in terms of its point (x,y) on the two dimensional grid

x 0 1 2 3

0

dimensional grid. • The displacement of a person or object from (x0,y0) to (xf,yf) can be drawn as

an arrow. This is the graphical representation of the displacement vector. – The length of the vector (arrow) is the distance moved if the displacement

i i i h liis in a straight line. – The direction of the vector (arrow) indicates the direction of the

displacement.

Vector representations• Graphical representation has the

displacement vector as an arrow O l t t b itdisplacement vector as an arrow starting at (x0,y0) and ending at (xf,yf). It is defined by its length and direction

• One can also represent a vector by its components in a Cartesian coordinate system.

• The Cartesian components of theThe Cartesian components of the displacement vector are:

y 0xxxr fx −=Δ=Δx component:

• Alternatively we can write them in

y3

2 (x0,y0)(xf,yf)rΔ 0yyyr fy −=Δ=Δy component:

Alternatively, we can write them in terms of the length of the displacement vector and its angle with respect some fixed axis, e.g. the

i

21

0 x axis.x 0 1 2 3

0

rΔ

ΔxΔyθ

Example

• What are the x- and y-components of this vector?

3 1a. 3, -1b. 3, -4b. 4,3d. 4, -3e. -3, -4

Length and angle of a vector

A

• The following vector has lengthAAyθ

The following vector has length 4.0 units. What are the x- and y-components?a. 3.5, 2.0

• By the Pythagorean theorem, the length of this vector is:

Ax

a. 3.5, .0b. 2.0, 3.5c. -3.5, 2.0d 2 0 3 5the length of this vector is:

22yx AAA +≡

d. -2.0, -3.5e. -3.5, -2.0

A 4 30 3 5°• The angle θ can be obtained from various trigonometric relations:

xA 4cos30 3.5= − ° = −

yA 4sin 30 2= − ° = −A

( ) ( ) xy

AA

AA

== θθ ;cos;sin

A

Ax

Ay

( )x

y

AA

=θtan andA

More about vectors

V t dditi• Vector addition:– Place the tail of one vector

(B) on the head of the other (A)Δ rΔ

These are all equivalent!

(A).– Draw the sum vector (C)

from the tail of the first vector to the head of the y

rΔ rΔ

second.

A B

C

y3

2 (x0,y0)(xf,yf)rΔ

A C

B A

21

0 rΔ

rΔ

– It doesn’t matter which order you do it.

• Addition by components:

x 0 1 2 3

0

rΔrΔ– Cx=Ax+Bx– Cy=Ay+By

rΔ

Other vector properties/operations

I A i d i th it S bt ti t b d b• Inverse: -A is drawn in the opposite direction as A. The x- component of -A is –Ax and the y – component is –Ay. Obviously, R=A+ (-A )= A-A

• Subtracting vectors can be done by subtracting components. If R=B-A,then Rx=Bx-Ax and Ry=By-Ay. Graphically: y y, ( )

=0, both from the drawing and also by adding components.

p y

A

AA−

B

A

A

−

R

A

Example:

A f tb ll l th tt i i th d i b th th• A football player runs the pattern given in the drawing by the three displacement vectors A, B and C. The magnitudes of these vectors are A=5 m, B=15 m, and C=18 m. Using the component method, find the magnitude and direction of the resultant vector A+B+Cmagnitude and direction of the resultant vector A+B+C.

label x-comp y-compA 0 5 mB 15 m 0C 14.7 -10.3mR 29 7 5 3R 29.7m -5.3,m

0x

0

C 18m cos(35 ) 14.7m;

C 18msin(35 ) 10.3m;

= =

= = −R=

yC 18msin(35 ) 10.3m;

2 2R 29.7 5.3 30.2mtan( ) 5 3 / 29 7θ

= + == −

0

tan( ) 5.3 / 29.710.1

θθ = −

quiz

B b t l 3 k d t d th t l 4 k d th R l ti t• Bob travels 3 km due east and then travels 4 km due north. Relative to his starting position what is the magnitude of his total displacement and the angle with respect to due east?

) 7 k 35 d t th th f d t– a) 7 km, 35 degrees to the south of due east.– b) 7 km, 53 degrees to the north of due east– c) 5 km, 35 degrees to the south of due east– d) 5 km, 53 degrees to the north of due east

vector x comp. y comp.

BR

A 3 km 0

B 0 4 km

3 k 4 k

A

R 3 km 4 km

543 2222yx kmRRR =+=+=

east due of north ,53333.1tan o

x

y

RR

===θ

Projectile motion in two dimensions

ya g= − xa 0=

( )

y

y y0 y y0

0 y yo

v v a t v gt

1y y y v v t2

= + = −

Δ ≡ − = + ( )

x

x x0 x x0

0 x xo x0

v v a t v1x x x v v t v t2

= + =

Δ ≡ − = + =

y

2 2y0 y y0

2 2y0 y

21 1y v t a t v t gt2 2

v v 2a y 2g y

Δ = + = −

− = Δ = − Δ x

2x0 x x0

2 2x0 x

1x v t a t v t2

v v 2a x 0

Δ = + =

− = Δ =y y0 y y g y 0

• This means that the x-component of the velocity remains constant. The y-component reflects the gravitational acceleration.

Projectile MotionThe horizontal motion isThe horizontal motion is constant; the vertical motion is free fall:

The horizontal and vertical components of the motion are independent.

Slide 3-37

example

A b ll t i fi d f ifl th t i h ld 1 6 b th d i• A bullet is fired from a rifle that is held 1.6 m above the ground in a horizontal position. The initial speed of the bullet is 1100 m/s. Find (a) the time it takes for the bullet to strike the ground and (b) the horizontal distance traveled by the bullethorizontal distance traveled by the bullet.

If upward is the direction of positive y:

x0x v t;Δ =1 1

vx0 1100 m/s

vy0 0

direction of positive y:2 2

y01 1y v t gt gt ;2 2

Δ = − = −

2 2 yΔ 2 yΔΔy -1.6 m

2 2 yt ;gΔ= − 2 yt 0.57s

gΔ= − =

( )x 1100m / s 0.57s 628mΔ = =

xΔ

( )

yΔ

clicker question• The diagram below shows two

successive positions of a particle; it'ssuccessive positions of a particle; it's a segment of a full motion diagram. Which of the following vectors best represents the acceleration betweenrepresents the acceleration between

)

i fv and v

a)

b)

c)

d)d)

example

A ll i h i f th ili A ifl i i d di tl t th• A small can is hanging from the ceiling. A rifle is aimed directly at the can, as the figure illustrates. At the instant the gun if fired, the can is released. Ignore air resistance and show that the bullet will always strike the can regardless of the initial speed of the bullet Assume thatstrike the can, regardless of the initial speed of the bullet. Assume that the bullet strikes the can before the can reaches the ground.

bullet 0x v cos tθΔ = line _ of _ sightx= Δ1

Δygravity Δycan

2bullet 0

1y v sin t gt2

θΔ = −

1bullet line _ of _ sight gravityy y yΔ = Δ + Δ

θ 2can

1y gt2

Δ = −

can gravityy y Δ = Δ g y

Therefore bullet will hit the can.