Two-Dimensional Motion and Vectors CHAPTER 3in.sjprep.org/science/Collins/Physics/vectors again and... · 2010-12-01 · VECTORS - AGAIN ´ ... ´This becomes a simple free-fall problem!

Two-Dimensional Motion and Vectors

CHAPTER 3

VECTORS - AGAINVECTORS AGAIN

Scalar- a quantity that has magnitude but no Scalar a quantity that has magnitude but no directionVector physical quantity that has both Vector- physical quantity that has both magnitude and directionR lt t th f d h ddi g t Resultant – the answer found when adding two or more vectorsEquilibrant- equal in magnitude to the resultant in the opposite direction (180˚)

VECTORS – AGAIN CONTVECTORS AGAIN, CONT.

Vectors can be added graphically Draw the vectors to scale

I.e. 1cm=2m

and in the correct directionAll vectors are drawn tip to tail All vectors are drawn tip to tail. The resultant is drawn from the tail of the fi h i f h lfirst to the tip of the last(I won’t make you do this again)But you will have to add vectors algebraically

SOME ADDITIONAL NOTES ON VECTORSSOME ADDITIONAL NOTES ON VECTORS

Vectors can be added in any orderVectors can be added in any orderWhen adding two or more vectors, the sum is independent of the order of the additionindependent of the order of the addition

See diagram on the bottom of pg 84

T bt t t dd it itTo subtract a vector you add its opposite.

COORDINATE SYSTEMCOORDINATE SYSTEMWe use compass directionsdirectionsAnd angles from a reference directionreference directionEx: 300m/s 30˚ North of eastNorth of eastEast is the reference directiondirectionOr: 300m/s 60˚ east of northeast of north

DETERMINING RESULTANT MAGNITUDE AND DIRECTION

Pythagorean theorem for right trianglesPythagorean theorem for right triangles222 bac +=

Tangent, sine, and cosine

hypotenuseopposite

=θsin

ihypotenuseadjacent

=θcos

adjacentopposite

=θtan

EXAMPLEEXAMPLEAn archaeologist climbs the Great Pyramid in Giza,

Egypt The pyramid’s height is 136m and its width Egypt. The pyramid s height is 136m and its width is 2.30x102 m. What is the magnitude and direction of the displacement of the archaeologist direction of the displacement of the archaeologist after he has climbed from the bottom of the pyramid to the top?

y Δy= 136 mΔx=1/2 width=115md=?

Δy=136md

θ=?

xΔx=115m

θ

RESOLVING VECTORS INTO COMPONENTSRESOLVING VECTORS INTO COMPONENTS

In the pyramid example, the horizontal and In the pyramid example, the horizontal and vertical parts that add up to give the total displacement are call componentsdisplacement are call componentsAll vectors can be broken up into x and y components using right triangle mathcomponents using right triangle math

2-D MOTION

Not all motion is along a straight path in one Not all motion is along a straight path in one dimensionMost motion involves two dimensionsMost motion involves two dimensionsVectors and vector components help to simplify th 2 d tithe 2-d motion

HORIZONTAL PROJECTILESHORIZONTAL PROJECTILES

Motion in a plane means motion in two Motion in a plane means motion in two directionsWe must consider not only the horizontal (x) We must consider not only the horizontal (x) direction, but also the vertical (y)W th ti d b f b t We use the same equations used before- but now have equations for both x and y

THE VERTICAL (Y) DIRECTIONTHE VERTICAL (Y) DIRECTIONAcceleration is constant “g”=-9.81 constant g 9.81 m/s2

Initial velocity (viy) is ZERO

yZEROFinal velocity (vfy) is given or calculatedgiven or calculatedTime (t) to strike the ground is given or can be calculatedHeight of release (dy) is given or can be given or can be calculated

THE VERTICAL (Y) EQUATIONSTHE VERTICAL (Y) EQUATIONS

Consider this equation from before but for the Consider this equation from before but for the y-direction

21 tatvd +2

tatvd yiyy +=

Remember that viy = 0 and ay=g, therefore-This becomes a simple free-fall problem!dy=1/2gt2

THE HORIZONTAL (X) DIRECTIONTHE HORIZONTAL (X) DIRECTION

Velocity is considered CONSTANT in the x-Velocity is considered CONSTANT in the xdirection! vix=vfx

This means that acceleration is ZEROThis means that acceleration is ZEROHorizontal distance (dx) traveled (range) is given

b l l t dor can be calculatedTime (t) must be calculated from the vertical direction equations, then substituted in

THE HORIZONTAL (X) EQUATIONSTHE HORIZONTAL (X) EQUATIONS

Consider this equation but for the x-directionConsider this equation but for the x direction

21 tatvd +2

tatvd xixx +=

Remember that vx is constant, and ax =0The equation simplifies to this-The equation simplifies to thisDx=vixt

AN EXAMPLEAN EXAMPLE

An airplane traveling at 100.m/s drops a An airplane traveling at 100.m/s drops a package from a height of 3000. meters. Calculate the time it takes to reach the ground Calculate the time it takes to reach the ground and how far in front of the target must the package be dropped?package be dropped?

THE SOLUTION!THE SOLUTION!

Given: vi =0.0m/s vi =100.m/sGiven: viy 0.0m/s vix 100.m/sdy=3000. m ay=g=-9.81m/s2

T k th d d di ti iti t Take the downward direction as positive to eliminate negative signFind: (1) time to reach the ground

(2) distance from the drop point to targetgSolution: divide the problem into two portions, vertical and horizontalvertical and horizontal

(1) VERTICAL(1) VERTICAL

At the beginning, the plane was moving At the beginning, the plane was moving horizontally, therefore – viy =0

1 2

21 tad yy =

3000.m = ½(9.81m/s2)t2

24.7 s =t

(2) HORIZONTAL(2) HORIZONTAL

The plane and package were traveling at a The plane and package were traveling at a constant speed, therefore-V =100 m/s ax=0 m/s2Vix =100.m/s ax=0 m/s2

dx=vixtdx=(100. m/s)(24.7s) dx=2470mx

The package should be released 2470m before the targetthe target

WASN’T THAT EASY!WASN T THAT EASY!

Th f h j t b g !The fun has just begun!