Chapter 3

Vectors and Two Dimensional Motion

• Vectors

• Operations with Vectors – Graphical Method

• Operations with Vectors – Component Method

• Two Dimensional Motion

• Application: Projectile Motion

That is, describing how objects

move in a plane

Vectors – Properties

Scalars are physical quantities completely described only by their magnitude.

Ex: time, mass, temperature, etc.

Vectors describe physical quantities having both magnitude and direction.

Ex: position, displacement, velocity, acceleration, force, etc.

magnitude

θ

θ or direction

direction

• The direction of a vector depends on the arbitrary system of coordinates

• However, the magnitude does not depend on how you choose to span the space

V

V

Vor

y

x

y

x

Operations with Vectors – Multiplication with a number

• A set of vectors can be added or subtracted in any order.

• If the vectors added represent physical quantities, they must have the same nature.

• Multiplying a vector by a positive number multiplies its magnitude by that number

(if the number is negative the vector flips in the opposite direction):

V

V

2V

2V

2V

2V

Operations with Vectors – We’ve already encountered them…

• In order to add or subtract 1D vectors, it suffices to add the vectors algebraically: if

the signs are correctly specified, the resultant vector will have the correct direction

and magnitude

1x 2x

1 2 10 km 4 km 14 kmx x

One dimensional case:

• Position, displacement, velocity and acceleration are all vectors

• In the 1D case they are confined along a straight line, so their direction is given by

the sign: positive x or negative x

x v

a0 x

Particle

Ex: The total displacement is the sum of the successive displacements

0

x (km)

10 km 4 km

Resultant = 14 km (east)

East 1x

2x

0

x (km)

Resultant =

6 km (east)

East

1 2 10 km 4 km 6 kmx x

• In general, even if the vectors are not along the same axis, they can be added

graphically by using the “tail-to-tip” method:

Operations with Vectors – Graphical addition: tail to tip method

• The method offers a qualitative idea about the resultant: in order to obtain the

resultant numerically (magnitude and direction), one has to use scaled grid paper

which is a rather cumbersome technique

The vector sum can be obtained graphically by chaining the vectors each with the

tail to the tip of the previous: then the vector resultant connects the tail of the first

vector to the tip of the last one. The operation can be done in any order.

Ex: Say that we have 3 arrows (vectors) in a plane (2D) and we want to add them up:

1v + 2v 3v+ = R

1v2v

3v

1 2 3R v v v

Notice that in 2D, the arrows above the vector symbols cannot be skipped since a vector can

have an infinity of directions not only two as in the 1D case: the operation between the arrows

cannot be reduced to an immediate algebraic addition or subtraction

Ex: Physical example: Successive 2D displacements can still be added to obtain the

total displacement

initial

final

netd

1d

2d

3d

• An application of vector

summation in mechanics is

calculating the net

displacement of an object

traveling from an initial

position to a final one via

several successive partial

displacements

• If we denote d1, d2 and d3

three successive displacements

the net displacement is

• It is given by the vector sum

(or resultant) of the partial

displacements

• Notice that adding the partial

displacement follows the logic

of tail-to-tip method

1 2 3netd d d d

Operations with Vectors – Subtraction

• In order to subtract vectors, we can still use the addition procedure by adding the

negative of the arrow being subtracted

• We define the negative of a vector to be a vector with the

same magnitude but pointing in the opposite direction.

v v

1v

_

2v

= R

1v

2v

1 2 1 2R v v v v

Ex: Say that we have 2 arrows (vectors) in a plane and we want to subtract them:

=

1v

+

2v

Ex: Physical example: linear displacement is defined as the final position minus the initial

position

reference

2 1r r r

initial

1r

final

2r

r

2 1r r r

1r

r

• If we denote r1 and r2 two

positions successively

occupied by a moving

objects, the displacement is

Operations with Vectors – Vector components

• Note that, in order to obtain magnitudes and

directions, the graphical methods should be

used on grid paper.

• A more computational way to get magnitudes

and directions is by using vector components

in arbitrary systems of coordinates:

y

x

V

xV

yV

2 2

1

cos Components from

sin direction and magnitude

Direction and magnitude

from componentstan

x y

x

y

x y

y

x

V V V

V V

V V

V V V

V

V

θ

Notation: ,x yV V V

II I

III IV

x

y

A

A

I II III IV

+ - - +

+ + - -

Caution: The components are not are not vectors or vector magnitudes. They are

positive or negative if the corresponding vector components point in the negative

direction of the respective axis.

Operations with Vectors – Addition using vector components

• The addition and subtraction of vectors can be reduced to the algebraic addition and

subtractions of components

• Given n vectors in the same plane, the addition can be solved in 2D as following:

1 2

1 2

1 2

......

...

x x

y y

x

y

V VV V

V V

RR

R

2 2

x yRR R

1tan y xR R

magnitude:

direction:

1. Calculate the components of the

resultant by adding the vector

components in the respective

directions:

2. Calculate the magnitude and direction

of the resultant from its components:

Ex: The procedure can be visualized graphically:

the components (Rx, Ry) of the resultant R are

aligned with the components of the vectors

involved so they can be added as numbers

R A B

Problem:

1. Operating with vectors: Given the two vectors in the figure, find the vector resultants

where and are vectors with magnitudes 4 and 5 units respectively, by using

a) Graphical method

b) Vector components

1 2R A B

A

B

30

4

5

A B

2 2R A B

2D Kinematics – Vector based approach

• Any 2D motion can be reduced to two one dimensional motions in an arbitrary

system of xy-coordinates since the kinematic quantities are vectors in the plane

• Thus, all vector quantities characterizing the 2D motion can be split in components

along x-axis and y-axis

• The projections of the motion along x-axis and y-axis are 1D, involving the

components along the respective directions, such that, in some conditions, we can

work using the knowledge from the 1D kinematics

• Once the information is solved in 1D, the 2D motion is characterized by vector-

adding the components

210 0 2 xxx x v t ta 21

0 0 2 yyy y v t ta

0x x xv v ta 0y y yv v ta

22

yx vvv

1tanv y xv v

22

yx aaa

Ex: If an object moves in a plane with a constant acceleration, the components ax,y of the

acceleration are constant both in x and y-directions. Therefore the components of the

position and velocity obey the equations of uniformly accelerated motion:

At any moment t, the 2D vector quantities can be obtained from their respective

components:

1tana y xa a

Projectile Motion – Main idea

• A typical example of 2D motion is the motion of a projectile launched with a

certain initial velocity while only its weight acts on it (so, we neglect air resistance)

• One way to model this motion is to project it along vertical and horizontal axes:

then the velocity in the x-direction is constant while in the y-direction the kinematics

is with constant acceleration g

Ex: This photograph shows two balls that start to fall

at the same time. The one on the right has an initial

speed in the x-direction. It can be seen that vertical

positions of the two balls are identical at identical

times, while the horizontal position of the yellow ball

increases linearly.

Projectile motion – Systematic Approach

• Since only gravity acts, the acceleration of the projectile is only the gravitational

acceleration g pointing vertically downward

Problem setup:

y

x Projectile

0v

trajectory

0

000 sinvv y

000 cosvv x

, 0,x ya g a a g

0x

y

a

a g

Ground

Time t

y-components:

Time t

x-components:

t0

t

210 2yy v t gt

0y yv v gt

Time t0

Initial velocity

yv

xv

v

0x xv v0xx v t

y (t)

x (t)

0 yv

Trajectory Shape, Total time ttot, Range R, Maximum height ymax

0 2 2

0 0

2tan2 cos

gy

vx x

0 02 sintot

vt

g

2

0 0sin 2vR

g

2 2

0 0sin

2max

vy

g

x 0

0v

0

R

ymax

y

0xv

Problems:

2. Trajectory shape: Show that the trajectory of a projectile acted only by gravity is parabolic

given by the following relationship:

3. Other trajectory characteristics: Demonstrate the following relationships for the total time

ttot, the horizontal range R, and maximum height ymax on the trajectory shown above:

Problem

4. Projectile motion: A quarterback throws a football with an initial upward velocity

component of 16.0 m/s and a horizontal velocity component of 20.0 m/s.

a) Sketch the trajectory and the initial velocity with components.

b) Represent and calculate the position and velocity of the ball at a time equal to three quarters

the total time the football needs to come back on the ground.

5. Hitting a target: A plane flies horizontally with a

constant speed v at an altitude h above the ground.

Say that a melon is to be dropped to hit a smug Kim

Kardashian who does whatever it is that she does

somewhere on the ground below. Neglecting air

resistance, find an expression in terms of v, h and g

for the horizontal distance d ahead of Kim where the

melon is supposed to be dropped in order to make

things right?

0v

h

d

Take for instance the 2001 war drama Pearl Harbor which is marred not only by historical

inaccuracies, but also by completely bogus airplane maneuvers and bomb trajectories. In the

adjacent clip you can follow the trajectory of a bomb that hits USS Arizona apparently after

being dropped from an airplane almost vertically above the ill-fated deck. Let’s analyze it:

Exercise: Bad Physics in “Pearl Harbor”. When it comes to Physics, if ignorance is bliss, Hollywood is

the garden of Eden. Their movies are a bottomless fountain of references to how subculture seems to offer

more respect to the rules of baseball than to the laws of nature…

Knowing that Arizona was bombed from about 3000 m altitude and the service speed of a

B5N Kate bomber is about 300 km/h, estimate the horizontal distance before the battleship

where the bomb must have been dropped in order to hit it (neglecting air drag).