Freely Falling Bodies
Conceptual Example: Acceleration Versus Velocity
There are three parts to the motion of the coin. On the way
up, the coin has a vector velocity that is directed upward and
has decreasing magnitude. At the top of its path, the coin
momentarily has zero velocity. On the way down, the coin
has downward-pointing velocity with an increasing magnitude.
In the absence of air resistance, does the acceleration of the
coin, like the velocity, change from one part to another?
Graphical Analysis of Velocity and Acceleration
Object moving with constant velocity (zero acceleration)
x = vot + 12 at 2 = vot
x = vot
Δx + 8 m
Slope =
=
= +4 m s = vo
Δt
2s
Graphical Analysis of Velocity and Acceleration
Changing velocity during a bike trip
v = 0 m/s
v = 2 m/s
v = -1 m/s
Graphical Analysis of Velocity and Acceleration
Object moving with changing velocity
Slope of the tangent line
is the instantaneous
velocity at the t = 20 s
point:
2
1
o
2
x = v t + at
Slope = Δx/Δt
= (26 m)/(5 s)
= 5.2 m/s = v
Graphical Analysis of Velocity and Acceleration
Object moving with constant acceleration
v = vo + at
Δv + 12 m s
Slope =
=
= +6 m s 2 = a
Δt
2s
Chapter 3
Kinematics in 2 Dimensions
Trigonometry and Vectors
Trigonometry
Right triangle – triangle that has one angle 90o
Trigonometry
Trig functions related to a right triangle
ho
sin θ =
h
ha
cos θ =
h
ho
tan θ =
ha
Trigonometry
Find the height of a building which casts a shadow
of 67.2 m when the angle of the Sun’s rays with respect
to the ground is 50.0o.
ho
tan θ =
ha
ho
tan 50 =
67.2m


ho = tan 50 (67.2m ) = 80.0m
Trigonometry
Inverse trig functions
& ho #
θ = sin $ !
%h"
−1
& ha #
θ = cos $ !
%h"
−1
& ho #
θ = tan $$ !!
% ha "
−1
Trigonometry
At what angle does the lakefront drop off
if the depth of the lake at a distance of 14.0 m
is 2.25 m?
& ho #
θ = tan $$ !!
% ha "
−1
& 2.25m #

θ = tan $
! = 9.13
% 14.0m "
−1
Trigonometry
Pythagorean theorem:
2
2
o
2
a
h =h +h
Scalars and Vectors
A scalar quantity is one that can be described
by a single number:
temperature, speed, mass
A vector quantity deals inherently with both
magnitude and direction:
velocity, force, displacement
Scalars and Vectors
Example of a two-dimensional vector
Vector Addition and Subtraction
Often it is necessary to add one vector to another.
=
+
Vector Addition and Subtraction
3m
5m
8m
Vector Addition and Subtraction
=
+
Vector Addition and Subtraction
Find the magnitude! and direction
of the sum vector R
2.00 m
6.00 m
Vector Addition and Subtraction
2
2
2
R=
2
2
R = ( 2.00 m ) + ( 6.00 m )
(2.00 m ) + (6.00 m )
= 6.32m
R
2.00 m
6.00 m
Vector Addition and Subtraction
tan θ = 2.00 6.00
θ = tan
−1
(2.00 6.00)= 18.4

R = 6.32 m
2.00 m
θ
6.00 m
Vector Addition and Subtraction
When a vector is multiplied
by -1, the magnitude of the
vector remains the same, but
the direction of the vector is
reversed.
Vector Addition and Subtraction

B
 
A+B

A
Comparing the sum of two vectors
with the difference between the same
two vectors

A
 
A−B

−B
The Components of a Vector


x and y are called the x vector component

and the y vector component of r.
The Components of a Vector

The vector components of A are two perpendicular


vectors A x and A y that are parallel to the x and y axes,
 

and add together vectorially so that A = A x + A y .
The Components of a Vector
It is often easier to work with the scalar components
rather than the vector components.
Ax and Ay are the scalar components

of A.
xˆ and yˆ are unit vectors with magnitude 1.

A = Ax xˆ + Ay yˆ
The Components of a Vector
Example
A displacement vector has a magnitude of 175 m and points at
an angle of 50.0 degrees relative to the x axis. Find the x and y
components of this vector.
sin θ = y r
y = r sin θ = (175 m ) (sin 50.0! ) = 134 m
cos θ = x r
!
x = r cosθ = (175 m ) ( cos50.0 ) = 112 m
θ
!
r = (112 m ) x̂ + (134 m ) ŷ
Addition of Vectors by Means of Components
  
C= A+B

A = Ax xˆ + Ay yˆ

B = Bx xˆ + B y yˆ
Addition of Vectors by Means of Components
!
C = Ax x̂ + Ay ŷ + Bx x̂ + By ŷ
= ( Ax + Bx ) x̂ + ( Ay + By ) ŷ
C x = Ax + Bx
C y = Ay + B y