Part One
We have taken a look at some basic
quantities: time and location.
We have taken a look at how to deal with 3-D
quantities – vectors.
We have taken a look at the basics of motion:
position, velocity and acceleration.
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Two Dimensional Motion - General Physics I - Lecture Slides, Slides of Physics
Following points are the summary of these Lecture Slides : Two Dimensional Motion, Time, Location, Vectors, Quantities, acceleration, Control, Predict Motion, Fundamental Property, Basic Forces
Typology: Slides
2012/2013
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Part One
We have taken a look at some basic
quantities: time and location. We have taken a look at how to deal with 3-D
quantities – vectors. We have taken a look at the basics of motion:
position, velocity and acceleration.
Part Two
In Part Two, we will look at:
how to work with
2-D motion
combine
motion and vectors); how we
control or
predict motion
(what
causes
motion);
how a third fundamental property:
mass
, is
involved; what are the
basic forces
and how do they
work.
Two Dimensional Motion
How do we work with two dimensions when
we consider motion?
- We work with vectors by working with the
components
Two Dimensional Motion
Does it matter which form we use:
rectangular or polar
Two Dimensional Motion
But if we work in components, how are the
components “
connected
Two Dimensional Motion
But if we work in components, how are
the components “connected”?
•^
By time!
Time is the same for the x
and for the y. For each time, there isonly one x and one y.
If we wanted, we could eliminate the time and
express y in terms of x: convert x(t) intot(x), and then convert y(t) into y(x). This issometimes used, but we do lose informationabout time if we do this.
Special Case:
Trajectory Motion
If we again consider no air resistance, there
should be zero acceleration in the xdirection. This leads to the followingequations for x:
x = x
o^
+ v
xo
t + (1/2)0t**
2
v
x^
= v
xo
+ 0t*
which become simply: v
x^
= v
xo
and
x = x
o^
+ v
tx
Special Case:
Trajectory Motion
- In effect, then, we have
three
equations:
x = x
o^
+ v
tx
y = y
o^
+ v
yo
t + (1/2)a
ty
2
v
y^
= v
yo
+ a
ty
- In these three equations we have
nine
quantities:
x
, x, yo
, y, vo
, vx
yo
, v
, ay^
y^
and t
- This means that we
need to know 6
of these
in order to
solve for the other 3
Example of Trajectory Motion
- The first thing we recognize is that this is a
trajectory
problem, so we have the three
equations:
x = x
o^
+ v
tx
y = y
o^
+ v
yo
t + (1/2)a
ty
2
v
y^
= v
yo
+ a
ty
- This also means that there is no acceleration
in the x, so v
xo
= v
.x
Example of Trajectory Motion
- The second thing we recognize is that this
takes place on the earth, so that
a
y^
m/s
- Next let’s draw the diagram and put our
information from the problem on thatdiagram:
(go to the next slide for the diagram)
Example of Trajectory Motion
Let’s look at the information given in the
problem and see how that specifies some ofthe remaining unknown quantities:
- Problem: If you are at the top of a building
12 meters
above the ground, ...
We can identify this value as the initial y
value, so
y
o^
= +12 meters
We can also choose to measure from the
building, so this means
x
o^
= 0 meters
Example of Trajectory Motionx^ o
= 0 m
a
y^
= -9.8 m/s
2
y
o^
= 12 m v
xo
t =
v
yo
x =y =v
x^
= v
xo
v
y^
Example of Trajectory Motion
- v
yo
= v
o^
sin (
) = 33 m/s sin (40o
o^ )
21.21 m/s
, and
- v
xo
= v
o^
cos (
) = 33 m/s cos (40o
o^ )
25.28 m/s
Example of Trajectory Motionx^ o
= 0 m
a
y^
= -9.8 m/s
2
y
o^
= 12 m v
xo
= 25.28 m/s
t =
v
yo
= 21.21 m/s
x =y =v
x^
= v
xo
v
y^