Kinematics
Motion in 1-Dimension
Lana Sheridan
De Anza College
Jan 19, 2016
Last time
• 1-D kinematics
• quantities of motion
Overview
• graphs of kinematic quantities vs time
Scientific Notation
An alternate way to write numbers is in scientific notation.
This is especially useful when numbers are very large or very small.
Scientific Notation
An alternate way to write numbers is in scientific notation.
This is especially useful when numbers are very large or very small.
For example, the speed of light in a vacuum is roughly:
300,000,000 m/s
In scientific notation, we could write this as:
3.0 × 108 m/s
Scientific Notation
An alternate way to write numbers is in scientific notation.
This is especially useful when numbers are very large or very small.
For example, the speed of light in a vacuum is roughly:
300,000,000 m/s
In scientific notation, we could write this as:
3.0 × 108 m/s
This is the same thing.
108 = 100, 000, 000
so,
3.0 × 100,000,000 = 300,000,000 m/s
Scientific Notation: One digit only before decimal!
One reason to use scientific notation is to clearly convey the
number of significant figures in a value.
When a number is in scientific notation, there is one digit,
followed by a decimal point, followed by more digits, if there is
more than one significant figure.
Here there are two significant figures:
3.0 × 108 m/s
Here there are 4 significant figures:
2.998 × 108 m/s
↑
one digit
one digit before the decimal + 3 digits after the decimal = 4 s.f.s
Graphing Kinematic Quantities
One very convenient way of representing motion is with graphs
that show the variation of these kinematic quantities with time.
Time is written along the horizontal axis – we are representing
time passing with a direction in space (the horizontal direction).
Equation 2–
Average Velocity
EXERCISE
IF the acceleration of an object is constant, then the velocity-time
graph is a straight line,
A ball is throw
of the ball is
v
a. 0.50 s, a
v
Solution
vav=
1
2 (v0
a. Substitu
+ v)
v0
t
O
t
b. Similarl
(a)
And,
v
v
vavg
1
= (v0 + v)
2
Next, ho
stant? To an
PROBLEMS
Velocity vs Time Graphs
49
The
under a velocity-time
a specialsegments
interpretation:
thearea
motorcycle
during each graph
of thehas
following
of the
it is
the
displacement
of
the
object
over
the
time
interval
motion: (a) A, (b) B, and (c) C.
considered.
Velocity, v (m/s)
15
B
10
C
5
O
A
5
15
10
Time, t (s)
20
25
▲ FIGURE 2–31 Problem 32
33. •• A person on horseback moves according to the velocity-
PROBLEMS
Velocity vs Time Graphs
49
The
under a velocity-time
a specialsegments
interpretation:
thearea
motorcycle
during each graph
of thehas
following
of the
it is
the
displacement
of
the
object
over
the
time
interval
motion: (a) A, (b) B, and (c) C.
considered.
Velocity, v (m/s)
15
B
10
C
A
5
O
5
15
10
Time, t (s)
20
25
▲ FIGURE 2–31 Problem 32
∆x = (25 m + 100 m + 75 m)i = 200 m i
33. •• A person on horseback moves according to the velocity-
at an
out t
mod
gene
If
t, we
Acceleration in Velocity vs Time Graphs
a
vx
Slope ! ax
axt
vxi
vx i
t
b
vxf
t
or
Acceleration vs Time Graphs
ax
Slope ! 0
ax
t
c
aavg =
∆v
∆t
Rearrange:
aavg (∆t)
Figure 2.11∆vA=particle
under
t
This
t if
velo
The
slop
tive,
slop
stan
slop
B
Equ
b
Acceleration vs Time Graphs
This
t if
velo
Slope ! 0
The
slop
ax
tive,
slop
t
t
stan
slop
∆v = aavg (∆t)
c
The area under an acceleration-time graph is the change in velocity
B
over that time interval.
Figure 2.11 A particle under
Equ
ax
Example 2.5
A boat moves slowly inside a marina (so as not to leave a wake)
with a constant speed of 1.50 m/s. As soon as it passes the
breakwater, leaving the marina, it throttles up and accelerates at
2.40 m/s2 .
(a) How fast is the boat moving after accelerating for 5.00 s?
(b) How far has the boat traveled in this time?
1
Walker, pg 31.
2–10
Example 2.5
constant acceleration. The utility
A boat moves slowly inside
xt Example.
in the problem, such as initial position,
initial velocity, acceleration, and so on.
These preliminaries will help in producing a mathematical representation of the
aproblem.
marina (so as not to leave a wake)
with a constant speed of 1.50 m/s. As soon as it passes the
breakwater, leaving the marina, it throttles up and accelerates at
2.40 m/s2 .
(a) How fast is the boat moving after accelerating for 5.00 s?
ave a wake) with a constant speed of 1.50 m/s. As soon as it passes the break(b) How
has the
boat
traveled
in this
rates at 2.40
(a) How
fast is
the boat
moving
after time?
accelerating for 5.00 s?
m>s2. far
Sketch:
reakwater,
of motion.
the initial
ons 2–7 to
so we use
dge of the
on to time 1
Breakwater
a = 2.40 m/s2
x
Walker, pg 31.
Example 2.5
A boat moves slowly inside a marina (so as not to leave a wake)
with a constant speed of 1.50 m/s. As soon as it passes the
breakwater, leaving the marina, it throttles up and accelerates at
2.40 m/s2 .
(a) How fast is the boat moving after accelerating for 5.00 s?
(b) How far has the boat traveled in this time?
Hypothesis:
1
Walker, pg 31.
Example 2.5
A boat moves slowly inside a marina (so as not to leave a wake)
with a constant speed of 1.50 m/s. As soon as it passes the
breakwater, leaving the marina, it throttles up and accelerates at
2.40 m/s2 .
(a) How fast is the boat moving after accelerating for 5.00 s?
(b) How far has the boat traveled in this time?
Hypothesis: 10 m/s. (Must be bigger than 1.50 m/s, but there’s
only so fast a boat can go.)
1
Walker, pg 31.
Example 2.5
A boat moves slowly inside a marina (so as not to leave a wake)
with a constant speed of 1.50 m/s. As soon as it passes the
breakwater, leaving the marina, it throttles up and accelerates at
2.40 m/s2 .
(a) How fast is the boat moving after accelerating for 5.00 s?
1
Walker, pg 31.
Example 2.5
A boat moves slowly inside a marina (so as not to leave a wake)
with a constant speed of 1.50 m/s. As soon as it passes the
breakwater, leaving the marina, it throttles up and accelerates at
2.40 m/s2 .
(a) How fast is the boat moving after accelerating for 5.00 s?
Given: a = 2.40 m/s2 , v0 =1.50 m/s, t = 5.00 s
Want: vf
1
Walker, pg 31.
Example 2.5
A boat moves slowly inside a marina (so as not to leave a wake)
with a constant speed of 1.50 m/s. As soon as it passes the
breakwater, leaving the marina, it throttles up and accelerates at
2.40 m/s2 .
(a) How fast is the boat moving after accelerating for 5.00 s?
Given: a = 2.40 m/s2 , v0 =1.50 m/s, t = 5.00 s
Want: vf
∆v
v − v0
=
∆t
∆t
Acceleration is constant, so a avg = a
a avg =
v = v0 + at
1
Walker, pg 31.
Example 2.5
A boat moves slowly inside a marina (so as not to leave a wake)
with a constant speed of 1.50 m/s. As soon as it passes the
breakwater, leaving the marina, it throttles up and accelerates at
2.40 m/s2 .
(a) How fast is the boat moving after accelerating for 5.00 s?
Given: a = 2.40 m/s2 , v0 =1.50 m/s, t = 5.00 s
Want: vf
∆v
v − v0
=
∆t
∆t
Acceleration is constant, so a avg = a
a avg =
v = v0 + at
= 13.5 m/s i
1
Walker, pg 31.
Example 2.5
(b) How far has the boat traveled in this time?
Example 2.5
(b) How far has the boat traveled in this time?
∆x
= vavg t
Example 2.5
(b) How far has the boat traveled in this time?
∆x
= vavg t
=
=
1
(v0 + v)t
2
1
(1.50 m/s i + 13.5 m/s i)(5.00 s )
2
= 37.5 m i
Example 2.5
L KINEMATICS
(b)
How far has the boat traveled in this time?
e
al
v
v = 13.5 m/s
Total area = 30.0 m + 7.50 m = 37.5 m
Area of triangle
= 1– (!v)t
2
!v
v0 = 1.50 m/s
t
5s
O
= 1–2 (13.5 m/s – 1.50 m/s) (5.00 s)
= 30.0 m
Area of rectangle
= v0 t
= (1.50 m/s)(5.00 s)
= 7.50 m
t
The velocity of the boat in Example 2–5 is plotted as a function of time
Figure 2–14, with the acceleration starting at time t = 0 and ending at t = 5.00
∆x
= 37.5 m i
We will now show that the distance traveled by the boat from t = 0 to t = 5.00 s
equal to the corresponding area under the velocity-versus-time curve. This is a gener
Relating Position, Velocity, Acceleration graphs
For a single moving object, the graphs of its position, velocity, and
acceleration are not independent!
The slope of the position-time graph is the velocity.
The slope of the velocity-time graph is the acceleration.
2.6
Analysis Mod
If the acceleration of a parti
to analyze.Constant
A very common
Under
Ac
2.6 Analysis
Mod
x
Constant Acceleration
Graphs
xx
i
xi
Slope ! vxf
Slope ! vx i
Slope ! vxf
t
aSlope ! vx i
xi
vSlope
x
! vxi
a
vx
a
vx i
vx
Slope ! ax
vxi
t
t
t
t
axt
Slope ! ax
avxtx i
Slope ! ax
vx i
t
b
t
t
abx
vaxxit
vx i
Slope ! 0
ax
Slope ! 0
vx f t
vx f t
t
t
ax
b
vx f
ax
Slope ! 0 t ax
t
that in which the accelerat
If the acceleration of a particl
a
over any time interval
to x,avg
analyze. A very common a
at any instant within the int
that
which the acceleratio
Ifout
thein
acceleration
of a situat
partic
the
motion.
This
ato
over any
interval isa
analyze.
A time
very common
x,avg
model:
the particlethe
under
at
anyininstant
inter
that
whichwithin
the acceleratio
generate
several
equations
out
the
motion.
This
situatio
ax,avg over any time interval i
If we
replace
ax,avg
by aco
xi
model:
the
particle
under
at any instant
within
the inte
t, we find
that equations th
generate
several
out the motion. This situatio
If we the
replace
ax,avgunder
by ax inc
model:
particle
t,generate
we find several
that equations th
orIf we replace ax,avg by ax in
t, we find that
v
or
This powerful expression
vxf e
t if we know the object’s
or
velocity–time
graph for thi
This
powerful expression
en
vxf
a straight
t The
if wegraph
knowisthe
object’sline
in
velocity–time
graph
for this
c
slopepowerful
is consistent
with
axen
5
This
expression
The
graph
is
a
straight
line,
t
tive,
which
indicates
a
pos
t if we know the object’s in
slope
with
athis
slopeisofconsistent
the line
in
Figure
x 5
velocity–time
graph
for
tive,
which
positi
stant,
the indicates
graph
of aacceler
The
graph
is a straight
line,
Under Constant A
CHAPTER 2 | KINEMATICS
Falling Objects
Figure 2.40 Vertical position, vertical velocity, and vertical acceleration vs. time for a rock thrown vertically up at the edge of a cliff. Notice that velocity changes linearly
with
1 time and that acceleration is constant. Misconception Alert! Notice that the position vs. time graph shows vertical position only. It is easy to get the impression that
the graph shows some horizontal motion—the shape of the graph looks like the path of a projectile. But this is not the case; the horizontal axis is time, not space. The
OpenStax Physics
65
Acceleration vs. Time Graphs
tangent
and t 5
mly, so
re 2.8b.
is cont " and
value of
he slope
at that
t graph
ase unihe slope
to zero.
, mean-
the tanof accelure 2.8c.
n 0 and
x
a
t!
t" t&
t#
t$ t%
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t" t&
t#
t$ t%
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t
Summary
• graphing velocity vs time
• acceleration vs time
Homework
Walker Physics:
• PREV: Ch 2, onward from page 47. Probs: 9, 13, 21, 23, 37,
41
• NEW: Ch 2. Probs: 43, 47