Chapter 2: Motion in One Dimension
For now, our “particle” or “object” or “body”
is represented by a moving single point in space
(Later in the course, we will consider “systems” of
particles, and extended bodies)
Our questions are, for now, simple:
• Where is the body?
• When is it there?
Where: position! (in meters, m)
When: time! (in seconds, s)
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Curvilinear motion can take place in 2 or 3
Dimensions, but for now we assume the motion
is strictly along a straight line!
• Where is the body?
• Where it is: position x, where x is
a space displacement from the space origin x = 0
• When is the body there?
When it’s there: time t, where t is
a time interval from the origin of time t = 0
Soon, we’ll ask much more interesting
questions about the body’s motion!!
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What kind of ‘representations’ of motion?
Motion diagram: qualitative and pretty, like a
movie of the motion
Graphical Representation of x as a function of
t: x(t): quantitative, visually informative to the
trained eye, if labeled carefully
{show active figure 0201}
Tabular Representation of t and x: include
sufficient number of pairs of values to be
informative
Explicit functional form for x(t): nice for
the theorists, and very handy for calculating
numbers. Not always obtainable!
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Note: this is a piecewiseContinuous function: OK!
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A tabular representation of this motion
Position x [km]
10.0
0.0
–10.0
–20.0
–20.0
–15.0
–10.0
Clock time
10:00 am
10:10 am
10:20 am
10:30 am
10:40 am
10:50 am
11:00 am
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Time t [min]
0.0
10.0
20.0
30.0
40.0
50.0
60.0
An explicit function for x(t)
x(t )  20 m  (10 m/s) t OR x(t )  20  10t
x [m]
80
60
40
20
0
-6
-4
-2
0
2
-20
-40
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4
6
t [s]
 (45 /s) t 
x(t )  5 m sin 

 90 
6
x [m]
4
2
0
Series1
0
200
400
-2
-4
-6
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600 t [s]
A grab-bag of mathematical considerations
• if two things in an expression are added or subtracted,
they are called terms and their dimensions must agree
• you must make their units agree too when it comes to
putting in actual numbers
• the result will have the same units
• if two things in an expression are multiplied or divided,
they are called factors and their dimensions may differ
• the result will have units that obey the same algebra of
multiplication/division
• in complicated unit algebra, whatever is on the top of the
top, and the bottom of the bottom, is actually on the top
• whatever is on the bottom of the top, or the bottom of the
top, is actually on the bottom
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The velocity concept answer new questions…
• How fast does object move during a time interval?
• How fast does the object move at any time t?
Crucial notion of the CHANGE in a quantity:
The Delta operator D
• Time interval Dt := t2 – t1 or Dt := tlater – tearlier or
Dt := tfinal – tinitial or Dt := tB – tA ….
• Time interval has the same dimensions as t
• Often, tinitial = 0 s and tfinal = t (‘present’ time)
so if that is the case then Dt = t
• Change in position Dx is called displacement
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Definition of the average velocity
average velocity during a time interval Dt is defined as
change in position displaceme nt Dx
vavg :


change in time
time interval Dt
• vavg is the slope of the line connecting A and B!!
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Definition of the (instantaneous) velocity
• Suppose we want to know the velocity v AT A
• Allow the time
interval to shrink
by letting tB  tA
• The displacement
also shrinks, as
xB := x(tB)
approaches
xA:=x(tA)
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Definition of the
(instantaneous) velocity
• In the limit of
infinitesimal time
interval Dt and
displacement Dx, we
have the
derivative of x(t):
 Dx  dx(t )
v(t )  lim   :
Dt 0 Dt 
dt
 the derivative of x with respect to t
[Show active figure 0203]
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Position–time
graph for a
particle having
an x coordinate
that varies with t
(quadratically) as
x = –4.0 t + 2.0t2
dx
v
 4.0  4.0t
dt
• Note that v(t) is a
linear function if x(t) is
quadratic!
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Properties of the position and the velocity
function
• v(t) is the slope of the tangent line to the graph
of x(t) at time t
• x(t) must be a (piecewise) continuous function
(no gaps or holes, passes vertical line test). Why?
• x(t) may not have any sharp corners, either. If
it had them, what would that imply about v(t)?
• v(t) may be graphed as well. Can it have sharp
corners? Must it be continuous?
• If there is an explicit function x(t), then all of
the tricks from calculus may be used to get v(t)
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Speed as distinct from velocity
distance traveled
average speed :
time for trip
• Average speed does not include information about
direction of motion
• Distance is a loose concept: if you walk to the store
which is 1km away ‘as the crow flies’, and then return
home, your distance may be anywhere from 2 km on up
(depending how you wander), but your displacement is
ZERO!!
speed : magnitude of velocity  v
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Three motion diagrams are shown.
Which is a dust particle settling to the
floor at constant speed, which is a ball
dropped from the roof of a building, and
which is a descending rocket slowing to
make a soft landing on Mars?
A. (a) is ball, (b) is dust, (c) is rocket
B. (a) is ball, (b) is rocket, (c) is dust
C. (a) is rocket, (b) is dust, (c) is ball
D. (a) is rocket, (b) is ball, (c) is dust
E. (a) is dust, (b) is ball, (c) is rocket
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Example Problem
A bicycle moves on a straight path so that its position
as a function of time is x(t) = – 200 t+ 50 t2, where all
quantities are in m and s.
a) What is the average velocity
between t = 2 s and t = 4 s?
b) What are the instantaneous velocities
at those two times?
c) Make graphs of x(t) and v(t)
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The next question: how is the velocity changing?
• Velocity is ‘the rate of change with time’ of position
• So, acceleration is the rate of change with time of velocity
• Average acceleration (during a time interval) is defined as
aavg
change in velocit y Dv


time interval
Dt
• The dimensions of acceleration: Length per time, per time
m
m
m 1 m
s
s
aavg 

   2
s
s
s s s
1
• aavg is slope of line on a graph of v(t) between two points


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The (instantaneous)
acceleration
• At any instant of time t,
the acceleration is the slope
of the tangent to the graph of v(t)
 Dv  dv
a (t ) : lim   
Dt 0 Dt 
dt
d dx d 2 x

 2
dt dt dt
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The full set
of x(t), v(t)
and a(t)
graphs,
obtained by
eyeballing
the slopes
Note how a(t)
may BE
discontinuous!
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Special case: constant acceleration I
• there are two ways to work out the set of equations
that I call ‘the big five’ for this case.
• we assume initial time zero and present time t
• initial position x0 and present position x(t) := x
• initial velocity v0 and present velocity v(t) := v
• acceleration does not change: call it simply a(t): = A
• therefore we have (duh!!) a = A (constant)
(#1)
Dx x  x0
1
 vavg :

but for this case vavg  v0  v 
Dt
t
2
1
so we easily get x  x0  v0  v  t (#2)
2
• This came about because of the LINEAR relation
between
v and t in this special case of constant a
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Special case: constant acceleration II
• From the definition of the acceleration we have
Dv v  v0
 aavg :

 A so we easily get v  v0  A t (#3)
Dt
t
• Now insert #3 for v into #2 for x:
1
 x  x0  ((v0  At )  v0 )t so we easily get
2
1 2
x  x0  v0t  At
(#4)
2
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Special case: constant acceleration III
• Finally, it is nice to get an equation that does not
involve t at all. To get it, solve #3 for t:
v  v0
t
and now insert tha t into #3 in two places :
A
 v  v0  A  v  v0 
x  x0  v0 
 

 A  2 A 
2
• Expanding the squared term and combining gives




1
A 2
2
2
 x  x0  vv0  v0 
v

2
vv

v
0
0
2
A
2A
2
2
v
1 
v
1 2

vv0  v02 
 vv0  0  
v  v02
A 
2
2  2 A


 Rearranged , we see that v 2  v02  2 A( x  x0 )
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(#5)
Extra-special case: zero acceleration
• This is synonymous with constant v = v0 := V
• the big 5 collapse down into simplicity itself:
#1 : A  0
#2 : Since v  v0 : V we get x  x0  Vt
or x  x0  Vt (' distance  rate  time' )
#3 : merely says v  V
#4 : same as #2
#5 : same as #3
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A different book’s version of these equations
• Note that v(t) is a linear function
• Note that x(t) is a quadratic function
• Note that v(x) is a square-root function
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{show active figures 0210-0212
Example problem: Two railroad engines on
parallel tracks are both moving at 25 m/s.
Engine A applies its brakes and comes to a
stop in 100 seconds (assume constant
acceleration), while engine B continues at
constant velocity.
a) What is the acceleration of both engines?
b) How far do both engines travel during that
time interval?
c) Make three graphs of the three kinematical
quantities, using color to distinguish the
two engines’ graphs.
d) Convert 25 m/s to mi/hr.
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Example problem: the position function of
an object is given by
x(t) = – 12.5 + 5t – 15 ln (2t)
where all units are m and s.
a) Find an expression for v(t) by taking a
derivative.
b) Find an expression for a(t) by taking
another derivative.
c) Graph all three functions, for the time
domain 0.5 s< t < 10.0 s.
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x(t )  12.5  5t  15 ln 2t 
position
120
100
80
60
position
40
20
0
0
2
4
6
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8
10
12
dx(t ) d
d (t )
d (ln( 2t ))
15 d (2t )
15
v(t ) 
 (12.5)  5
 15
 05
5
dt
dt
dt
dt
2t dt
t
velocity
40
35
30
25
20
velocity
15
10
5
0
0
2
4
6
8
10
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12
dv(t ) d
d (t 1 )
15
15
a(t ) 
 (5)  15
0 2   2
dt
dt
dt
t
t
acceleration
0
0
2
4
6
8
10
12
-10
-20
-30
acceleration
-40
-50
-60
-70
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Free fall in gravity near the Earth
It is known (thanks to Galileo and his contemporaries)
That near the Earth, IN THE ABSENCE OF ANY OTHER
FORCES BUT GRAVITY, ALL OBJECTS
ACCELERATE DOWNWARD AT THE SAME RATE:
A = – 9.81 m/s2
Our convention is to say that A = – g, where
g = 9.81 m/s2
Thus, our 1d coordinate system, called y now, has
the positive y direction pointing UP
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What happens to the big five?
We insert a subscript on velocity, to indicate y
A  g
v y  voy  gt
1 2
y  y0  voyt  gt
2

2
v 2y  voy
 2 g  y  y0 
There is nothing fundamentally different here from
motion with uniform acceleration along x!!!


1
y  y0  v y  voy t
2

 9.81 m/s 2 3.28 ft/m   32.2 ft/s 2
g 
  10 m/s 2 (close enuf....)
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A Table of Free-Fall Kinematics
Here, Dynow = ylater – ynow
Dy [m]
vy [m/s]
Dvy [m/s]
ay[m/s2]
50.0
– 1.25
0.0
– 5.0
–g
1 0.50 48.75
– 3.75
– 5.0
– 5.0
–g
2 1.0
45.0
– 6.25
– 10.0
– 5.0
–g
3 1.5
38.75
– 8.75
– 15.0
– 5.0
–g
4 2.0
30.00
– 11.25
– 20.0
– 5.0
–g
5 2.5
18.75
– 13.75
– 25.0
– 5.0
–g
6 3.0
5.0
X
– 30.0
– 5.0
–g
t [s] y [m]
0 0.0
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Example problem: An object is thrown
directly upward from the ground at 15 m/s. It
climbs vertically, stops for an instant, and
returns to the ground.
a) How high does it climb? How much time
Is required to reach maximum height?
b) What are the times for it to be at half of
the maximum height? How fast is it
moving when it is at that height?
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To find how high it climbs, note that vy = 0 at the top.
Use #5 (and for simplicity take g = 10 m/s2):
v 2y  vo2  2 g  y  y0   0  vo2  2 g ( y )
v02 (15 m/s) 2 225 m 2 /s 2
y


 11.25 m
2
2
2g
20 m/s
20 m/s
To find the time needed, one could use the above
result and plug into #4… but that requires one to solve
a quadratic equation for t. Instead, just use #3:
v y  v0 y  gt  0  v0 y  gt
t 
v0 y
g

15 m/s
10 m/s
2
 1.5 s
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• #2 also can be
used with the
above result!!
• total flight time
is 3 seconds!
To find how the time to be at half of the maximum height,
(which is y = 5.625 m exactly), one can either first
find how fast it is moving at that height (using #5) , or
resort to #4 and bite the quadratic bullet:
y  y0  v0 y t 
1 2
1
gt  gt 2  v0 y t  y  0
2
2
g
t 
 (voy )  ( v0 y ) 2  4( 2 )( y )
g
2( 2 )
15  ( 15) 2  4(5)(5.625 )

2(5)
15  225  112 .5 15.00  10.61  .44 s



10
10
2.56 s
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