Ch 2. Motion in a Straight Line
One Dimensional (x-axis only)
“Dinophysics : VelocityRaptor”
Definitions
1. Kinematics - Motion Motion in 3-D space can be complicated
Kinetic Energy - Energy associated with motion
2. Motion in physics is broken down into categories
Straight Line
a.) Translational Motion - motion such that an object moves
from one position to another along a straight line.
Spinning
b.) Rotational Motion - motion such that an object moves from
one position to another along a circular path.
Up and Back
c.) Vibrational Motion - motion such that an object moves back
and forth in some type of periodicity.
Example: Diatomic Molecule Moving Through Space.
Vibrational
f
Rotational
i
X - Dir Translational
Note: In this chapter all objects are going to be considered POINT
PARTICLES – No Spatial Extent – No Rotations – No Vibrations
Speed
1.
Speed - How fast an object is moving regardless of what direction it is
moving.
≡ Distance Traveled
Speed =
Change in time
Equality by Definition
Example 1 Traveling from your parking space at Conestoga to New York
City and back to Conestoga. Find your average speed for the round trip.
y(mi)
One way travel = 130 mi.
Total Distance Traveled = 260 mi.
Total time elapsed = 5.2 hrs. or (5 hrs 12 min)
back
NY
up
Speed Calculations are EASY
Always distance / time
Round Trip Average Speed
vspeed
260miles

 50 miles
hr
5.2hr
0 Conestoga
97 x(mi)
Displacement - Change in position (straight line distance with direction)
Must specify a coordinate system.
Example: Cartesian coordinate system
y(m)
xi = xinitial= Initial Position
xf = xfinal = Final Position
up
back
∆ “Delta”
xi= 2m
xf= 6m
x1
x2
x(m)
x = Change in x = x  x  6 m- 2 m = + 4 m x
f
i
Delta x is the displacement or change in the x position
Mathematical
Notation for
Direction
Average Velocity
Avg. Velocity - How fast an object is moving and in what direction it is moving.
Average Velocity =≡ Change in position
Change in time
Equality by Definition

vavg
x x f  xi


t t f  ti
Notation for Displacement & Velocity
x “hat”, and has a value of one.
x̂ =direction
The sole purpose of
x̂ is to indicate the
Example Problem:
A particle initially at position x = 5 m at time t= 2 s moves to position x = -2 m and
arrives at time t = 4 s.
a.) Find the displacement of the particle.
b.) Find the average speed and velocity of the particle.
Given:
xi  5m @ ti  2s
x f  2m @ t f  4s
b.) vspeed 
vspeed
a.) x  x f  xi
x  [2 xˆ  (5 xˆ )]m  7 m xˆ
distance
t
7m

 3.5 m
s
2s
y(m)
xi
xf
vavg
vavg
x

t
7 m xˆ

 3.5 m xˆ
s
2s
x
x(m)
Example Problem 1 revisited
Example 1. Traveling from your parking space at Conestoga to New York City
and back to Conestoga. The straight line distance from Conestoga to Y is 97 mi.
y(mi)
One way travel = 130 mi.
back
Total Distance Traveled = 260 mi.
NY
up
Travel time Con. to NY = 2.6 hrs.
0 Conestoga
Travel time NY to Con. = 2.6 hrs.
a.) What was the avg speed
from Conestoga to NY?
vspeed 
130miles
2.6hr
97 x(mi)
 50 miles
hr
ˆ
b.) What was the avg velocity from Conestoga to NY?
c.) What was the avg speed for the round trip?
d.) What was the avg velocity for the round trip?
vavg  972.6mihrx  37.3mi xˆ
vspeed 
vavg 
hr
260miles
5.2hr
0mi xˆ
5.2hr
 50 miles
hr
 0 mi
Speed in PATH DEPENDENT.
Velocity is PATH INDEPENDENT. It only depends on the initial and final
positions.
hr
Scalar vs. Vector Quantities
Scalar - Quantity that has magnitude only.
- Mass
- Length
A number (with units) that
describes how big or small
- Speed
- Energy
Vector - A quantity that has both magnitude and direction.
- Position
- Acceleration
- Velocity
- Forces
Example: Length vs. Position
Scalar
Vector
Pt. B
Pt. A
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6
A  3m xˆ
A  3m
x(m)
B  4m xˆ
B  4m
x̂ = x “hat”, and is called a unit vector in the x-direction.
It has a
magnitude of one (hence the name unit) and is used solely to specify
direction.
Concepts Check – The Negatives
Q. Can speed be negative?
A. NO! – The least speed an object can have is zero – it is at rest
Q. Can velocity be negative?
A . YES! Negative velocity means an object is moving backwards.
E.g. An object moving -10 m s xˆ is moving backwards with a speed of 10 m/s
Q. Can distance be negative?
A. NO! – The least distance an object can move is zero – it is at rest
Q. Can displacement be negative?
A . YES! Negative displacement means an object moved backwards.
E.g. An object with a displacement ∆x of -10 m xˆ moved backwards 10m.
Both of these movements
describe an object moving in
one dimension along the x-axis!
NOT up and to the right!
Position vs. Time Graph
x
Y1
Y2
Time (s)
Position
(m)
Position
(m)
0
0
0
1
1
5
30
2
4
10
25
3
9
15
4
16
20
5
25
25
X - Position (m)
Position vs. Time
20
10
x=rise
5
t=run
0
0
Movement 1
1s 2s
●
●
0
1
2
3
4
5
6
Time (sec)
3s
●
4s
●
5s
●
5m
10m
15m
20m
25m
●
1s
●
2s
●
3s
●
4s
●
5s
Movement 2
Movement 1
Movement 2
15
For any time interval
vavg 
x
t

rise
run
 slope
SLOPE is Avg. Velocity
Position vs. Time Graph for a Complete Trip
x
y
Time (s)
Position (m)
0
0
10
200
20
200
25
150
45
-100
60
0
300
250
B
C
Position (m)
200
D
150
100
50
F
A
0
Find the average velocity as the
object moves from:
a.) A to B
b.) B to C
c.) C to D
d.) A to E
-50
-100
E
-150
0
a.) vAB 
10
x
t
c.) vDE 

x
t
20
 2000m
100s

30
Time (s)
40
50
60
 20 m  Slope b.) vBC 
150 200m
 25 20s
s
 10 m
s
x
t
d.) vAE 

x
t
 0  0 m
 2010s

 0, Stopped
 1000m
 450s
 2.22 m
s
Velocity vs. Time (Constant Velocity)
Position Function
t run
8
7
x(m)
6
x
rise
v


 Slope
avg
t run
5
4
x rise
3
2
1
0
0
1
2
3
4
5
t(sec)
Slope
Area
Velocity Function
4
x  v(t )  height (base)  area
v (m/s)
3
∆t = base
2
x
1
v = height
0
0
1
2
3
t(sec)
4
5
Velocity vs. Time Graph for a Complete Trip
300
25
250
B
Position (m)
200
20
C
150
100
50
F
A
0
B
Area =
200m
10
5
E
B
0
C
-5
Area =
-250m
-50m
C
D
E
D
-15
-100
F
Area =
-10
-50
− X -150
A
15
D
Velocity (m/s)
+X
-20
E
-25
0
10
20
30
Time (s)
40
50
60
0
10
20
30
Time (s)
40
50
60
Instantaneous Velocity
vavg 
recall:
x x  t f   x  t i 

xˆ
t
tf  ti

m 2 
m 4
x t  3 m +  10
t -  0.5

 t


s2 
s4 
Consider the function x(t):
(Average velocity)
A.
vavg 
50.5m - 35.0m
 10.3 m xˆ
s
3.5s - 2.0 s
v avg 
39.7 m - 35.0m
 23.5 m xˆ
s
2.2s - 2.0 s
B.
∆t = 1.5 sec
∆t = 0.2 sec
60
●
50
x(m)
40
●
●
30
20
10
0
0
1
2
3
4
t(sec)
The instantaneous velocity at the time t = ti is the limiting value we get by letting the upper
value of the tf approach ti.
Mathematically this is expressed as:

 X t f   X t i  
dX t

v t 
 lim 

dt
t f  ti
t f  ti 

The velocity function v t is the time derivative of the position function X t .
Differentiation (Calculus)
Acceleration
When the instantaneous velocity of a particle is changing with time, the particle is accelerating
v v

v
aavg   f i xˆ
t tf  ti
Units:
(Average Acceleration)
a avg  m/s  m
s s2
Example: If a particle is moving with a velocity in the x-direction given by


m 
v(t)   3 3  t 2


s 
a.) What is the average acceleration over the time interval 6 s  t  12 s

v f  vi
aavg 
v
t
aavg 
3(12)2 3(6)2
(12  6)
t f ti

324 m s
6s
 54 m 2
s
Example: Instantaneous Acceleration
time (s) vel. (m/s)
Velocity vs. Time
0
1
2
3
4
5
6
7
8
35
velocity (m/s)
30
25
20
15
10
5
0
-5
-10
Stopped
-15
0
2
4
6
8
10
time (s)
a.) Find aavg. over the time interval 5  t  8
v
3014
m
avg
t
8 5
2
a


 5.3
s
b.) What is the acceleration at time t = 6 s ?
Slope of tangent – pick 2 points on the tangent line.
Answer will we smaller than the answer to part a, (5.3 sm2 ).
c.) What is the acceleration when the velocity of the particle is
zero? Slope  0
Objects with zero velocity can be accelerating!
-10
-2
-5
5
12
14
12
21
30
Positive and Negative Accelerations
v(m/s)
D
C
Moving
Forward
Stopped 0
Moving
Backward
B
E
A
t (s)
F
A→B: Slowing Down, Moving Backward, Pt. B=Stop
v  ()negative
a  slope  () positive
B→C:
Speeding Up, Moving Forward
C→D:
Constant Speed, Moving Forward
D→E:
Slowing Down, Moving Forward, Pt. E=Stop
E→F:
v  () Positive
v  () positive
v  () Positive
a  slope  () positive
a  slope  ZERO
a  slope  ()negative
Speeding Up, Moving Backward
v  ()negative
a  slope  ()negative
Special Case: Constant Acceleration
We make the assumption that the acceleration does not change.
Near the surface of the earth, (where most of us spend most of our
time) the acceleration due to gravity is approximately constant ag =
9.8 m/s2
a(m/s2)
a
a
v
0
ti = 0
v f  vi  a(t f  ti )
tf = t
t (s)
vf
x1
 x2
tf = t
1.
y  b  mx
x  x1  x2
x  vi t  12 t (v f  vi )
t (s)
Area!
Slope!
x  vi t  12 t (at )
x(m)
xf
xf = xi + v i t
xi
ti = 0
vf = vi + a t
Slope!
v(m/s)
0
ti = 0
 v  at  height (base)
0
Area!
vi
v
t
tf = t
t (s)
+
1 2
at
2
2.
Solving for the 3rd constant acceleration equation
Solve equation 1 for t and substitute t into equation 2 to get the following equation.
v 2f  vi2  2 a x
v f  vi
a
x 
v f vi vi2
a
x  vi t  at
t
x  vi
3.
1
2
 
v f  vi
a

1
2a
 12 a
v
2
f
 
v f  vi
2
2
a
 2v f vi  v
2
i

2ax  2v f vi  2v  v  2v f vi  v
2
i
2
f
2ax  v  v
2
i
v  v  2ax
2
f
2
i
2
i
2
f
FREE-FALL ACCELERATION
(9.8 m/s2 = 32 ft/s2)
Consider a ball is thrown straight up.
It is in “Free Fall” the moment it leaves you hand.
Plot y(t) vs. t for the example above.
Why? Because
y(t)
y f  yi  viy t  12 at 2
a  9.8 sm2
from the moment it leaves
your hand.
v0 t 
t/2
top
Moving
up
Plot v(t) vs. t
tf
2
tf
v f  vi  at
Stopped 0
t/2
Moving
down
tf
 vi
vi
Ground
Level
FINAL NOTES ON CH 2.
Remember , when going between the following graphs
x(t)
Slope
v(t)
Area
Under
Curve
a(t)
Problem Solving with the constant acceleration equations
1.Write down all three equations in the margin
2.a =  9.8 m/s2 for free fall problems
3.Analyze the problem in terms of initial and final sections.