Physics I
Class 01
1D Motion
Rev. 04-Jan-07 GB
01-1
Definitions
Scalar:
Magnitude:
Vector:
A number – positive, negative, or 0.
Absolute value – positive or 0.
Magnitude (or length) and direction
in space.
Time:
Position:
Displacement:
t(scalar)
x (vector)
 
x  x  x 0
Time interval: t  t  t 0
Average or mean velocity is defined as follows:

v avg
 

x  x 0 x


t  t0
t
01-2
Definitions (Continued)
Instantaneous velocity or just “velocity”:


x d x

v  lim

t 0 t
dt
Example: When you take a car trip, you get the

magnitude of v avg by dividing the change in the
odometer (or distance)
 by the hours you drove. You
get many values of v during the trip by checking the
speedometer moment by moment.



v
v

v
If is constant: avg
01-3
Definitions (Continued)
Average acceleration is defined as follows:

a avg
 

v  v 0 v


t  t 0 t
Instantaneous acceleration or just “acceleration”:


2 
v d v d x

a  lim


t  0  t
dt dt 2
01-4
Definitions (Continued)
Beware: The English word “acceleration” does not
have the same meaning as the physics word. In
physics, any change in the velocity vector is an
acceleration!
Some Additional Physics I Terms:
Speed:
Speed Up:
Slow Down:
Magnitude of velocity vector.
Any time the velocity vector’s
magnitude increases.
Any time the velocity vector’s
magnitude decreases.
01-5
Components of Vectors
Any vector can be written in component form:

a  aî  bĵ  ck̂
where î , ĵ, k̂ are unit vectors in the X, Y, and Z
directions respectively. a, b, c are components.
(Sometimes you will see x̂ , ŷ, ẑ unit vectors.)
The components of a vector are scalars. They can be
positive, negative, or zero.

a  aî
In one dimension:
The magnitude of a one-dimensional vector is the
absolute value of its component: |a|.
If a is negative, the vector points in the negative X
direction.
01-6
Velocity and Acceleration
We will start with 1D motion. We will deal with the X components
of displacement, velocity and acceleration: x, v and a.
“When in doubt, draw a graph of velocity versus time.”
slope = a
v
area = x
(displacement)
t0
t1
t
01-7
Constant Acceleration
For the special case of constant acceleration, the
graph of v versus t is a straight line. The equation is
v  v 0  a t  t 0 
This is the same equation you had in math class for a
line – [ y  m x  b] – but with different symbols.
v
slope = a
v0
t0
t
01-8
Displacement with
Constant Acceleration
Math Fact: Because velocity is the derivative of
displacement, displacement is the area (integral)
under the graph of v versus t.
displacement = area = rectangle + triangle
rectangle:
height  base  v 0 ( t  t 0 )
triangle:
1
height  base 
2
1
(v  v0 )  (t  t 0 ) 
2
1
[a ( t  t 0 )]  ( t  t 0 )
2
( x  x 0 )  v 0 ( t  t 0 )  12 a ( t  t 0 ) 2
v
v0
t0
t
Important:
Check Section 2-10
in the book (pg. 27).
x  x 0  v 0 ( t  t 0 )  12 a ( t  t 0 ) 2
01-9
Class #1
Take-Away Concepts
1D Equations of Motion for Constant Acceleration
Basic Equations
1. v f  v 0  a t f
 t0 
1
2
2. x f  x 0  v 0 ( t f  t 0 )  2 a ( t f  t 0 )
Derived Equations
1
x

x

( v 0  v f )( t f  t 0 )
3. f
0
2
4. x f  x 0  v f ( t f  t 0 )  2 a ( t f  t 0 ) (compare with 2.)
1
2
2
v

v
5. f
0  2a  x f  x 0 
2
01-10
Class #1
Problems of the Day
_______1.
Which one graph below represents a motion for which it would be incorrect to
use equations 1-5 to solve a one-dimensional motion problem – even if you
broke the motion into two time intervals? Note: The respective graphs are
straight line segments and v = velocity, a = acceleration.
v
A)
v
t
B)
a
C)
t
a
t
D)
t
01-11
Class #1
Problems of the Day
2. The Faster and the Furiouser
Two teams of students, one from RPI and one from MIT, agree
to a special kind of drag race. The cars will begin by
accelerating in a straight line at a constant acceleration, but at a
certain point (which they separately calculate) they will use
engine braking to slow down at a constant acceleration. The
rules state that each car must reach the finish line with exactly
zero speed. The RPI car begins at +5.0 m/s2 and slows down at
–2.5 m/s2. The MIT car begins at +6.0 m/s2 and slows down at
–2.0 m/s2. They both start at the same time. The team whose
car reaches the finish line first wins the other team’s car. The
finish line is 2000 m from the start. Who wins?
01-12
Activity #1
Software Loading / 1D Motion
Objectives of the Activity:
1. Making sure Physics I software is installed and
working correctly on your laptop.
2. Understanding the basic operation of the motion
detector, cart, and track. (We will use these a lot.)
3. Learning general rules and guidelines that will
apply to all Physics I activities.
4. Review of 1D motion.
01-13
Optional Material
at the End of the Lecture Notes
At the end of most lecture notes, there will
be a section of extra material. This is for the
interest of students who would like to get
some additional depth from the course.
We will not be testing on this material and
you are free to skip it.
01-14
Class #1 Optional Material
Deriving the Other Equations
Equation 3:
3a. Solve 1 for a:
vf  v0
a
tf  t0
3b. Substitute 3a into 2 and simplify.
Equation 4:
4a. Solve 1 for v0: v 0  v f  a ( t f  t 0 )
4b. Substitute 4a into 2 and simplify.
Equation 5:
5a. Solve 1 for (vf–v0):
vf  v0  a (t f  t 0 )
xf  x0
5b. Solve 3 for (vf+v0): v f  v 0  2
tf  t0
2
v
5c. Multiply 5a by 5b and bring 0 to r.h.s.
01-15