Motion
If you want to keep something from
moving…
…you need to understand what would
make it move!
Physical Principles of Design
Class 02, Thursday, 15 January, 2004
Motion in One Dimension
•
•
•
•
•
Position(x) : Measured in Meters.
Velocity(v):Measured in m/sec.
Acceleration(a):Measured in m/sec.sec.
Be Careful of Signs.
Note:”Speed”=Absolute value of “v”.
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
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.
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
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
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.
Velocity and Acceleration
We will start with 1D motion. We will deal with the
X components of velocity and acceleration, v and a.
a is the slope of the graph of v versus t (time).
slope = a
v
t
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
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
x  x 0  v 0 ( t  t 0 )  12 a ( t  t 0 ) 2
v
v0
t0
t
Class #2
Take-Away Concepts
1D Equations of Motion for Constant Acceleration
Basic Equations
1. v  v 0  a  t  t 0 
2. x  x 0  v 0 ( t  t 0 )  2 a ( t  t 0 )
1
2
Derived Equations
1
x

x

( v 0  v)(t  t 0 )
3.
0
2
1
2
x

x

v
(
t

t
)

a
(
t

t
)
4.
(compare with 2.)
0
0
0
2
5. v  v 0  2a x  x 0 
2
2
Class #2
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
Class #2
Problems of the Day
2. A construction worker on a new high-rise building drops a
wrench and it falls 78.4 m to the ground. How fast is it moving
when it hits, assuming no air resistance? Use g = 9.8 m/s2.
Class #2 Optional Material
Deriving the Other Equations
Equation 3:
3a. Solve 1 for a:
v  v0
a
t  t0
3b. Substitute 3a into 2 and simplify.
Equation 4:
4a. Solve 1 for v0: v 0  v  a ( t  t 0 )
4b. Substitute 4a into 2 and simplify.
Equation 5:
5a. Solve 1 for (v-v0):
v  v0  a (t  t 0 )
x  x0
5b. Solve 3 for (v+v0): v  v 0  2
t  t0
2
v
5c. Multiply 5a by 5b and bring 0 to r.h.s.