Announcements
• Make up “Math Skills Test” Wednesday 7:30 am in
HL G59.
• No class after 5pm Thursday April 1.
– Instead of Lab that week, Statistics Quiz.
– Can be taken during any scheduled lab time except after
5pm Thursday.
• Reminder do Pre-Lab for lab 1 Before Lab.
• Mastering Physics assignment due Monday (1/18),
next due Saturday (1/23).
• Bring your “Student Workbook” to your discussion
section.
Kinematics in One Dimension
Uniform Motion
In one dimension there is only one direction (+,-) and so
We can drop the “Vector” and I will call that direction x.
 avg
x

t
Some distance traveled
= The time interval to travel that distance
Time
Odometer
0.0 s
60.0 s
120. s
180. s
240. s
66214.3mi SW
66215.4mi SW
66216.3mi SW
66217.2mi SW
66218.3mi SW
Average Velocity
 avg 
x

t
One
Dimension
66218.10mi  66215.35mi 2.75mi
233s  63s


∆x
∆t
170s
 0.0162 mis
Instantaneous Velocity
2
7 mi 3
x s (t)  66214.3mi  0.0211 mis t  5.5 105 mi

1.54
10
t
2 t
s
s3
v s (t)  0.0211 mis 11.0 105 mi
t  4.62 107 mi
t2
s2
s3

x dx
 s  lim

dt
t 0 t


x avg (t)  66214.3mi  0.0162 mis t
v avg (t)  0.0162 mis
Special Cases
Constant Position
x s (t)  3.4m
dx
v s (t) 
 0.0 ms
dt
d d 2 x
as (t) 
 2  0.0 sm2
dt dt
Special Case
Constant Velocity
x s (t)  0.0m  0.2125 ms t
dx
 s (t) 
 0.2125 ms
dt
d d 2 x
as (t) 
 2  0.0 sm2
dt dt
Special Case
Constant Acceleration
Not a straight Line
x s (t)  0.0m  0.2125 ms t  0.002125 sm2 t 2
dx
 s (t) 
 0.2125 ms  0.00425 sm2 t
dt
d d 2 x
as (t) 
 2  0.00425 sm2
dt dt
One dimensional motion
problem
a dna ™em iT k ciuQ
ro s se rpmo ced
.eru t cip sih t ee s ot dedeen e ra
x a (t)  70mi  1.0 mis t
QuickTi me™ and a
decompressor
are needed to see t his pict ure.
Xa=-70 mi
x b (t)  6mi  0.7 mis t
Xb=+6 mi
a=+1.0mi/s
At t  t then x a ( t )  x b ( t )
b=-0.7mi/s
70mi  1.0 mis t  6mi  0.7 mis t 
1.7 mis t  76mi
76mi
t 
 44.7s or
mi
1.7 s
x a ( t )  25.29mi  x b ( t )
General Relations
x(t)  x 0   0 t  12 at 2
dx(t)
 (t) 
  0  at
dt
d (t) d 2 x(t)
a(t) 

a
2
dt
dt

x(t)  x 0  x1  x 2 
a(t)  a
 (t) 
x(t) 
t
 a(t )dt 
t0
t
  (t )dt 

a(t)  a

 (t)   0  at
 x(t)  x 0   0 t  12 at 2
t0
n
t
 x n   s (t k )t 
  s ( t )dt 
k 0
lim t 0
t0