PHY 113 A General Physics I
9-9:50 AM MWF Olin 101
Plan for Lecture 3:
Some announcements
Chapter 2 – Motion in one dimension
1. Acceleration
2. Relationships between position,
velocity, and acceleration
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Some updates/announcements
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Starting September 3, 2012
Schedule for Physics 113 Tutorials
5-7 PM in Olin 101
Sunday
Monday
Tuesday
Wednesday
Thursday
Friday
Loah Stevens
Jiajie Xiao
Jiajie Xiao
Stephen Baker
Stephen Baker
Loah Stevens
First Webassign sets “due” on Friday, Sept. 7th
PHY 113 Labs start September 3, 2012
(Please see Eric Chapman in Olin 110
[email protected] for all of your
laboratory concerns)
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Velocity
Instanteous velocity :
dx
v(t ) 
dt
Average velocity :
v
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B
A
x(t B )  x(t A )

tB  t A
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Instantaneous velocity using calculus
Suppose:
x(t )  5  6t  7t 2  2t 4
dx
v(t ) 
 6  14t  8t 3
dt
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Anti-derivative relationship
Constant velocity motion
dx
Suppose :
 v0
dt
Then :
x(t )  x0  v0t
Example - - suppose x0  0 and v0  0.3 m/s :
x
t
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Acceleration
Instanteous acceleration :
dv d dx d 2 x
 2

a (t ) 
dt dt dt dt
Average acceleration :
a
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B
A
v(t B )  v(t A )

tB  t A
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Rate of acceleration
Instanteous rate of acceleration :
da d dv d 2 v d d dx d 3 x

 2 
 3
dt dt dt dt
dt dt dt dt
Instanteous rate of rate of acceleration :
d da d 2 a d 2 dv d 3v d 3 dx d 4 x
 2  2
 3  3
 4
dt dt dt
dt dt dt
dt dt dt
iclicker exercise
How many derivatives of position are useful for describing motion:
A. 1 (dx/dt)
B. 2 (d2x/dt2)
C. 3 (dx3/dt3)
D. 4 (dx4/dt4)
E. 
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Anti-derivative relationship
Constant acceleration motion
dv
Suppose :
 a0
and v( 0 )  v0 , x( 0 )  x0
dt
Then :
v(t )  v0  a0t
x(t )  x0  v0t  12 a0t 2
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Examples
v(t )  v0  a0t
x0  0
x(t)  x0  v0t  12 a0t 2
v0  5m/s
a0  9.8 m/s 2
v(t)
x(t)
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Examples
v(t )  v0  a0t
x0  0
x(t)  x0  v0t  12 a0t 2
v0  5m/s
a0  9.8 m/s 2
v(t)
x(t)
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Summary
dx
v(t ) 
dt
dv
a (t ) 
dt
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t

x(t )   v(t ' )dt '
t0
t

v(t )   a (t ' )dt '
t0
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Another example
x(t)
v(t)
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From webassign:
velocity at t  5s :
v(5s )  2m / s 2  5s  10m / s
position at t  5s :
1
2
2
x(5s )  2m / s  5s   25m
2
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Special case: constant velocity due to earth’s gravity
In this case, the “one” dimension is the vertical
direction with “up” chosen as positive:
a = -g = -9.8 m/s2
y(t) = y0 + v0t – ½ gt2
y0 = 0
v0 = 20 m/s
At what time tm will the ball hit the ground
ym = -50m ?
Solve: ym = y0 + v0tm – ½ gtm2
quadratic equation
physical solution: tm = 5.83 s
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