kinema tic 2
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KinemaTicTRANSCRIPT
Physics II: Lecture 1, Pg 1
AP Physics AP Physics “Mechanics for Physicists and Engineers”“Mechanics for Physicists and Engineers”
Agenda for TodayAgenda for Today 1-D Kinematics (review).1-D Kinematics (review).
Average & instantaneous velocity and accelerationMotion with constant acceleration
Introduction to calculus applicationsderivatives and slopesIntegrals and area
Physics II: Lecture 1, Pg 2
Kinematics ProblemsKinematics Problems 1-D Kinematics 1-D Kinematics
Average & instantaneous velocity (Chapter2 1,4,5,11-13,15-17) and acceleration (18,21)
Motion with constant acceleration(23,24,27,31,35,37,39,40-(23,24,27,31,35,37,39,40-1,43)1,43)
Free Fall (44,47,49,51,53,56,61,63)Free Fall (44,47,49,51,53,56,61,63) Motion Graphs (66,67,69,70) Review Phun!! Review Phun!!
Physics II: Lecture 1, Pg 3
KinematicsKinematics Location and motion of objects is described using
Kinematic Variables: Some examples of kinematic variables.
position rr vectorvelocity vv vector
Kinematic VariablesKinematic Variables: : Measured with respect to a reference frame. (x-y axis)Measured using coordinates (having units).Many kinematic variables are VectorsVectors, which means
they have a directiondirection as well as a magnitudemagnitude.Vectors denoted by boldface VV or arrow
v
Physics II: Lecture 1, Pg 4
Motion in 1 dimensionMotion in 1 dimension In general, position at time t1 is usually denoted rr(t1).
In 1-D, we usually write position as x(t1 ). Since it’s in 1-D, all we need to indicate direction is + or .
Displacement in a time t = t2 - t1 is x = x(t2 ) - x(t1 ) = x2 - x1
t
x
t1 t2
x
t
x1
x2some particle’s trajectory
in 1-D
See text : 2-1
Physics II: Lecture 1, Pg 5
1-D kinematics1-D kinematics
vx t x tt t
xtav
( ) ( )2 1
2 1
t
x
t1 t2
xx1
x2trajectory
Velocity v is the “rate of change of position” Average velocity vav in the time t = t2 - t1 is:
t
Vav = slope of line connecting x1 and x2.
See text : 2-1
Physics II: Lecture 1, Pg 6
Instantaneous velocity v is defined as:
1-D kinematics...1-D kinematics...
v t dx tdt
( ) ( )
t
x
t1 t2
xx1
x2
t
so V(t2 ) = slope of line tangent to path at t2.
See text : 2-2
Physics II: Lecture 1, Pg 7
1-D kinematics...1-D kinematics...
a v t v tt t
vtav
( ) ( )2 1
2 1
Acceleration a is the “rate of change of velocity” Average acceleration aav in the time t = t2 - t1 is:
And instantaneous acceleration a is defined as:
a t dv tdt
d x tdt
( ) ( ) ( ) 2
2
See text : 2-3
Physics II: Lecture 1, Pg 8
RecapRecap If the position x is known as a function of time, then we can find both velocity v and acceleration a as a function of time!
a dvdt
d xdt
2
2
v dxdt
x x t ( )
x
a
v t
t
t
Physics II: Lecture 1, Pg 9
More 1-D kinematicsMore 1-D kinematics
We saw that v = x / t so therefore x = v t ( i.e. 60 mi/hr x 2 hr = 120 mi )
In “calculus” language we would write dx = v dt, which we can integrate to obtain:
x t x t v t dtt
t( ) ( ) ( )2 1
1
2
Graphically, this is adding up lots of small rectangles:
v(t)
t
+ +...+
= displacement
v
t1 2
60
Physics II: Lecture 1, Pg 10
High-school calculus:
Also recall that
Since a is constant, we can integrate this using the above rule to find:
Similarly, since we can integrate again to get:
1-D Motion with constant acceleration1-D Motion with constant acceleration
t dtn
t constn n
11
1
a dvdt
v adt a dt at v 0
v dxdt
x vdt at v dt at v t x ( )02
0 012
See text : 2-4
Physics II: Lecture 1, Pg 11
RecapRecap So for constant acceleration we find:
v v at 0
x x v t at 0 021
2
a const
x
a
v t
t
tv v a x x
v v vav
22
12
2 1
1 2
212
( )
( )
From which we can derive:
See text : Table 2-1 (p. 33)
Physics II: Lecture 1, Pg 12
Problem 1Problem 1
A car traveling with an initial velocity vo. At t = 0, the driver puts on the brakes, which slows the car at a rate of ab
x = 0, t = 0ab
vo
Physics II: Lecture 1, Pg 13
Problem 1...Problem 1...
A car traveling with an initial velocity vo. At t = 0, the driver puts on the brakes, which slows the car at a rate of ab. At what time tf does the car stop, and how much farther xf does it travel ??
x = xf , t = tf
v = 0x = 0, t = 0ab
vo
Physics II: Lecture 1, Pg 14
Problem 1...Problem 1...
Above, we derived: (a)
(b)
Realize that a = -ab
Using (b), realizing that v = 0 at t = tf :
find 0 = v0 - ab tf or tf = vo /af
Plugging this result into (a) we find the stopping distance:
x x v t at 0 021
2v v at 0
x vva
ava
vaf o
o
bb
o
b
o
b
1
212
2 2
Physics II: Lecture 1, Pg 15
Problem 1...Problem 1...
So we found that
Suppose that vo = 65 mi/hr x .45 m/s / mi/hr = 29 m/s Suppose also that ab = |g| = 9.8 m/s2.
Find that tf = 3 s and xf = 43 m
tva
xvaf
o
bf
o
b , 1
2
2
Physics II: Lecture 1, Pg 16
Tips:Tips: Read !
Before you start work on a problem, read the problem statement thoroughly. Make sure you understand what information in given, what is asked for, and the meaning of all the terms used in stating the problem.
Watch your units !Always check the units of your answer, and carry the units
along with your numbers during the calculation.
Understand the limits !Many equations we use are special cases of more general
laws. Understanding how they are derived will help you recognize their limitations (for example, constant acceleration).
Physics II: Lecture 1, Pg 17
Recap of kinematics lecturesRecap of kinematics lectures 1-D Kinematics 1-D Kinematics
Average & instantaneous velocity (Chapter3- 1,3,7,9,11) and and acceleration
Motion Graphs (14,15,17,19) Motion with constant acceleration(Ch3 (Ch3 21,23,27,29,31,35,37
41)) Free Fall (Ch3-Free Fall (Ch3-41,43,47,49,51,52)) Review Phun!! (Review Phun!! (67,69,70 ))