Physics 121C Mechanics Lecture 3 Position

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Physics 121C - MechanicsLecture 3Position
VelocityOctober 4, 2004

• John G. Cramer
• Professor of Physics
• B451 PAB
• cramer_at_phys.washington.edu

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Announcements

• Homework Assignment 1 is now posted on Tycho.
  For full credit, your assignment must be entered
  in full by 900 PM on next Wednesday, October
  6.(Homework up to 24 hours late will receive 70
  credit.)
• The H-ITT clickers should work now, and we will
  try them later. (On Friday the program had the
  wrong port assignments.)
• The H-ITT clickers for this class may be
  registered on the Web athttp//faculty.washingto
  n.edu/jcramer/ph121c/Clicker . Note it is not
  necessary to register your clicker before using
  it, but it is necessary to register it (soon) in
  order for your class participation to be applied
  and credited.

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Mastering Physics
The Mastering Physics web site was advertised
during the 1st lecture, but people have had
difficulty in using it. A set of Physics 121
assignments in Mastering Physics has now been
created. Students can login and do these
problems. The computer will grade them and keep
a record, but this will not count in any way
toward course grades. There is an assignment for
each chapter, and each assignment consists of
about 5 or 6 "tutorial" style questions and 5 or
6 end of chapter problems from Knight. Here is
more information Logon at http//www.mastering
physics.com/ Course ID MPPEDIGO0002 Access
Code supplied in the packet that came with your
textbook Do the problems, then hit Submit.
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Lecture Schedule (Part 1)
You are here!
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Uniform Motion
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The Math of Uniform Motion
Important Calculus Concepts The slope of a
position-timegraph is the first derivative
ofposition s with respect to time,ds/dt, and it
is also the velocityin the s direction. If
this slope changes withtime, the object is
accelerating(d2s/dt2 ¹ 0). Otherwise,
themotion is uniform.
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Skating with Constant Velocity
(vx)A DxA/DtA rise/run
(2.0 m)/(0.40 s) 5.0 m/s
DxB (0.0 m) - (1.0 m) -1.0
(vx)B DxB/DtB (-1.0 m)/(0.50 s)
-2.0 m/s
Reality check (vx)A 5.0 m/s 10 mph (vx)B
-2.0 m/s -4 mph Reasonable.
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Instantaneous Velocity
Position
A jet plane accelerates for takeoff
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Using Motion Diagrams
Instantaneous velocity is thelocal slope of the
curve.
Observation Any smooth curve becomes
linearat a sufficiently high magnification.
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Using Motion Diagrams (2)
Position
Velocity
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Velocity from Graphical Position
aygt0
aylt0
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Time Derivatives
Position
Turning point 1st derivative goesto zero,
indicating thatthe change in positionreverses
direction.
Velocity
Maxima and Minima 1st derivative goes to
zero. 2nd derivative is gt0 for minimum.
2nd derivative is lt0 for maximum.
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Example
A particle has position x(t) (3t - t3) where x
is in m and t is in s.

• What is the particles position and velocity at
  t2 s?x 3(2) (23) 6 m 8 m -2 mdx/dt
  (3 - 3t2) 3 - 3(22) 3 m/s 12 m/s
  -9 m/s
• Plot position and velocity for -3 s lt t lt 3 s.
• Draw a diagram to illustrate the motion.

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Clicker Question 1
Which of the green velocity vs.time graphs goes
with this blueposition vs. time graph?
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Finding Position from Velocity
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Drag Racers Displacement
The figure shows thevelocity of a drag
racer.How far does the racermove during the
first 3.0 s?
Solution The net distance traveledis the
area under the velocitycurve shown in blue.
This isa triangle with sides 12 m/sand 3.0 s.
The area of thistriangle is A ½(12 m/s)(3
s) 18 m. Thus, the drag racer moves18 m in the
first 3 seconds.
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Drag Racers Position

• Find an algebraic expression for the drag
  racers position. Assumethat si 0 m and ti
  0 s.
• Plot a position vs. time graph.

Solution (a) The speed of the racer
increaseslinearly with t, with v(t) 4 t m/s.
The position iss(t) si 0?t v(t1) dt1 0
0?t 4 t1dt1 2 t120t 2 t2 m (b)
The graph is a parabola from theorigin, as shown.
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Finding the Turning Point
The figure shows the velocity of a
particle that starts at xi 30 m at time ti0 s.
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• Draw a motion diagram forthe particle.
• Where is the particles turning point?
• At what time does the particlereach the origin?

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10
10
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• Solution
• The figure shows the motion.
• The particle has zero velocityat t2 s, which
  must be its turning point. Its position is x
  x0 0?2 v dt 30 m area of triangle from 0 to
  2 s 30 m ½(10 m/s)(2 s) 40 m.
• To get to the origin, the particle must move -40
  m from the turning point. This occurs at 6 s
  (see diagram).

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Time Integrals
Previous example x x0 0?2 v(t) dt x0
0?2 (10 5 t) dt x0 100?tdt 50?tt dt
x0 10t0t (5/2)t20t 30 10 t
(5/2)t2 m Solve quadratic equation. At
x0, t -2 s or 6 s. Only the positive
root is relevant, and it is the same solution
wefound graphically in the previous slide.
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Clicker Question 2
Which of the blue position vs.time graphs goes
with this greenvelocity vs. time graph?
Theparticles position at ti 0 s isxi -10
m.
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End of Lecture 3

• Before the next lecture, read Knight,Chapters
  2.5 through 2.8
• Lecture Homework 1 is due at900 PM next
  Wednesday, Oct. 6.