Physics 2211 Mechanics Lecture 3 Knight: 2'4 to 2'8 Constant Acceleration

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Physics 2211 - MechanicsLecture 3 (Knight 2.4
to 2.8)Constant Acceleration

• Dr. John Evans
• jevans_at_agnesscott.edu

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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.Assume that 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 particle reach 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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Calculus of Time Integrals
From the 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 Solving the
quadratic equation, we find that x0 at 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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Uniformly Accelerated Motion
Rate of velocity change Dvs/Dt (30 m/s)/(6.0
s) 5.0 (m/s)/s 5.0 m/s2
aavg º Dvs/Dt (average acceleration), so aPor avg
5.0 m/s2 (constant)
aVW avg º Dvs/Dt (10 m/s)/(5.0 s) 2.0 m/s2
(constant)
This is uniformly accelerated motion, i.e. motion
with constant acceleration.
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Running the Court

• A basketball player starts at theleft end of the
  court and moveswith the velocity shown below.
• What does his motion look like?
• What is his acceleration?

Dv1 (6 m/s) (0 m/s) 6 m/sa1 Dv1/Dt1 (6
m/s)/(6 s) 1.0 m/s2 Dv2 (-6 m/s) (6 m/s)
-12 m/sa2 Dv2/Dt2 (-12 m/s)/(6 s) -2.0
m/s2
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Kinematic Equations
Motion with constant acceleration as as
Dvs/Dt (vfs vis)/Dt therefore, vfs vis
asDt. sf si (area under velocity curve vs
between ti and tf) si visDt
½as(Dt)2
Dt (vfs vis)/as , so, substituting sf si
vis(vfs vis)/as ½as(vfs vis)/as2 si
(visvfs/as) - (vfs2/as) (vis2/2as)
- (visvfs/as) (vis2/2as) si (vfs2 -
vis2)/as or vfs2 vis2 2asDs
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Kinematic Equations
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Motion Summary
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Rocket Sled
A rocket sled acceleratesat 50 m/s2 for 5.0
s, coastsfor 3.0 s, then deploys abraking
parachute anddecelerates at 3.0 m/s2until
coming to a halt.

• What is the maximumvelocity reached bythe
  rocket sled?
• v1x v0x a0x(t1 t0) a0xt1
• (50 m/s2)(5.0 s) 250 m/s

(b) What is the total distance traveled bythe
rocket sled? x1 x0 v0x(t1 t0) ½a0x(t1
t0)2 ½a0xt12 ½(50 m/s2)(5.0 s)2 625
m x2 x1 v1x(t2 t1) 625 m (250 m/s)(3.0
s) 1,375 m v3x2 v2x2 2a2xDx v2x2 2a2x(x3
x2), so x3 x2 (v3x2 - v2x2)/2a2x
1,375 m 0 (250 m/s)2/2(-3.0 m/s2)
11,800 m
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Clicker Question 1
Which velocity-vs-time graph goeswith the
acceleration-vs-time graphshown at the right?
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Free Fall
An object moves in free fall when it
movesunder the influence of uniform gravity, and
noother forces. Strictly speaking, this
happens only in vacuum, but in practice we can
often neglect air resistance and observe this
behavior. a constant acceleration due to
gravity g 9.80 m/s2 ay -g ( -9.80
m/s2) vy - gt y -½g t2
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A Falling Rock

• A rock is released fromrest at the top of a 100
  mtall building.
• How long does to takefor the rock to fall to
  theground?
• y1 y0 - ½gt12 , so
• t1 2(y1 - y0)/g½ 2(100
  m)/(9.80 m/s2)½ 4.52 s 4.52 s
• What is the impact velocity?
• v1 v0 - gt1 0 (9.80 m/s2)(4.52 s) -44.3
  m/s

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Motion on an Inclined Plane
An frictionless inclined plane can be usedto
dilute the acceleration due to gravity. The
component of g perpendicular to the surface (g
Cosq) is cancelled by the normal forcefrom the
plane. The component of g parallel to the
surface (g Sinq) acts to accelerate the object
down the plane. as g Sinq (depending on
the orientation of the inclined plane).
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Downhill Skiing
A skiers speed at the bottom of a 100 m slope
is 20 m/s. What is the angleof the slope? v2s2
v1s2 2 as Dx 2 g Sinq Dx, so Sinq v2s2/(2
g Dx) (20 m/s)2/2(9.80 m/s2)(100 m)
0.204 q Sin-1(0.204) 11.8O
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Ball Rolling on Track
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Instantaneous Acceleration
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Velocity from Acceleration
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Nonuniform Acceleration
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Clicker Question 2
Rank in order, fromlargest to smallest,the
accelerations at points A, B, and C.
(a) aA gt aB gt aC (b) aC gt aA gt aB (c)
aC gt aB gt aA (d) aB gt aA gt aC
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Important Concepts
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End of Lecture 3

• Before the next lecture, read Knight,Chapters
  3.1 through 3.4

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End of Lecture 3

• Before the next lecture, read Knight,Chapters
  3.1 through 3.4

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Football
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Football
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Vertical Cannonball
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Amusement Park Car
Dx
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Interpreting Motion Graphs