velocity and acceleration
TRANSCRIPT
Physics - Mechanics
Velocity and AccelerationSeptember , 2013
Dr. Muhammad KamranDepartment of Electrical Engineering
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Lecture Schedule (Part 1)
Physics - Introduction to Mechanics - -2013
Textbook: Physics, Vol. 1
Week Date L# Lecture Topic PagesSlide
s Reading HW Due Lab
17-Jan-13 1 Introduction to Physics 12 21 Chapter 1
No Lab 1st week8-Jan-13 2 Position & Velocity 8 22 2-1 to 2-3
10-Jan-13 3 Velocity & Acceleration 10 25 2-4 to 2-5 No HW
11-Jan-13 4 Equations of Motion 9 21 2-6 to 2-7
214-Jan-13 5 Vectors 8 24 3-1 to 3-3
1-D Kinematics15-Jan-13 6 r, v & a Vectors 5 24 3-4 to 3-5
17-Jan-13 7 Relative Motion 3 18 3-6 HW1
18-Jan-13 8 2D Motion Basics 5 19 4-1 to 4-2
321-Jan-13 H2 MLK Birthhday Holiday
Free Fall & Projectiles22-Jan-13 9 2D Examples 13 22 4-3 to 4-5
24-Jan-13 R1 Review & Extension - 49 - HW2
25-Jan-13 E1 EXAM 1 - Chapters 1-4
We are here.
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Motion with Constant Acceleration
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Motion with Constant AccelerationVelocity: (2-7)
Average velocity: (2-9)
Position as a function of time:(2-10)(2-11)
Velocity as a function of position: (2-12)
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Motion with Constant Acceleration, Other Variables
(1)0
0v v vv v at ta a
00 av
av av
x x xx x v t tv v
vat
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21 0 02
0 0 2
2v a x vx x v t at t
a
Motion with Constant Acceleration, Other Variables
(2)
02
2( )x v ta
t
2 20
2v va
x
2 22 2 0
0 22
v vv v a x xa
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Motion with Constant AccelerationThe relationship between position and time
follows a characteristic curve.
Parabola
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Practice Clicker Question Assume that the brakes on your car create a constant deceleration, independent of how fast you are going. Starting at a particular speed, you apply the brakes and note the stopping distance D and the stopping time T. Now you double the speed of the car. How does this change affect the stopping distance D and the stopping time T?
(a) D and T remain the same.(b) D and T both double (i.e., x2).(c) D doubles and T quadruples (i.e., x4).(d) D quadruples and T doubles.(e) D and T both quadruple.
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Freely Falling ObjectsAn object falling in air is subject to air resistance (and therefore is not freely falling).
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Freely Falling ObjectsFree fall is the motion of an object subject only to the influence of gravity. The acceleration due to gravity is a constant, g.
We will normally use the valueg = 9.81 m/s2.
+
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Free Fall and g One important example of constant acceleration is the “free fall” of an object under the influence of the Earth’s gravity. The picture shows an apple and a feather falling in vacuum with identical motions. The magnitude of this acceleration, designated as g, has the approximate value of a = g = 9.81 m/s2 = 32.2 ft/s2. If downward is designated as the +y direction, then a = +g; if downward is designated as the y direction, then a = g. (Note that g is always positive., but a may have either sign.)
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Freely Falling ObjectsFree fall from rest (with x = down):
0( )v t v gt
1 20 0 2( )x t x v t gt
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Freely Falling ObjectsTrajectory of a projectile:
Position
Velocity
Acceleration
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Upon graduation, a joyful student throws her cap straight up in the air with an initial speed of 14.7 m/s. Given that its acceleration has a magnitude of 9.81 m/s2 and is directed downward (we neglect air resistance),(a) When does the cap to reach its highest point?(b) What is the distance to the highest point?(c) Assuming the cap is caught at the same height it was released, what is the total time that the cap is in flight?
Example: The Flying Cap
1. Draw the cap (as a dot) in its various positions.2. (a) Use the time, velocity and acceleration relation.
(b) Use average velocity: vav = v0/2 = 7.35 m/s; x = vav t = (7.35 m/s)(1.5 s) = 11.0 m(c) Up time = down time, so total time is 3.0 s. (see text for a more complicated method.)
3. The answers have the right units and seem reasonable.
00 2
(0 m/s) (14.7 m/s); 1.5 s9.81 m/s
x xx x x
x
v vv v a t t
a
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Example: The Flying Cap (continued)
The height of the cap vs. time has the form of a parabola (since x ~ t2). It is symmetric about the midpoint (but would not be, if air resistance were present).
The velocity of the cap vs. time has the form of a straight line (since v ~ t). The velocity crosses zero at the midpoint and is negative thereafter, because the cap is moving downward.
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An electron in a cathode-ray tube accelerates from rest with a constant acceleration of 5.33 x 1012 m/s2 for 0.150 ms, then drifts with a constant velocity for 0.200 ms, then slows to a stop with a negative acceleration of2.67 x 1013 m/s2. (Note: 1 ms = 10-6 s) How far does the electron travel?
Example: A Traveling Electron
1. Draw the electron (as a dot)in its various positions xi.
2. Calculate the displacement xi and velocity vi for each part of the path:1 12 -6 12 2 -6 2
1 0 2 2(0 m/s)(0.150 10 s)+ (5.33 10 m/s )(0.150 10 s) 0.06 mxx v t a t
3. The answers have the correct units and appear to be reasonable.
12 2 -6 51
5 -62 1 2
(5.33 10 m/s )(0.150 10 s) 8.0 10 m/s
(8.0 10 m/s)(0.200 10 s) 0.16 mxv a t
x v t
2 2 2 5 2
2 2 00 3 13 2
(0 m/s) (8.0 10 m/s)2 ; 0.012 m2 2( 2.67 10 m/s )
x xx x x
x
v vv v a x xa
1 1 1 (0.06 m) (0.16 m) (0.012 m) 0.232 m =23.2 cmx x x x
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Example: Speed of a Lava Bomb
A volcano shoots out blobs of molten lava (lava bombs) from its summit. A geologist observing the eruption uses a stopwatch to time the flight of a particular lava bomb that is projected straight upward. If the time for it to rise and fall back to its launch height is 4.75 s, what is its initial speed and how high did it go? (Use g = 9.81 m/s2.)
1 120 0 02 20 ( )x x v t gt x t v gt
10 2Either 0 or 0t v gt
1 1 20 2 2 (9.81 m/s )(4.75 s) 23.3 m/sv gt
2 20At maximum height, 0 2v v g x
2 20
2
(23.3 m/s) 27.7 m2 2(9.81 m/s )vxg