kinematics ap physics c: mechanics mr. wright lhs 2015-2016
TRANSCRIPT
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KINEMATICSAP Physics C: MechanicsMr. Wright LHS 2015-2016
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The Kinematic Equations
There are 3 major kinematic equations than can be used to describe motion. All are used when the acceleration is
CONSTANT.
2
2 2
1
2
2 ( )
o
o o
o o
v v at
x x v t at
v v a x x
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KEY VARIABLES
Kinematics: describing the motion of objects without describing the causes, in words, math, pictures and graphs.
Symbol Variable Units
t Time s
a Acceleration m/s/s
x or y Displacement m
v0 Initial velocity m/s
v Final velocity m/s
g or agAcceleration
due to gravitym/s/s
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Kinematic Equation #1
o o
o
o
o
v v v vva
t t t t
v v at
v v at
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Kinematic #1Example: A boat moves slowly out of a marina (so as to not
leave a wake) with a speed of 1.50 m/s. As soon as it passes the breakwater, leaving the marina, it throttles up and accelerates at 2.40 m/s/s.
What do I know?
What do I want?
vo= 1.50 m/s v = ?
a = 2.40 m/s/s
t = 5 s
v
v
atvv o
)5)(40.2()50.1(
a) How fast is the boat moving after accelerating for 5 seconds?
13.5 m/s
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Kinematic #22
21 attvxx oxo
x
x
attvxx oxo
)5)(40.2(21)5)(5.1(0
21
2
2
b) How far did the boat travel during that time?
37.5 m
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Does all this make sense?
mA
AbhA
50.7
)5.1)(5(
mA
bhA
30
)12)(5(2
1
2
1
mA
AbhA
50.7
)5.1)(5(
1.5 m/s
13.5 m/s
Total displacement = 7.50 + 30 = 37.5 m = Total AREA under the line.
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Interesting to Note
vttvxx
atttvxx
attvxx
oxo
oxo
oxo
21
21
21 2A = HB
A=1/2HB
Most of the time, xo=0, but if it is not don’t forget to ADD in the initial position of the object.
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Kinematic #3)(222
oo xxavv
What do I know?
What do I want?
vo= 12 m/s x = ?
a = -3.5 m/s/s
V = 0 m/s
x
x
x
xxavv oo
7144
)0)(5.3(2120
)(22
22
Example: You are driving through town at 12 m/s when suddenly a ball rollsout in front of your car. You apply the brakes and begin decelerating at 3.5 m/s/s.
How far do you travel before coming to a complete stop?
20.57 m
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Common Problems Students HaveI don’t know which equation to choose!!!
Equation Missing Variable
x
v
t
atvv o
2
21 attvxx oxo
)(222oo xxavv
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Kinematics for the VERTICAL DirectionAll 3 kinematics can be used to analyze one dimensional motion in either the X direction OR the y direction.
)(2)(2
21
21
2222
22
ooyyoox
oyooxo
oyyo
yygvvxxavv
gttvyyattvxx
gtvvatvv
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“g” or ag – The Acceleration due to gravityThe acceleration due to gravity is a special constant that exists
in a VACUUM, meaning without air resistance. If an object is in FREE FALL, gravity will CHANGE an objects velocity by 9.8 m/s every second.
2/8.9 smag g
The acceleration due to gravity:•ALWAYS ACTS DOWNWARD•IS ALWAYS CONSTANT near the surface of Earth
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ExamplesA stone is dropped at rest from the top of a cliff. It is observed to hit the ground 5.78 s later. How high is the cliff?
What do I know?
What do I want?
voy= 0 m/s y = ?
g = -9.8 m/s2
yo=0 m
t = 5.78 s
2
21 gttvyy oyo
Which variable is NOT given andNOT asked for?
Final Velocity!
y
y 2)78.5(9.4)78.5)(0(
-163.7 m
H =163.7m
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ExamplesA pitcher throws a fastball with a velocity of 43.5 m/s. It is
determined that during the windup and delivery the ball covers a displacement of 2.5 meters. This is from the point behind the body when the ball is at rest to the point of release. Calculate the acceleration during his throwing motion.
What do I know?
What do I want?
vo= 0 m/s a = ?
x = 2.5 m
v = 43.5 m/s
)(222oo xxavv
Which variable is NOT given andNOT asked for?
TIME
a
a )05.2(205.43 22
378.5 m/s/s
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ExamplesHow long does it take a car at rest to cross a 35.0 m
intersection after the light turns green, if the acceleration of the car is a constant 2.00 m/s/s?
What do I know?
What do I want?
vo= 0 m/s t = ?
x = 35 m
a = 2.00 m/s/s
t
t 2)2(21)0(035
Which variable is NOT given andNOT asked for?
Final Velocity
2
21 attvxx oxo
5.92 s
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ExamplesA car accelerates from 12.5 m/s to 25 m/s in 6.0 seconds. What was the acceleration?
What do I know?
What do I want?
vo= 12.5 m/s a = ?
v = 25 m/s
t = 6satvv o
Which variable is NOT given andNOT asked for?
DISPLACEMENT
a
a )6(5.1225
2.08 m/s/s
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Kinematics and CalculusLet’s take the “derivative” of kinematic #2 assuming the object started at x = 0.
20 0
20
0
1
21
( )2
( )
x
x
o
x x v t at
d v t atdxv
dt dtv v at
d v atdva a
dt dt