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Galileo’s Motion Studies
0
0
2
0
2 2
0
, , 2
1
2
2
f
f
f
v vx vv v a
t t
v v at
x v t at
v v a x
Constant Acceleration…
Kinematic Equations
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Translational Motion Circular Motion
Projectile Motion Rotational Motion
Types of Motion
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Natural Motion
•Objects have a proper place
•Objects seek their natural place
•External forces must be constantly
applied to moving objects in order
to keep them going.
•The heavier the object,
the faster it falls.
•Did not experiment to test theories.
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Galileo Challenged The Dogma
Of Natural Motion with
Experiments
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The natural motion of
a body is to remain in
whatever state of
motion it is in unless
acted upon by net
external forces.
Galileo Challenged The Dogma
Of Natural Motion
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Galileo Challenged Aristotle Physics
In a vacuum, all objects fall with the same
acceleration due to gravity: 9.80 m/s2,
independent of their weight.
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Galileo’s Motion Studies
0
2
f
xv
t
v vv
va
t
gave us…
Definitions:
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Distance and Displacement
(delta) means "change in"
= 'final - initial'
The total distance traveled relative to an origin.
Distance is a scalar.
Displacement is a vector. The unit is the meter.
0fx x x
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Average Speed &Velocity
Speed is how fast something moves.
The average speed is the total distance per time.
The average velocity is the the total displacement per time.
Velocity is a vector. The unit is m/s.
total displacement
total time
xv
t
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Newton’s Calculus
will give us INSTANTEOUS motion…
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Acceleration: Changing Velocity
acceleration: v
at
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Acceleration: Changing Velocity
va
t
• Speeding up: a is positive
• Slowing down: a is negative
• a is a vector: magnitude and direction!
• Can v be positive but a be negative?
+x
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Acceleration How fast How fast is changing.
The rate at which the speed is changing.
Speeding up
Slowing down
Constant speed, changing direction.
change in velocity
change in time
va
t
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Acceleration is in the direction of
the net Force but not necessarily
in the direction of velocity. Velocity is always in the direction of the motion!
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va
t
A car travels speeds up from zero to 339m/s in 4.75s.
What is acceleration?
339 / 0
4.75
m s
s
271.3 /m s
339 /v m s
Acceleration: Changing Velocity
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Galileo’s Motion Studies
0 , ,
2
fv vx vv v a
t t
gave us…
Kinematic Equations
With a little al-jbr….
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0 , ,
2
fv vx vv v a
t t
va
t
Start:
0fv v a t
Assume constant acceleration!
0fv v a t
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0 , ,
2
fv vx vv v a
t t
va
t
Start:
0fv v a t
Assume constant acceleration!
0fv v a t
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0 , ,
2
fv vx vv v a
t t
0
2
fv vx
t
Start:
0 0( )
2
f ix x v v a t
t
2
0
1
2f ix x v t a t
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0 , ,
2
fv vx vv v a
t t
0
2
fv vx
t
0
2
f
t xv v
va
t
Start:
Combine &
Eliminate t:
0 =
fv vt
a
2 0
0
v vf
t xv v a
f
2 2
0 2fv v a x Algebra:
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Galileo’s Motion Studies
0
0
2
0
2 2
0
, , 2
1
2
2
f
f
f
v vx vv v a
t t
v v at
x v t at
v v a x
gave us…
Kinematic Equations
WARNING!!!
Average
Velocity is NOT
Instantaneous
Velocity!!
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Sense of Speed
1 / 3.6 / 2.24 /
10 / 36 / 22.4 /
20 / 72 / 44.8 /
30 / 108 / 67.2 /
m s km hr mi hr
m s km hr mi hr
m s km hr mi hr
m s km hr mi hr
1 / 2.25 /m s mi hr
1 / 0.62 /km hr mi hr
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Quicky Question
An automobile enters a freeway on-ramp at 15.0m/s and accelerates
uniformly up to 25.0 m/s in a time of 10.0s.
a) What is the automobile’s average velocity?
0
2
f
ave
v vv
15 / 25 /20 /
2
m s m sm s
Which equation?
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An automobile enters a freeway on-ramp at 15.0m/s and accelerates
uniformly up to 25.0 m/s in a time of 10.0s.
b) What is the automobile’s average acceleration?
f iv vva
t t
225 / 15 /1 /
10
m s m sm s
s
Which equation?
Quicky Question
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An automobile enters a freeway on-ramp at 15.0m/s and accelerates
uniformly up to 25.0 m/s in a time of 10.0s.
c) What is the distance traveled in this amount of time?
Which equation?
Quicky Question
2
0
1
2x v t at 2
2
1
215 / (10 ) 1 (10 )
mm s s s
s
200x m
21 /a m s
(you could also use vave equation.)
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Draw the Motion Diagram
An automobile enters a freeway on-ramp at 15.0m/s and accelerates
uniformly up to 25.0 m/s in a time of 10.0s.
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Acceleration: Changing Velocity
QUESTION:
From t = 0, how long does it take the car to come to a full stop?
+x
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Acceleration: Changing Velocity
2
Knowns
5 /
28 /
0
?
i
f
a m s
v m s
v
t
f iv vt
a
2
0 28 /5.6
5 /
m ss
m s
f iv v at
Which equation to use?
Solve for t:
General Solution
5.6t s
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Acceleration: Changing Velocity
From t = 0, to t = 5.6s, how far does the car travel before it
comes to a stop?
+x
2
Knowns
5 /
28 /
0
5.6
i
f
a m s
v m s
v
t s
Which equation? 2
0
1
2x v t at
2 2128 5.6 ( 5 / )(5.6 ) 78.4
2
mx s m s s m
s
78.4x m
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Stopping Distance
2 2
0 2fv v a x
If you are traveling 70 mph, how far does it take to stop?
(Assume the same deceleration in each case)
2
0
2
vx
a If v doubles,
d quadruples!!!
70 mph = 2x35mph, so stopping distance is 4x105ft = 420 ft!
1 ½ FOOTBALL FIELDS!!!!
0
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Brake Question You are driving a car going 80 km/hr (50mph) when a head-on
collision happens 25 meters ahead of you. If you can brake at 6 m/s2,
will you hit the crash or stop before it?
2 2
0 2fv v a x 2
0
2
vx
a
2
2
(22 / )40.3 25
2( 6 / )
m sx m m
m s
2
0: 80 / 22 / , 0, 6 . fKnowns v km hr m s v a m s
: ?Unknown x
CRASH!
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HW: Speedy Sally
Speedy Sally, driving at 30.0 m/s, enters a one-lane tunnel. She then observes a slow-moving van 155 m ahead traveling at 5.00 m/s. Sally applies her brakes but can accelerate only at 2.00 m/s2 because the road is wet. Will there be a collision? If yes, determine how far into the tunnel and at what time the collision occurs. If no, determine the distance of closest approach between Sally's car and the van. Sketch the x-t graphs for both the vehicles. What does it mean?