2053_Lecture_01-17-13
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1-d Motion: Position & Displacement
• The x-axis:
We locate objects by specifying their
position along an axis (in this case x-axis).
The positive direction of an axis is in the
direction of increasing numbers. The
opposite is the negative direction.
• Displacement:
x(t)
The change from position x1 to position x2
is called the displacement, Dx.
Dx = x2 –x1
The displacement has both a magnitude,
|Dx|, and a direction (positive or negative).
time t
• Graphical Technique:
A convenient way to describe the motion
of a particle is to plot the position x as a
function of time t (i.e. x(t)).
R. Field 1/17/2013
University of Florida
PHY 2053
Page 1
1-d Motion: Average Velocity
• Average Velocity
v v ave
The average velocity is defined to be
the displacement, Dx, that occurred
during a particular interval of time, Dt
(i.e. vave = Dx/Dt).
Dx
Dt
x ( t 2 ) x ( t1 )
t 2 t1
x 2 x1
t 2 t1
• Average Speed
The average speed is defined to be the
magnitude of total distance covered
during a particular interval of time, Dt
(i.e. save = (total distance)/Dt).
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University of Florida
PHY 2053
Page 2
1-d Motion: Instantaneous Velocity
x
x(t+Dt)
x(t)
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shrink Dt
t
t+Dt
t
Dt
PHY 2053
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1-d Motion: Instantaneous Velocity
x
x(t+Dt)
x(t)
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University of Florida
shrink Dt
t+Dt
t
t
Dt
PHY 2053
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1-d Motion: Instantaneous Velocity
x
x(t+Dt)
x(t)
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University of Florida
t+Dt
shrink Dt
t
t
Dt
PHY 2053
Page 5
1-d Motion: Instantaneous Velocity
x
x(t+
D
t)
x(t)
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University of Florida
shrink Dt
t t+Dt
Dt
t
PHY 2053
Page 6
1-d Motion: Instantaneous Velocity
x
tangent line at t
x(t)
v lim
Dt 0
Dx
Dt
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University of Florida
dx
dt
Instantaneous velocity v(t) is
slope of x-t tangent line at t
t
t
The velocity is the derivative of x(t)
with respect to t.
PHY 2053
Page 7
1-d Motion: Acceleration
• Acceleration
a a ave
When a particles velocity changes, the
particle is said to undergo acceleration
(i.e. accelerate).
Dv
Dt
v ( t 2 ) v ( t1 )
t 2 t1
v
v 2 v1
t 2 t1
v(t)
v2
Hello
• Average Acceleration
Dv “rise”
v1
The average acceleration is defined to
be the change in velocity, Dv, that
occurred during a particular interval of
time, Dt (i.e. aave = Dv/Dt).
• Instantaneous Acceleration
a
Dv
v
The acceleration is the derivative of
v(t) with respect to t.
a lim
Dt 0
R. Field 1/17/2013
University of Florida
Dv
Dt
v(t)
dv
Instantaneous acceleration a(t) is
slope of v-t tangent line at t
dt
PHY 2053
Page 8
Equations of Motion: a = constant
(constant acceleration)
• v is a linear
function of t
a (t ) a
Acceleration a (in m/s2)
• Special case!
Acceleration as a function of time t: a(t)
5
v at t = 0
v v 0 at
4
3
a = 2 m/s2
2
1
0
0
1
2
3
4
5
6
7
8
9
10
time t (in s)
Velocity as a function of time t: v(t)
x x0 v0t
1
2
at
2
Velocity v (in m/s)
• x is a quadratic
function of t
20
a = 2 m/s2
v0 = 0
15
10
5
0
0
x at t =0
2
3
4
5
6
7
8
9
10
9
10
time t (in s)
Position as a function of time t: x(t)
• Note also that
120
a = 2 m/s2
v0 = 0
x0 = 0
2
0
Position x (in m)
100
v v 2 a ( x x0 )
2
1
80
60
40
20
0
0
1
2
3
4
5
6
7
8
time t (in s)
R. Field 1/17/2013
University of Florida
PHY 2053
Page 9
Acceleration Due to Gravity
• Experimental Result
y-axis
Near the surface of the Earth all objects fall toward the
center of the Earth with the same constant acceleration,
g ≈ 9.8 m/s2, (in a vacuum) independent of mass, size,
shape, etc.
• Equations of Motion
a y g 9 .8 m / s
2
h
The acceleration due to
gravity is almost constant
and equal to 9.8 m/s2
provided h << RE!
x-axis
RE
v y ( t ) v y 0 gt
y (t ) y 0 v y 0 t
1
2
gt
Earth
2
v y v y0 2 g ( y y0 )
2
2
R. Field 1/17/2013
University of Florida
PHY 2053
Page 10
Equations of Motion: Example Problem
y-axis
• Example Problem
A ball is tossed up along the y-axis (in a vacuum on
the Earth’s surface) with an initial speed of 49 m/s.
vy0 = 49 m/s
Earth
How long does the ball take to reach its maximum
height?
v
t
y0
49 m / s
g
9 .8 m / s
2
5s
Velocity as a function of time t: v(t)
60
a = -9.8 m/s2
v0 = 49 m/s
Velocity v (in m/s)
40
What is the ball’s maximum height?
2
h
v y0
2g
( 49 m / s )
2
2
122 . 5 m
20
0
-20
-40
2 (9 .8 m / s )
-60
0
1
2
3
4
5
6
7
8
9
10
time t (in s)
How long does it take for the ball to get back to its
starting point?
2v
y0
10 s
120
g
What is the velocity of the ball when it gets back to
its starting point?
v y v y 0 49 m / s
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University of Florida
PHY 2053
Position y (in m)
t
Position as a function of time t: y(t)
140
100
80
a = -9.8 m/s2
v0 = 49 m/s
y0 = 0
60
40
20
0
0
1
2
3
4
5
6
7
8
9
10
time t (in s)
Page 11
2-d Motion: Constant Acceleration
• Kinematic Equations of Motion (Vector Form)
Acceleration Vector (constant)
Velocity Vector (function of t)
Position Vector (function of t)
The velocity vector and position
vector are a function of the time t.
a a x xˆ a y yˆ
v (t ) v 0 a t
r ( t ) r0 v 0 t
Warning! These
equations are only
valid if the
acceleration is
constant.
1
2
2
at
Velocity Vector at
time t = 0.
v 0 v x 0 xˆ v y 0 yˆ
Position Vector at
time t = 0.
r0 x 0 xˆ y 0 yˆ
The components of the acceleration vector, ax and ay, are constants.
The components of the velocity vector at t = 0, vx0 and vy0, are constants.
The components of the position vector at t = 0, x0 and y0, are constants.
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University of Florida
PHY 2053
Page 12
2-d Motion: Constant Acceleration
• Kinematic Equations of Motion (Component Form)
ax
ay
constant
v x (t ) v x 0 a x t
x (t ) x 0 v x 0 t
Warning! These
equations are only
valid if the
acceleration is
constant.
constant
v y (t ) v y 0 a y t
1
2
a xt
2
y (t ) y 0 v y 0 t
1
2
a yt
2
The components of the acceleration vector, ax and ay, are constants.
The components of the velocity vector at t = 0, vx0 and vy0, are constants.
The components of the position vector at t = 0, x0 and y0, are constants.
• Ancillary Equations
Valid at any time t
v x ( t ) v xo 2 a x ( x ( t ) x 0 )
2
2
v y ( t ) v yo 2 a y ( y ( t ) y 0 )
2
2
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University of Florida
PHY 2053
Page 13
Example: Projectile Motion
y-axis
• Near the Surface of the Earth (h = 0)
In this case, ax= 0 and ay= -g, vx0 = v0cos, vy0 = v0sin, x0 = 0, y0 = 0.
v x ( t ) v 0 cos
v y ( t ) v 0 sin gt
x ( t ) ( v 0 cos ) t
y ( t ) ( v 0 sin ) t
v0
x-axis
1
2
gt
2
• Maximum Height H
The time, tmax, that the projective reaches its maximum
height occurs when vy(tmax) = 0. Hence,
t max
v 0 sin
g
H y ( t max )
( v 0 sin )
2
2g
• Range R (maximum horizontal distance traveled)
The time, tf, that it takes the projective reach the
ground occurs when y(tf) = 0. Hence,
2 v 0 sin
t
2
f
0 y ( t f ) ( v 0 sin ) t f 12 gt f
g
2 v 0 sin cos
2
R x ( t f ) ( v 0 cos ) t f
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v 0 sin 2
2
g
g
PHY 2053
Page 14
Example: Projectile Motion
y-axis
• Near the Surface of the Earth (h = 0)
In this case, ax= 0 and ay= -g, vx0 = v0cos, vy0 = v0sin, x0 = 0, y0 = 0.
v x ( t ) v 0 cos
v y ( t ) v 0 sin gt
x ( t ) ( v 0 cos ) t
y ( t ) ( v 0 sin ) t
v0
x-axis
1
2
gt
2
• Maximum Height H
The time, tmax, that the projective reaches its maximum
height occurs when vy(tmax) = 0. Hence,
t max
v 0 sin
g
H y ( t max )
( v 0 sin )
2
2g
For a fixed v the largest
R
• Range R (maximum horizontal
distance
traveled)
0
occurs when = 45o!
The time, tf, that it takes the projective reachthe ground
occurs when y(tf) = 0. Hence,
2 v 0 sin
2
tf
1
0 y ( t f ) ( v 0 sin ) t f 2 gt f
g
2 v 0 sin cos
2
R x ( t f ) ( v 0 cos ) t f
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v 0 sin 2
2
g
g
PHY 2053
Page 15
Exam 1 Fall 2012: Problem 11
v
• Near the surface of the Earth a projectile
d
is fired from the top of a building at a
height h above the ground at an angle
h
building
relative to the horizontal and at a distance
d from the edge of the building as shown
ground
o
in the figure. If = 20 and d = 20 m, what
2
is the minimum initial speed, v0, of the
v 0 sin 2
d R
projectile such that it will make it off the
g
building and reach the ground? Ignoring
air resistance.
0
Answer: 17.5 m/s
% Right: 35%
v0
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University of Florida
dg
sin 2
2
( 20 m )( 9 . 8 m / s )
sin( 2 20 )
PHY 2053
17 . 46 m / s 17 . 5 m / s
Page 16
Exam 1 Spring 2012: Problem 12
• A beanbag is thrown horizontally from a dorm room
window a height h above the ground. It hits the ground a
horizontal distance d = h/2 from the dorm directly below
the window from which it was thrown. Ignoring air
resistance, find the direction of the beanbag's velocity
just before impact.
Answer: 76.0° below the horizontal
% Right: 22%
v x (t ) v 0
x (t ) v 0 t
t
2
h
v y ( t ) gt
y (t ) h
1
2
gt
v0
2
2h
y-axis
g
d
h
d
x-axis
th
Let th be the time the beanbag hits the ground.
y (t h ) 0 h
1
2
gt
2
h
tan
v x (t h )
x (t h ) d v 0 t h
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v y (t h )
gt h
v0
76
PHY 2053
2
gt h
d
2h
4
d
Page 17
Example Problem: Projectile Motion
• A suspension bridge is 60.0 m above the level base of a gorge. A stone
is thrown or dropped off the bridge. Ignore air resistance. At the
location of the bridge g has been measured to be 9.83 m/s2. If you drop
the stone how long does it take for it to fall to the base of the gorge?
In this case, ax= 0 and ay= -g, vx0 = 0, vy0 = 0, x0 = 0, y0 =h. Hence,
y-axis
h
x-axis
v y ( t ) gt
v x (t ) 0
x (t ) 0
y (t ) h
1
2
gt
2
The time, tf, that the it takes the stone to reach the ground occurs when y(tf) = 0.
Hence,
2
1
2h
2 ( 60 m )
y (t f ) 0 h
2
gt
f
tf
2
g
3 . 494 s
( 9 . 83 m / s )
• If you throw the stone straight down with a speed of 20.0 m/s, how long
before it hits the ground?
Have to use the
In this case, ax= 0 and ay= -g, vx0 = 0, vy0 = -v0, x0 = 0, y0 =h. Hence,
y (t ) h v 0 t
tf
v0
1
2
gt
y (t f ) 0 h v 0 t f
2
( v o ) 2 gh
2
g
R. Field 1/17/2013
University of Florida
20 m / s
1
2
gt
Quadratic Formula!
2
f
( 20 m / s ) 2 ( 9 . 83 m / s )( 60 m )
2
9 . 83 m / s
PHY 2053
2
2
2 . 01 s
Page 18