Transcript 2053_Lecture_01-17-13

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)).
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PHY 2053
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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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PHY 2053
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1-d Motion: Instantaneous Velocity
x
x(t+Dt)
x(t)
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
shrink Dt
t
t+Dt
t
Dt
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1-d Motion: Instantaneous Velocity
x
x(t+Dt)
x(t)
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
shrink Dt
t+Dt
t
t
Dt
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1-d Motion: Instantaneous Velocity
x
x(t+Dt)
x(t)
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t+Dt

shrink Dt
t
t
Dt
PHY 2053
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1-d Motion: Instantaneous Velocity
x
x(t+
D
t)
x(t)
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
shrink Dt
t t+Dt
Dt
t
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1-d Motion: Instantaneous Velocity
x
tangent line at t
x(t)
v  lim
Dt  0
Dx
Dt
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
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.
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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
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Dv
Dt

v(t)
dv
Instantaneous acceleration a(t) is
slope of v-t tangent line at t
dt
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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)
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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
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PHY 2053
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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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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)
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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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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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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
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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
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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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dg
sin 2
2

( 20 m )( 9 . 8 m / s )
sin( 2  20 )

PHY 2053
 17 . 46 m / s  17 . 5 m / s
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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

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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
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
 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
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