1-d Motion: Position &amp; Displacement The x-axis:

```1-d Motion: Position &amp; 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, Δx.
Δx = x2 –x1
The displacement has both a magnitude,
|Δx|, 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 1/14/2014
University of Florida
PHY 2053
Page 1
1-d Motion: Average Velocity
• Average Velocity
v = vave
The average velocity is defined to be
the displacement, Δx, that occurred
during a particular interval of time, Δt
(i.e. vave = Δx/Δt).
Δx x(t 2 ) − x(t1 ) x2 − x1
=
=
=
Δt
t 2 − t1
t 2 − t1
• Average Speed
The average speed is defined to be the
magnitude of total distance covered
during a particular interval of time, Δt
(i.e. save = (total distance)/Δt).
R. Field 1/14/2014
University of Florida
PHY 2053
Page 2
1-d Motion: Instantaneous Velocity
x
x(t+Δt)
x(t)
R. Field 1/14/2014
University of Florida
{
shrink Δt
t
t+Δt
t
Δt
PHY 2053
Page 3
1-d Motion: Instantaneous Velocity
x
x(t+Δt)
x(t)
R. Field 1/14/2014
University of Florida
{
shrink Δt
t+Δt
t
t
Δt
PHY 2053
Page 4
1-d Motion: Instantaneous Velocity
x
x(t+Δt)
x(t)
R. Field 1/14/2014
University of Florida
t+Δt
{
shrink Δt
t
t
Δt
PHY 2053
Page 5
1-d Motion: Instantaneous Velocity
x
x(t) x(t+Δt)
R. Field 1/14/2014
University of Florida
{
shrink Δt
t t+Δt
Δt
t
PHY 2053
Page 6
1-d Motion: Instantaneous Velocity
x
tangent line at t
x(t)
Δx dx
=
v = lim
dt
Δt →0 Δt
R. Field 1/14/2014
University of Florida
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
a = aave
• Acceleration
When a particles velocity changes, the
particle is said to undergo acceleration
(i.e. accelerate).
Δv v(t 2 ) − v(t1 ) v2 − v1
=
=
=
Δt
t 2 − t1
t 2 − t1
v
v(t)
v2
Hello
Δv “rise”
• Average Acceleration
v1
The average acceleration is defined to
be the change in velocity, Δv, that
occurred during a particular interval of
time, Δt (i.e. aave = Δv/Δt).
• Instantaneous Acceleration
a
Δv
v
The acceleration is the derivative of
v(t) with respect to t.
Δv dv
a = lim
=
Δt →0 Δt
dt
R. Field 1/14/2014
University of Florida
v(t)
Instantaneous acceleration a(t) is
slope of v-t tangent line at t
PHY 2053
Page 8
Equations of Motion: a = constant
(constant acceleration)
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(t ) = v0 + at
• v is a linear
function of t
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 is a quadratic
function of t
x(t ) = x0 + v0t + at
1
2
2
Velocity v (in m/s)
20
a = 2 m/s2
v0 = 0
15
10
5
0
0
1
2
3
4
5
6
7
8
9
10
9
10
time t (in s)
x at t =0
Position as a function of time t: x(t)
• Note also that
120
2
0
Position x (in m)
v = v + 2a ( x − x0 )
2
a = 2 m/s2
v0 = 0
x0 = 0
100
80
60
40
20
0
0
1
2
3
4
5
6
7
8
time t (in s)
R. Field 1/14/2014
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 ≈ −9.8m / s 2
v y (t ) = v y 0 − gt
h
The acceleration due to
gravity is almost constant
and equal to 9.8 m/s2
provided h &lt;&lt; RE!
y (t ) = y0 + v y 0t − 12 gt 2
x-axis
RE
Earth
v y2 = v y20 − 2 g ( y − y0 )
R. Field 1/14/2014
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?
vy0
49m / s
t=
=
= 5s
2
g
9.8m / s
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?
v y20
(49m / s) 2
h=
=
= 122.5m
2 g 2(9.8m / s 2 )
20
0
-20
-40
-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
g
= 10 s
120
What is the velocity of the ball when it gets back to
its starting point?
v y = −v y 0 = −49m / s
R. Field 1/14/2014
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
Final Exam Fall 2010: Problem 9
• Near the surface of the Earth a startled armadillo leaps vertically
upward at time t = 0, at time t = 0.5 s it is a height of 0.98 m above the
ground. At what time does it land back on the ground?
% Right: 47%
y (t ) = vot − 12 gt 2 = t (vo − 12 gt )
Let tf be the time the armadillo
lands back on the ground.
v0
y (t f ) = 0 = t f (vo − gt f )
1
2
tf =
2vo
g
y (t h ) = h = vot h − 12 gt h2
y-axis
Let th be the time the armadillo is
at the height h.
h + 12 gt h2
vo =
th
2vo 2(h + 12 gt h2 ) 2h
2(0.98m)
tf =
=
=
+ th =
+ (0.5s ) = (0.4 s ) + (0.5s ) = 0.9 s
2
g
gt h
gt h
(9.8m / s )(0.5s )
R. Field 1/14/2014
University of Florida
PHY 2053
Page 12
Example Problem: 1d 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,
v x (t ) = 0
x(t ) = 0
y-axis
h
x-axis
v y (t ) = − gt
y (t ) = h − 12 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(60m)
y (t f ) = 0 = h − 2 gt f
tf =
g
=
2
(9.83m / s )
≈ 3.494 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 − v0t − 12 gt 2
y (t f ) = 0 = h − v0t f − 12 gt 2f
− v0 &plusmn; (vo ) 2 + 2 gh − 20m / s + (−20m / s ) 2 + 2(9.83m / s 2 )(60m)
tf =
=
≈ 2.01s
g Take the positive root!
9.83m / s 2
R. Field 1/14/2014
University of Florida
PHY 2053
Page 13
Final Exam Fall 2011: Problem 7
• Near the surface of the Earth a suspension bridge is a height H
above the level base of a gorge. Two identical stones are
simultaneously thrown from the bridge. One stone is thrown straight
down with speed v0 and the other is thrown straight up a the same
speed v0. Ignore air resistance. If one of the stones lands at the
bottom of the gorge 2 seconds before the other, what is the speed v0
(in m/s)?
ydown (t ) = H − vot − 12 gt 2
yup (t ) = H + vot − 12 gt 2
2
% Right: 36% yup (tup ) = 0 = H + vo tup − 12 gtup
votup = 12 gtup2 − H
vot down = − gt
1
2
v0 = g
1
2
2
down
(t − t
2
up
+H
2
down
)
(tup + t down )
R. Field 1/14/2014
University of Florida
2
ydown (t down ) = 0 = H − vot down − 12 gt down
v0 (tup + t down ) = 12 g (t − t
2
up
y-axis
2
down
)
H
x-axis
= 12 g (tup − t down ) = 12 (9.8m / s 2 )(2s ) = 9.8m / s
PHY 2053
Page 14
Exam 1 Fall 2010: Problem 4
• A motorist drives along a straight road at a constant
speed of 80 m/s. Just as she passes a parked motorcycle
police officer, the officer takes off after her at a constant
acceleration. If the officer maintains this constant value
of acceleration, what is the speed of the police officer
when he reaches the motorist?
% Right: 18%
acar = 0
vcar (t ) = v0
xcar (t ) = v0t
R. Field 1/14/2014
University of Florida
Let tc be the time it takes for officer
to reach the motorist.
xcar (tc ) = xcop (tc )
acop = a
vcop (t ) = at
v0tc = at
1
2
xcop (t ) = 12 at 2
2
c
2v0
tc =
a
⎛ 2v0 ⎞
vcop (tc ) = atc = a⎜
⎟ = 2v0
⎝ a ⎠
= 2(80m / s ) = 160m / s
PHY 2053
Page 15
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