Constant Acceleration and Free Fall
L-4 constant acceleration
and free fall
R
E
V
I
E
W
•
Acceleration is the
change
in velocity with time
•
Galileo showed that
i
n the absence of air
resistance, all objects,
regardless of their mass,
fall to earth with the same acceleration g
•
g
1
0
m
/
s
2
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1
0
m
/
s
e
v
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r
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c
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n
d
•
Free fall is an example of motion with constant
acceleration
1
Motion with constant acceleration
•
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)
•
acceleration is measured in distance units
divided by (time)
2
, for example: m/s
2
, cm/s
2
, ft/s
2
•
We will see how the velocity of an object
changes when it experiences constant
acceleration.
•
First, we’ll consider the simplest case where the
acceleration is zero, so that the velocity is
constant.
2
Simplest case:
constant velocity
(a=0)
•
If a = 0, then the velocity v is constant.
•
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x
f
=
x
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+
v
t
(
f
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a
=
0
)
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x
i
is the starting (initial) position, and
x
f
is the final position.
3
Example: constant velocity (a = 0)
•
A car moves with a constant velocity of 25
m/s. How far will it travel in 4 seconds?
•
Solution:
Suppose we take the starting point
x
i
as zero. (We can put x = 0 wherever we
like).
•
Then, x
f
= 0 + vt = 0 + (25 m/s)(4 s) = 100 m
4
The 100 m dash
•
Usain Bolt in 2009 set a new
world record ( ) in the 100 m
dash at 9.58 s.
•
Did he run with constant
velocity, or was his motion
accelerated?
•
Initially at the starting line he
was not moving
(at rest),
then
he began moving when the
gun went off,
so his motion
was clearly
accelerated
•
Although his
average speed
was about 100 m/10 s = 10
m/s, he did not maintain this
speed during the entire race.
5
100 m dash -- Seoul 1988
6
Ben Johnson
Carl Lewis
Florence Griffith Joyner
How to calculate acceleration
Example:
Starting from rest, a car accelerates
up to 50 m/s (112 mph) in 5 sec. Assuming
that the acceleration was constant, compute
the acceleration.
Solution:
acceleration (a) = rate of change of
velocity with time
7
Motion with Constant acceleration
•
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=
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)
•
a t is the amount by which the velocity
increases
from v
i
to v
f
after a time t.
•
Note that if a = 0, v
f
= v
i
, i.e., velocity is constant.
8
Example: constant acceleration
•
A car moving initially at v
i
= 3 m/s begins
accelerating with a = 2 m/s
2
.
What is its velocity at t = 5 s?
•
Solution:
•
v
f
=
v
i
+
a
t
= 3 m/s + 2 m/s
2
5 s
= 3 m/s + 10 m/s
= 13 m/s
9
Example –
deceleration – slowing down
•
Deceleration means that the acceleration
is
opposite
in direction to the velocity
•
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1
5
m
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–
5
m
/
s
2
.
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?
•
v
f
= v
i
+ a t
•
0 = 15 m/s + (
–
5 m/s
2
) t
•
0 = 15 – 5t
5t = 15
t = 15/5 = 3 s
10
Free Fall:
Motion with constant acceleration
•
According to Galileo,
in the absence of air
resistance
, all objects fall to earth with a
constant
acceleration a = g
10 m/s
2
•
g is the special symbol we use for the
acceleration due gravity.
•
Since we know how to deal with constant
acceleration, we can also solve problems
involving free fall.
11
Free fall – velocity and distance
•
If we observe an object falling from the top of
a building we find that it gains speed as it falls
•
Every second, its speed increases by 10 m/s.
•
We also observe that it does not fall equal
distances in equal time intervals.
The formula
in the right column was discovered by Galileo
.
12
Ball dropped from rest
•
If the ball is dropped
from rest,
that means that
its initial velocity is zero, v
i
= 0
•
T
h
e
n
i
t
s
f
i
n
a
l
v
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l
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a
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t
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v
f
=
a
t
,
w
h
e
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e
a
=
g
1
0
m
/
s
2
s
o
,
v
f
=
g
t
•
Example
: What is the velocity of a ball 5 sec.
after it is dropped from rest from the top of the
Sears Tower (
Willis Tower
)?
Solution
: v
f
= g t = 10 m/s
2
5 s = 50 m/s
(about 112 mph)
13
Relationship between time
and distance in free fall
•
It would be useful to know how long it
would take for an object,
dropped from
rest,
to fall a certain distance
•
For example, how long would it take an
object to fall to the ground from the top of
the Sears Tower, a distance of 442 m?
•
Or, after a certain time, how far will an
object,
dropped from rest
, have fallen?
14
Falling distance
•
Suppose an object
falls from rest
so its
initial velocity v
i
= 0.
•
After a time t the ball will have fallen a
distance:
y
f
=
½
acceleration
time
2
•
y
f
=
½ g
t
2
(This is the formula Galileo found.)
•
For vertical motion, we use the symbol y for
distance; for horizontal motion we use x.
15
Falling from the Sears Tower
•
Example
: How far would a ball dropped
from rest at the top of the Sears Tower fall
in 5 seconds?
•
Solution:
y
f
=
½
10 m/s
2
(5 s)
2
= 5
25
= 125 m (about 410 feet)
16
Time to reach the ground
•
Another interesting question, is how long it
will take an object,
dropped from rest
from
the top of the Sears Tower (442 m) take to
reach the ground?
•
To answer this question we need to solve
the time-distance formula for t
17
Velocity as object hits the ground
•
Q: How fast will the object be moving when it
hits the ground?
•
A: We apply the velocity – time relation
v
f
= v
i
+ g t, with v
i
= 0.
v
f
= g
t = 10 m/s
2
9.4 s = 94 m/s
or
about 210 mph
(neglecting air resistance)
18
Time to go up
•
Suppose a ball is thrown straight up with a speed v
i
.
When does it reach its maximum height?
•
As it rises, it slows down (decelerates)
because gravity is pulling it down.
•
At its maximum height v
f
= 0
(We say that it is instantaneously at rest).
•
v
f
=
v
i
+
a
t
a
p
p
l
i
e
s
w
h
e
t
h
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,
s
o
a
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w
n
=
+
g
=
1
0
m
/
s
2
;
o
n
t
h
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w
a
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p
,
i
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s
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s
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a
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=
g
=
1
0
m
/
s
2
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S
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c
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f
=
0
a
t
t
h
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t
o
p
,
t
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n
w
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a
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:
v
f
=
0
=
v
i
+
(
g
)
t
,
s
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p
=
v
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/
g
(
t
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t
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a
x
.
h
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g
h
t
)
19
Example
•
A volleyball player can jump straight
up at 5 m/s.
How long is she in the air?
•
Solution
:
total time = t
total
= t
up
+ t
down
•
time for her to get to top = t
up
= v
i
/ g
where v
i
is her initial upward velocity
t
up
= 5 m/s / 10 m/s
2
= ½ sec.
•
It takes exactly the same amount of time to reach the
top as it does to return to the ground, or t
up
= t
down
, so
t
total
= ½ s + ½ s = 1 s (This is the amount of time
that she is in the air.)
20
Escape from planet earth
•
To escape from the gravitational pull of the earth
an object must be given a velocity larger than the
so called
escape velocity
•
For earth the escape velocity is 7 mi/sec or 11,000
m/s, 11 kilometers/sec or about 25,000 mph.
•
An object given at least this velocity on the earth’s
surface can escape from earth!
•
The Voyager 2 spacecraft, launched on Aug. 20,
1977, recently left the solar system and is the first
human-made object to reach interstellar space.
21
Acceleration is the rate of change in velocity with time. In free fall, all objects fall to Earth with the same acceleration regardless of mass. This review covers motion with constant acceleration, including examples and explanations of velocity changes and simple cases with zero acceleration.
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Presentation Transcript
L-4 constant acceleration and free fall REVIEW Acceleration is the change in velocity with time Galileo showed that in the absence of air resistance, all objects, regardless of their mass, fall to earth with the same acceleration g g 10 m/s2 the speed of a falling object increases by 10 m/s every second Free fall is an example of motion with constant acceleration 1
Motion with constant acceleration acceleration is the rate at which the velocity changeswith time (increases or decreases) acceleration is measured in distance units divided by (time)2, for example: m/s2, cm/s2, ft/s2 We will see how the velocity of an object changes when it experiences constant acceleration. First, we ll consider the simplest case where the acceleration is zero, so that the velocity is constant. 2
Simplest case: constant velocity (a=0) If a = 0, then the velocity v is constant. In this case the distance xf an object will travel in a certain amount of time t is given by distance = velocity x time xf = xi + v t (for a = 0) xi is the starting (initial) position, and xf is the final position. 3
Example: constant velocity (a = 0) A car moves with a constant velocity of 25 m/s. How far will it travel in 4 seconds? Solution: Suppose we take the starting point xi as zero. (We can put x = 0 wherever we like). Then, xf = 0 + vt = 0 + (25 m/s)(4 s) = 100 m 4
The 100 m dash Usain Bolt in 2009 set a new world record ( ) in the 100 m dash at 9.58 s. Did he run with constant velocity, or was his motion accelerated? Initially at the starting line he was not moving (at rest), then he began moving when the gun went off, so his motion was clearly accelerated Although his average speed was about 100 m/10 s = 10 m/s, he did not maintain this speed during the entire race. 5
100 m dash -- Seoul 1988 100 m dash Rome 1988 12 10 Ben Johnson Carl Lewis Florence Griffith Joyner 8 Ben Johnson Carl Lewis Fl-G-Joyner 6 4 2 0 0 20 40 distance (meters) 60 80 100 6
How to calculate acceleration Example: Starting from rest, a car accelerates up to 50 m/s (112 mph) in 5 sec. Assuming that the acceleration was constant, compute the acceleration. Solution: acceleration (a) = rate of change of velocity with time change in velocity time interval 50m/s-0m/s 5 s - 0 s final velocity- initial velocity final time-initial time 50 m/s = 10 m/s 5 s a= = 2 = = 7
Motion with Constant acceleration Suppose an object moves with a constant acceleration a. If at t = 0 its initial velocity is (vi), then we want to know what its final velocity (vf) be after a time t has passed. final velocity = initial velocity + acceleration time vf = vi + a t (for constant acceleration) a t is the amount by which the velocity increases from vi to vf after a time t. Note that if a = 0, vf = vi, i.e., velocity is constant. 8
Example: constant acceleration A car moving initially at vi = 3 m/s begins accelerating with a = 2 m/s2. What is its velocity at t = 5 s? Solution: vf = vi + a t = 3 m/s + 2 m/s2 5 s = 3 m/s + 10 m/s = 13 m/s 9
Example deceleration slowing down Deceleration means that the acceleration is opposite in direction to the velocity Suppose you are moving at vi 15 m/s and apply the brakes. The brakes provide a constant deceleration of 5 m/s2. How long will it take the car to stop? vf = vi + a t 0 = 15 m/s + ( 5 m/s2) t 0 = 15 5t 5t = 15 t = 15/5 = 3 s 10
Free Fall: Motion with constant acceleration According to Galileo, in the absence of air resistance, all objects fall to earth with a constant acceleration a = g 10 m/s2 g is the special symbol we use for the acceleration due gravity. Since we know how to deal with constant acceleration, we can also solve problems involving free fall. 11
Free fall velocity and distance time (s) 0 1 2 3 4 5 velocity (m/s) 0 10 20 30 40 50 distance y (m) 0 = 10 (0)2 5 = 10 (1)2 20 = 10 (2)2 45 = 10 (3)2 80 = 10 (4)2 125 = 10 (5)2 If we observe an object falling from the top of a building we find that it gains speed as it falls Every second, its speed increases by 10 m/s. We also observe that it does not fall equal distances in equal time intervals. The formula in the right column was discovered by Galileo. 12
Ball dropped from rest If the ball is dropped from rest, that means that its initial velocity is zero, vi = 0 Then its final velocity after a time t is vf = a t, where a = g 10 m/s2 so, vf = g t Example: What is the velocity of a ball 5 sec. after it is dropped from rest from the top of the Sears Tower (Willis Tower)? Solution: vf = g t = 10 m/s2 5 s = 50 m/s (about 112 mph) 13
Relationship between time and distance in free fall It would be useful to know how long it would take for an object, dropped from rest, to fall a certain distance For example, how long would it take an object to fall to the ground from the top of the Sears Tower, a distance of 442 m? Or, after a certain time, how far will an object, dropped from rest, have fallen? 14
Falling distance Suppose an object falls from rest so its initial velocity vi = 0. After a time t the ball will have fallen a distance: yf = acceleration time2 yf = g t2 (This is the formula Galileo found.) For vertical motion, we use the symbol y for distance; for horizontal motion we use x. 15
Falling from the Sears Tower Example: How far would a ball dropped from rest at the top of the Sears Tower fall in 5 seconds? Solution: yf = 10 m/s2 (5 s)2 = 5 25 = 125 m (about 410 feet) 16
Time to reach the ground Another interesting question, is how long it will take an object, dropped from rest from the top of the Sears Tower (442 m) take to reach the ground? To answer this question we need to solve the time-distance formula for t 1 y = gt 2y = gt 2 2y g 2y g 2 2 2 t = t = f f f f 2 442 10 884 10 So: t = = = 88.4 = 9.4 s. 17
Velocity as object hits the ground Q: How fast will the object be moving when it hits the ground? A: We apply the velocity time relation vf = vi + g t, with vi = 0. vf = g t = 10 m/s2 9.4 s = 94 m/s or about 210 mph (neglecting air resistance) 18
Time to go up Suppose a ball is thrown straight up with a speed vi. When does it reach its maximum height? As it rises, it slows down (decelerates) because gravity is pulling it down. At its maximum height vf = 0 (We say that it is instantaneously at rest). vf = vi + a t applies whether an object is falling or rising. On the way down it speeds up, so adown = +g = 10m/s2; on the way up, it slows down, so aup = g = 10m/s2 Since vf = 0 at the top, then we have: vf = 0 = vi + ( g) t, so tup = vi / g (time to max. height) 19
Example A volleyball player can jump straight up at 5 m/s. How long is she in the air? Solution: total time = ttotal = tup + tdown time for her to get to top = tup = vi / g where vi is her initial upward velocity tup = 5 m/s / 10 m/s2 = sec. It takes exactly the same amount of time to reach the top as it does to return to the ground, or tup = tdown, so ttotal = s + s = 1 s (This is the amount of time that she is in the air.) 20
Escape from planet earth To escape from the gravitational pull of the earth an object must be given a velocity larger than the so called escape velocity For earth the escape velocity is 7 mi/sec or 11,000 m/s, 11 kilometers/sec or about 25,000 mph. An object given at least this velocity on the earth s surface can escape from earth! The Voyager 2 spacecraft, launched on Aug. 20, 1977, recently left the solar system and is the first human-made object to reach interstellar space. 21
