XVA  Constant Acceleration 1
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Here we are looking at the motion of an
object which is accelerating at a constant value.
[1] Object is not moving. Starting position
is at the origin of the position number line, velocity is
constantly zero, acceleration is constantly zero.
Parameters: 
x_{o}
= 0.0 m 
v_{o}
= 0.0 m/s^{ } 
a =
0.0 m/s^{2 } 



The object's position
starts at x = 0.0 m, and it
stays at that location as time passes. 
The object's velocity
starts at v = 0.0 m/s. It's
velocity remains at that value as time passes. 
The constant acceleration
is a = 0.0 m/s^{2};
so, the velocity does not change as time
passes. 

[1] XVA
demonstration 
Comments:
 Our starting constant acceleration
example shows a 0.0 m/s^{2} acceleration with
a starting velocity of 0.0 m/s. The object stands
still.
 See
Constant Acceleration Animation 1.
[2]Starting from rest the object experiences a small constant
positive acceleration.
Back
Parameters: 
x_{o}
= 0.0 m 
v_{o}
= 0.0 m/s^{ } 
a =
0.5 m/s^{2 } 



The object's position
starts at x = 0.0 m. It moves
in a positive direction, and in each second
it passes through more meters than in the
previous second. 
The object's velocity
starts at v = 0.0 m/s. From
that value the object's velocity continues to
be added to at constant rate as time passes. 
The constant acceleration
is a = 0.5 m/s^{2}.
The velocity, therefore, has 0.5 m/s added to
it each second. 

[2] XVA
demonstration 
Comments:
 The position of the object starts at 0.0
m.
 The velocity of the object starts at 0.0
m/s.
 The constant positive acceleration
means that the velocity is added to. This causes the
velocity to become more and more positive as time
passes.
 The object speeds up in a positive
direction.
 Since the value of the velocity is
always positive, (except at v_{o} = 0.0 m/s),
and becomes increasingly more positive as time
passes, the slope of the x vs. t graph is positive
and becomes increasingly more positive as time passes.
This is because the slope of the x vs. t graph is the
velocity.
 Note that the x vs. t graph is curved since
its slope is ever increasing as the velocity ever increases.
 Since the value of the acceleration is
constant and positive, the slope of the v vs. t graph
is constant and positive.
This is because the slope
of the v vs. t graph is the acceleration.
 See
Constant Acceleration Animation 1.
[3]Starting from rest the object experiences a medium constant
positive acceleration.
Back
Parameters: 
x_{o}
= 0.0 m 
v_{o}
= 0.0 m/s^{ } 
a =
1.0 m/s^{2 } 



The object's position
starts at x = 0.0 m. It moves
in a positive direction, and in each second
it passes through more meters than in the
previous second. 
The object's velocity
starts at v = 0.0 m/s. From
that value the object's velocity continues to
be added to at constant rate as time passes. 
The constant acceleration
is a = 1.0 m/s^{2}.
The velocity, therefore, has 1.0 m/s added to
it each second. 

[3] XVA
demonstration 
Comments:
 This example is much like the one in
[2] except here the acceleration is a greater
positive value.
 The slope of the v vs. t graph here is
greater than in [2].
 See
Constant Acceleration Animation 1.
[4]Starting from rest the object experiences a large constant
positive acceleration.
Back
Parameters: 
x_{o}
= 0.0 m 
v_{o}
= 0.0 m/s^{ } 
a =
1.5 m/s^{2 } 



The object's position
starts at x = 0.0 m. It moves
in a positive direction, and in each second
it passes through more meters than in the
previous second. 
The object's velocity
starts at v = 0.0 m/s. From
that value the object's velocity continues to
be added to at constant rate as time passes. 
The constant acceleration
is a = 1.5 m/s^{2}.
The velocity, therefore, has 1.5 m/s added to
it each second. 

[4] XVA
demonstration 
Comments:
 The velocity of the object starts out
at 0.0 m/s.
 The positive acceleration means that
the velocity is added to. This causes the velocity to
become more and more positive as time passes.
 The addition to the velocity here is
greater than in [3].
 The object speeds up in a positive
direction.
 Since the value of the velocity is
always positive, (except v_{o} = 0.0 m/s),
and increasingly more positive, the slope of the x vs.
t graph is positive and increasingly more positive.
This is because the slope of the x vs. t graph is the
velocity.
 The increasing of the slope of the x
vs. t graph here proceeds more rapidly than in
[3].
 Since the value of the acceleration is
constant and positive, the slope of the v vs. t graph
is constant and positive.
This is because the slope
of the v vs. t graph is the acceleration.
 Since the acceleration here is greater
than in [3], the slope of the v vs. t graph here is
greater than in [3].
 See
Constant Acceleration Animation 1.
[5]Starting from rest the object experiences a small constant
negative acceleration.
Back
Parameters: 
x_{o}
= 0.0 m 
v_{o}
= 0.0 m/s^{ } 
a =
0.5 m/s^{2 } 



The object's position
starts at x = 0.0 m. It moves
in a negative direction, and in each second
it passes through more meters than in the
previous second. 
The object's velocity
starts at v = 0.0 m/s. From
that value the object's velocity continues to
be subtracted from at constant rate as time
passes. 
The constant acceleration
is a = 0.5 m/s^{2}.
The velocity, therefore, has 0.5 m/s
subtracted from it each second. 

[5] XVA
demonstration 
Comments:
[6]Starting from rest the object experiences a medium constant
negative acceleration.
Parameters: 
x_{o}
= 0.0 m 
v_{o}
= 0.0 m/s^{ } 
a =
1.0 m/s^{2 } 



The object's position
starts at x = 0.0 m. It moves
in a negative direction, and in each second
it passes through more meters than in the
previous second. 
The object's velocity
starts at v = 0.0 m/s. From
that value the object's velocity continues to
be subtracted from at constant rate as time
passes. 
The constant acceleration
is a = 1.0 m/s^{2}.
The velocity, therefore, has 1.0 m/s
subtracted from it each second. 

[6] XVA
demonstration 
Comments:
 The negative acceleration means that
the velocity is subtracted from. This causes the
velocity to become more and more negative as time
passes.
 The subtraction from the velocity here
is greater than in [5].
 The object speeds up in a negative
direction.
 Since the value of the velocity is
always negative, (except v_{o} = 0.0 m/s),
and increasing, the slope of the x vs. t graph is
negative and increasing.
This is because the slope of
the x vs. t graph is the velocity.
 The increasing of the slope of the x
vs. t graph here proceeds more rapidly than in
[5].
 Since the value of the acceleration is
constant and negative, the slope of the v vs. t graph
is constant and negative.
This is because the slope
of the v vs. t graph is the acceleration.
 Since the acceleration here is greater
than in [5], the slope of the v vs. t graph here is
greater than in [5].
 See
Constant Acceleration Animation 1.
[7] Starting from rest the object
experiences a large constant negative acceleration.
Parameters: 
x_{o}
= 0.0 m 
v_{o}
= 0.0 m/s^{ } 
a =
1.5 m/s^{2 } 



The object's position
starts at x = 0.0 m. It moves
in a negative direction, and in each second
it passes through more meters than in the
previous second. 
The object's velocity
starts at v = 0.0 m/s. From
that value the object's velocity continues to
be subtracted from at constant rate as time
passes. 
The constant acceleration
is a = 1.5 m/s^{2}.
The velocity, therefore, has 1.5 m/s
subtracted from it each second. 

[7] XVA
demonstration 
Comments:
 The negative acceleration means that
the velocity is subtracted from. This causes the
velocity to become more and more negative as time
passes.
 The subtraction from the velocity here
is greater than in [6].
 The object speeds up in a negative
direction.
 Since the value of the velocity is
always negative, (except v_{o} = 0.0 m/s),
and increasing, the slope of the x vs. t graph is
negative and increasing.
This is because the slope of
the x vs. t graph is the velocity.
 The increasing of the slope of the x
vs. t graph here proceeds more rapidly than in
[6].
 Since the value of the acceleration is
constant and negative, the slope of the v vs. t graph
is constant and negative.
This is because the slope
of the v vs. t graph is the acceleration.
 Since the acceleration here is greater
than in [6], the slope of the v vs. t graph here is
greater than in [6].
 See
Constant Acceleration Animation 1.
[8] Starting with a positive velocity the
object experiences zero acceleration.
Parameters: 
x_{o}
= 0.0 m 
v_{o}
= 4.0 m/s^{ } 
a =
0.0 m/s^{2 } 



The object's position
starts at x = 0 m. It moves in
a positive direction, and in each second it
passes through the same number of meters. 
The object's velocity
starts at v = 4.0 m/s. It's
velocity remains at that value as time passes. 
The constant acceleration
is a = 0 m/s^{2}; so,
the velocity does not change as time passes. 

[8] XVA
demonstration 
Comments:
[9] Starting with a positive velocity the
object experiences a small constant positive acceleration.
Parameters: 
x_{o}
= 0.0 m 
v_{o}
= 4.0 m/s^{ } 
a =
0.5 m/s^{2 } 



The object's position
starts at x = 0.0 m. It moves
in a positive direction, and in each second
it passes through more meters than in the
previous second. 
The object's velocity
starts at v = 4.0 m/s. From
that value the object's velocity continues to
be added to at constant rate as time passes. 
The constant acceleration
is a = 0.5 m/s^{2}.
The velocity, therefore, has 0.5 m/s added to
it each second. 

[9] XVA
demonstration 
Comments:
[10] Starting with a positive velocity the
object experiences a medium constant positive acceleration.
Parameters: 
x_{o}
= 0.0 m 
v_{o}
= 4.0 m/s^{ } 
a =
1.0 m/s^{2 } 



The object's position
starts at x = 0.0 m. It moves
in a positive direction, and in each second
it passes through more meters than in the
previous second. 
The object's velocity
starts at v = 4.0 m/s. From
that value the object's velocity continues to
be added to at constant rate as time passes. 
The constant acceleration
is a = 1.0 m/s^{2}.
The velocity, therefore, has 1.0 m/s added to
it each second. 

[10] XVA
demonstration 
Comments:
[11] Starting with a positive velocity the
object experiences a large constant positive acceleration.
Parameters: 
x_{o}
= 0.0 m 
v_{o}
= 4.0 m/s^{ } 
a =
1.5 m/s^{2 } 



The object's position
starts at x = 0.0 m. It moves
in a positive direction, and in each second
it passes through more meters than in the
previous second. 
The object's velocity
starts at v = 4.0 m/s. From
that value the object's velocity continues to
be added to at constant rate as time passes. 
The constant acceleration
is a = 1.5 m/s^{2}.
The velocity, therefore, has 1.5 m/s added to
it each second. 

[11] XVA
demonstration 
Comments:
[12]Starting with a positive velocity the object
experiences a small constant negative acceleration.
Back
Parameters: 
x_{o}
= 0.0 m 
v_{o}
= 4.0 m/s^{ } 
a =
0.5 m/s^{2 } 



The object's position
starts at x = 0.0 m. It moves
in a positive direction slowing down,
eventually stopping, and then it begins to
speed up in a negative direction. 
The object's velocity
starts at v = 4.0 m/s. From
that value the object's velocity continues to
be subtracted from at constant rate as time
passes. The velocity eventually becomes 0.0 m/s
and then goes negative. 
The constant acceleration
is a = 0.5 m/s^{2}.
The velocity, therefore, has 0.5 m/s
subtracted from it each second. 

[12] XVA
demonstration 
Comments:
 The original velocity is positive.
 The negative acceleration means that
the velocity is subtracted from. This causes the
velocity to become less and less positive until it
becomes zero, then, as it is further subtracted from,
the velocity becomes negative, (the object's motion
reverses direction).
 The object is originally positioned at
the origin of the position number line. It moves
forward slowing down, eventually stopping for a
moment. Then, it turns around, and the direction of
its motion reverses.
 Since the value of the velocity is at
first positive, then zero for a moment, and finally
negative, the slope of the x vs. t graph is positive
at first, zero for a moment, and ends up negative.
This is because the slope of the x vs. t graph is the
velocity.
 Since the value of the acceleration is
constant and negative, the slope of the v vs. t graph
is constant and negative.
This is because the slope
of the v vs. t graph is the acceleration.
 See
Constant Acceleration Animation 2.
[13]Starting with a positive velocity the object
experiences a medium constant negative acceleration.
Parameters: 
x_{o}
= 0.0 m 
v_{o}
= 4.0 m/s^{ } 
a =
1.0 m/s^{2 } 



The object's position
starts at x = 0.0 m. It moves
in a positive direction slowing down,
eventually stopping, and then it begins to
speed up in a negative direction. 
The object's velocity
starts at v = 4.0 m/s. From
that value the object's velocity continues to
be subtracted from at constant rate as time
passes. The velocity eventually becomes 0.0 m/s
and then goes negative. 
The constant acceleration
is a = 1.0 m/s^{2}.
The velocity, therefore, has 1.0 m/s
subtracted from it each second. 

[13] XVA
demonstration 
Comments:
[14] Starting with a positive velocity the
object experiences a large constant negative acceleration.
Parameters: 
x_{o}
= 0.0 m 
v_{o}
= 4.0 m/s^{ } 
a =
1.5 m/s^{2 } 



The object's position
starts at x = 0.0 m. It moves
in a positive direction slowing down,
eventually stopping, and then it begins to
speed up in a negative direction. 
The object's velocity
starts at v = 4.0 m/s. From
that value the object's velocity continues to
be subtracted from at constant rate as time
passes. The velocity eventually becomes 0.0 m/s
and then goes negative. 
The constant acceleration
is a = 1.5 m/s^{2}.
The velocity, therefore, has 1.5 m/s
subtracted from it each second. 

[14] XVA
demonstration 
Comments:
[15] Starting with a negative velocity the
object experiences zero acceleration.
Parameters: 
x_{o}
= 0.0 m 
v_{o}
= 4.0 m/s^{ } 
a =
0.0 m/s^{2 } 



The object's position
starts at x = 0 m. It moves in
a negative direction, and in each second it
passes through the same number of meters. 
The object's velocity
starts at v = 4.0 m/s. It's
velocity remains at that value as time passes. 
The constant acceleration
is a = 0 m/s^{2}; so,
the velocity does not change as time passes. 

[15] XVA
demonstration 
Comments:
[16]Starting with a negative velocity the object
experiences a small constant positive acceleration.
Back
Parameters: 
x_{o}
= 0.0 m 
v_{o}
= 4.0 m/s^{ } 
a =
0.5 m/s^{2 } 



The object's position
starts at x = 0.0 m. It moves
in a negative direction slowing down,
eventually stopping, and then it begins to
speed up in a positive direction. 
The object's velocity
starts at v = 4.0 m/s. From
that value the object's velocity continues to
be added to at constant rate as time passes.
The velocity eventually becomes 0 m/s and
then goes positive. 
The constant acceleration
is a = 0.5 m/s^{2}.
The velocity, therefore, has 0.5 m/s added to
it each second. 

[16] XVA
demonstration 
Comments:
 The original velocity is negative.
 The positive acceleration means that
the velocity is added to. This causes the velocity to
become less and less negative until it becomes zero,
then, as it is further added to, the velocity becomes
positive, (the object's motion reverses direction).
 The object is originally positioned at
the origin of the position number line. It moves
backward slowing down, eventually stopping for a
moment. Then, it turns around, and the direction of
its motion reverses.
 Since the value of the velocity is at
first negative, then zero for a moment, and finally
positive, the slope of the x vs. t graph is negative
at first, zero for a moment, and ends up positive.
This is because the slope of the x vs. t graph is the
velocity.
 Since the value of the acceleration is
constant and positive, the slope of the v vs. t graph
is constant and positive.
This is because the slope
of the v vs. t graph is the acceleration.
 See
Constant Acceleration Animation 3.
[17]Starting with a negative velocity the object
experiences a medium constant positive acceleration.
Back
Parameters: 
x_{o}
= 0.0 m 
v_{o}
= 4.0 m/s^{ } 
a =
1.0 m/s^{2 } 



The object's position
starts at x = 0.0 m. It moves
in a negative direction slowing down,
eventually stopping, and then it begins to
speed up in a positive direction. 
The object's velocity
starts at v = 4.0 m/s. From
that value the object's velocity continues to
be added to at constant rate as time passes.
The velocity eventually becomes 0 m/s and
then goes positive. 
The constant acceleration
is a = 1.0 m/s^{2}.
The velocity, therefore, has 1.0 m/s added to
it each second. 

[17] XVA
demonstration 
Comments:
[18] Starting with a negative velocity the
object experiences a large constant positive acceleration.
Parameters: 
x_{o}
= 0.0 m 
v_{o}
= 4.0 m/s^{ } 
a =
1.5 m/s^{2 } 



The object's position
starts at x = 0.0 m. It moves
in a negative direction slowing down,
eventually stopping, and then it begins to
speed up in a positive direction. 
The object's velocity
starts at v = 4.0 m/s. From
that value the object's velocity continues to
be added to at constant rate as time passes.
The velocity eventually becomes 0 m/s and
then goes positive. 
The constant acceleration
is a = 1.5 m/s^{2}.
The velocity, therefore, has 1.5 m/s added to
it each second. 

[18] XVA
demonstration 
Comments:
[19]Starting with a negative velocity the object
experiences a small constant negative acceleration.
Back
Parameters: 
x_{o}
= 0.0 m 
v_{o}
= 4.0 m/s^{ } 
a =
0.5 m/s^{2 } 



The object's position
starts at x = 0.0 m. It moves
in a negative direction, and in each second
it passes through more meters than in the
previous second. 
The object's velocity
starts at v = 4.0 m/s. From
that value the object's velocity continues to
be subtracted from at constant rate as time
passes. 
The constant acceleration
is a = 0.5 m/s^{2}.
The velocity, therefore, has 0.5 m/s
subtracted from it each second. 

[19] XVA
demonstration 
Comments:
[20]Starting with a negative velocity the object
experiences a medium constant negative acceleration.
Back
Parameters: 
x_{o}
= 0.0 m 
v_{o}
= 4.0 m/s^{ } 
a =
1.0 m/s^{2 } 



The object's position
starts at x = 0.0 m. It moves
in a negative direction, and in each second
it passes through more meters than in the
previous second. 
The object's velocity
starts at v = 4.0 m/s. From
that value the object's velocity continues to
be subtracted from at constant rate as time
passes. 
The constant acceleration
is a = 1.0 m/s^{2}.
The velocity, therefore, has 1.0 m/s
subtracted from it each second. 

[20] XVA
demonstration 
Comments:
 This example is much like the one in
[19] except here there is a greater negative
acceleration causing the velocity to be subtracted
from by a greater amount each second, and, thus,
causing the object to speed up more rapidly in a
negative direction than in
[19].
 See
Constant Acceleration Animation 3.
[21] Starting with a negative velocity the
object experiences a large constant negative acceleration.
Parameters: 
x_{o}
= 0.0 m 
v_{o}
= 4.0 m/s^{ } 
a =
1.5 m/s^{2 } 



The object's position
starts at x = 0.0 m. It moves
in a negative direction, and in each second
it passes through more meters than in the
previous second. 
The object's velocity
starts at v = 4.0 m/s. From
that value the object's velocity continues to
be subtracted from at constant rate as time
passes. 
The constant acceleration
is a = 1.5 m/s^{2}.
The velocity, therefore, has 1.5 m/s
subtracted from it each second. 

[21] XVA
demonstration 
Comments:
 This example is much like the ones in
[19] and
[20] except here there is a greater negative
acceleration causing the velocity to be subtracted
from by a greater amount each second, and, thus,
causing the object to speed up more rapidly in a
negative direction than in
[19] and
[20].
 See
Constant Acceleration Animation 3.
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