Derivative, Velcity, Acceleration and MIT Pumpkin
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A pumpkin is thrown-up from a MIT building as shown in 1st graph. 2nd graph is of its velocity (first derivative) and 3rd one is of acceleration (2nd derivative).
PROBLEM: I can't relate the reality of the motion of the pumpkin with the graphs.
Reality of Velocity: When we throw the pumpkin, its speed (velocity) starts to decrease and become zero at top and then speed starts to increase but since velocity is speed with direction and direction is downwards, hence velocity graph is going down. This means increase in velocity is being represented as negative number and a downward graph, I wonder how will one represent decreasing velocity, with a positive number and upward direction ?)
Reality of Acceleration: Going up pumpkin is de-accelerating (decreasing) and going down it is accelerating (increasing) but graph is constant, just straight line. By looking only at the graph of acceleration and without looking at velocity and the first graph, one can conclude acceleration was constant negative number. Hence when we throw anything up, it just continues to de-accelerate even when it falls down. Logic is wrong but that is what the graph says.
derivatives
asked Aug 24 at 10:42
Arnuld
7811
 |Â
show 3 more comments
up vote
0
down vote
favorite
A pumpkin is thrown-up from a MIT building as shown in 1st graph. 2nd graph is of its velocity (first derivative) and 3rd one is of acceleration (2nd derivative).
PROBLEM: I can't relate the reality of the motion of the pumpkin with the graphs.
Reality of Velocity: When we throw the pumpkin, its speed (velocity) starts to decrease and become zero at top and then speed starts to increase but since velocity is speed with direction and direction is downwards, hence velocity graph is going down. This means increase in velocity is being represented as negative number and a downward graph, I wonder how will one represent decreasing velocity, with a positive number and upward direction ?)
Reality of Acceleration: Going up pumpkin is de-accelerating (decreasing) and going down it is accelerating (increasing) but graph is constant, just straight line. By looking only at the graph of acceleration and without looking at velocity and the first graph, one can conclude acceleration was constant negative number. Hence when we throw anything up, it just continues to de-accelerate even when it falls down. Logic is wrong but that is what the graph says.
derivatives
asked Aug 24 at 10:42
Arnuld
7811
They are vectors and we usually choose the up direction to be the positive one. That's all.
– tst
Aug 24 at 10:44
Notice that if the number -2 decreases, its absolute value increases.
– tst
Aug 24 at 10:45
@tst Oh.. I need to learn Vectors then. 2nd, how |-2| is related here ?
– Arnuld
Aug 24 at 10:49
De-acceleration is an opinion and it is mostly used when the absolute value of the velocity decreases. You shouldn't think in this terms, there is only acceleration that measures the rate of change of the velocity.
– tst
Aug 24 at 10:52
You mean, while doing Math, I should not think of real-life experience of (de-)acceleration but more of as just a 2nd derivative like any other equation ?
– Arnuld
Aug 24 at 10:59
 |Â
show 3 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
A pumpkin is thrown-up from a MIT building as shown in 1st graph. 2nd graph is of its velocity (first derivative) and 3rd one is of acceleration (2nd derivative).
PROBLEM: I can't relate the reality of the motion of the pumpkin with the graphs.
Reality of Velocity: When we throw the pumpkin, its speed (velocity) starts to decrease and become zero at top and then speed starts to increase but since velocity is speed with direction and direction is downwards, hence velocity graph is going down. This means increase in velocity is being represented as negative number and a downward graph, I wonder how will one represent decreasing velocity, with a positive number and upward direction ?)
Reality of Acceleration: Going up pumpkin is de-accelerating (decreasing) and going down it is accelerating (increasing) but graph is constant, just straight line. By looking only at the graph of acceleration and without looking at velocity and the first graph, one can conclude acceleration was constant negative number. Hence when we throw anything up, it just continues to de-accelerate even when it falls down. Logic is wrong but that is what the graph says.
derivatives
asked Aug 24 at 10:42
Arnuld
7811
A pumpkin is thrown-up from a MIT building as shown in 1st graph. 2nd graph is of its velocity (first derivative) and 3rd one is of acceleration (2nd derivative).
PROBLEM: I can't relate the reality of the motion of the pumpkin with the graphs.
Reality of Velocity: When we throw the pumpkin, its speed (velocity) starts to decrease and become zero at top and then speed starts to increase but since velocity is speed with direction and direction is downwards, hence velocity graph is going down. This means increase in velocity is being represented as negative number and a downward graph, I wonder how will one represent decreasing velocity, with a positive number and upward direction ?)
Reality of Acceleration: Going up pumpkin is de-accelerating (decreasing) and going down it is accelerating (increasing) but graph is constant, just straight line. By looking only at the graph of acceleration and without looking at velocity and the first graph, one can conclude acceleration was constant negative number. Hence when we throw anything up, it just continues to de-accelerate even when it falls down. Logic is wrong but that is what the graph says.
derivatives
asked Aug 24 at 10:42
Arnuld
7811
asked Aug 24 at 10:42
Arnuld
7811
asked Aug 24 at 10:42
Arnuld
7811
asked Aug 24 at 10:42
Arnuld
7811
7811
They are vectors and we usually choose the up direction to be the positive one. That's all.
– tst
Aug 24 at 10:44
Notice that if the number -2 decreases, its absolute value increases.
– tst
Aug 24 at 10:45
@tst Oh.. I need to learn Vectors then. 2nd, how |-2| is related here ?
– Arnuld
Aug 24 at 10:49
De-acceleration is an opinion and it is mostly used when the absolute value of the velocity decreases. You shouldn't think in this terms, there is only acceleration that measures the rate of change of the velocity.
– tst
Aug 24 at 10:52
You mean, while doing Math, I should not think of real-life experience of (de-)acceleration but more of as just a 2nd derivative like any other equation ?
– Arnuld
Aug 24 at 10:59
 |Â
show 3 more comments
They are vectors and we usually choose the up direction to be the positive one. That's all.
– tst
Aug 24 at 10:44
Notice that if the number -2 decreases, its absolute value increases.
– tst
Aug 24 at 10:45
@tst Oh.. I need to learn Vectors then. 2nd, how |-2| is related here ?
– Arnuld
Aug 24 at 10:49
De-acceleration is an opinion and it is mostly used when the absolute value of the velocity decreases. You shouldn't think in this terms, there is only acceleration that measures the rate of change of the velocity.
– tst
Aug 24 at 10:52
You mean, while doing Math, I should not think of real-life experience of (de-)acceleration but more of as just a 2nd derivative like any other equation ?
– Arnuld
Aug 24 at 10:59
They are vectors and we usually choose the up direction to be the positive one. That's all.
– tst
Aug 24 at 10:44
They are vectors and we usually choose the up direction to be the positive one. That's all.
– tst
Aug 24 at 10:44
Notice that if the number -2 decreases, its absolute value increases.
– tst
Aug 24 at 10:45
Notice that if the number -2 decreases, its absolute value increases.
– tst
Aug 24 at 10:45
@tst Oh.. I need to learn Vectors then. 2nd, how |-2| is related here ?
– Arnuld
Aug 24 at 10:49
@tst Oh.. I need to learn Vectors then. 2nd, how |-2| is related here ?
– Arnuld
Aug 24 at 10:49
De-acceleration is an opinion and it is mostly used when the absolute value of the velocity decreases. You shouldn't think in this terms, there is only acceleration that measures the rate of change of the velocity.
– tst
Aug 24 at 10:52
De-acceleration is an opinion and it is mostly used when the absolute value of the velocity decreases. You shouldn't think in this terms, there is only acceleration that measures the rate of change of the velocity.
– tst
Aug 24 at 10:52
You mean, while doing Math, I should not think of real-life experience of (de-)acceleration but more of as just a 2nd derivative like any other equation ?
– Arnuld
Aug 24 at 10:59
You mean, while doing Math, I should not think of real-life experience of (de-)acceleration but more of as just a 2nd derivative like any other equation ?
– Arnuld
Aug 24 at 10:59
 |Â
show 3 more comments
1 Answer
1
active
oldest
votes
up vote
0
down vote
accepted
You’re using the word acceleration to mean two different, though related, things.
In colloquial usage (what you all “real lifeâ€Â), to accelerate means to increase speed, and to decelerate means to decrease speed. However, you’re working with velocities, which are directed—here signed—quantities. In this context, the word “acceleration†refers to any change in velocity, regardless of whether it increases or decreases the object’s speed, which is the velocity’s absolute value. This acceleration is also a signed quantity, but to determine whether it corresponds to an increase or decrease in speed, you also have to consider the sign of the velocity: when the signs are the same, the speed increases—an “acceleration†in the first sense above; when the signs differ, the speed decreases—a “deceleration.â€Â
answered Aug 24 at 21:34
amd
26.6k21046
I think I I understoond most of it. One last thing, in our pumpkin example., acceleration and velocity are both increasing and both have same sign (negative), then why is that only velocity graph is showing increase in velocity while acceleration graph shows constant ?
– Arnuld
Aug 25 at 4:10
Not at all. Until the pumpkin reaches its maximum height, the velocity and acceleration have different signs—the pumpkin slows down. Once the pumpkin starts dropping, they have the same sign—the pumpkin speeds up. The acceleration (due to gravity) is constant throughout.
– amd
Aug 25 at 4:28
Oh no, they are talking about acceleration due to gravity and I was confusing it with acceleration of the "throw". I was thinking this way: the force with which pumpkin is thrown up makes velocity up and acceleration up. That acceleration is not taken into account I see.
– Arnuld
Aug 25 at 9:08
@Arnuld The initial impetus is instantaneous in this simplified model. If you like, assume that the pumpkin had already been set in motion before you started tracking it. Either way, there is no upward component of acceleration during the time that is being shown.
– amd
Aug 25 at 18:08
understood. Thanks
– Arnuld
Aug 26 at 15:13
 |Â
show 1 more comment
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
You’re using the word acceleration to mean two different, though related, things.
In colloquial usage (what you all “real lifeâ€Â), to accelerate means to increase speed, and to decelerate means to decrease speed. However, you’re working with velocities, which are directed—here signed—quantities. In this context, the word “acceleration†refers to any change in velocity, regardless of whether it increases or decreases the object’s speed, which is the velocity’s absolute value. This acceleration is also a signed quantity, but to determine whether it corresponds to an increase or decrease in speed, you also have to consider the sign of the velocity: when the signs are the same, the speed increases—an “acceleration†in the first sense above; when the signs differ, the speed decreases—a “deceleration.â€Â
answered Aug 24 at 21:34
amd
26.6k21046
I think I I understoond most of it. One last thing, in our pumpkin example., acceleration and velocity are both increasing and both have same sign (negative), then why is that only velocity graph is showing increase in velocity while acceleration graph shows constant ?
– Arnuld
Aug 25 at 4:10
Not at all. Until the pumpkin reaches its maximum height, the velocity and acceleration have different signs—the pumpkin slows down. Once the pumpkin starts dropping, they have the same sign—the pumpkin speeds up. The acceleration (due to gravity) is constant throughout.
– amd
Aug 25 at 4:28
Oh no, they are talking about acceleration due to gravity and I was confusing it with acceleration of the "throw". I was thinking this way: the force with which pumpkin is thrown up makes velocity up and acceleration up. That acceleration is not taken into account I see.
– Arnuld
Aug 25 at 9:08
@Arnuld The initial impetus is instantaneous in this simplified model. If you like, assume that the pumpkin had already been set in motion before you started tracking it. Either way, there is no upward component of acceleration during the time that is being shown.
– amd
Aug 25 at 18:08
understood. Thanks
– Arnuld
Aug 26 at 15:13
 |Â
show 1 more comment
up vote
0
down vote
accepted
You’re using the word acceleration to mean two different, though related, things.
In colloquial usage (what you all “real lifeâ€Â), to accelerate means to increase speed, and to decelerate means to decrease speed. However, you’re working with velocities, which are directed—here signed—quantities. In this context, the word “acceleration†refers to any change in velocity, regardless of whether it increases or decreases the object’s speed, which is the velocity’s absolute value. This acceleration is also a signed quantity, but to determine whether it corresponds to an increase or decrease in speed, you also have to consider the sign of the velocity: when the signs are the same, the speed increases—an “acceleration†in the first sense above; when the signs differ, the speed decreases—a “deceleration.â€Â
answered Aug 24 at 21:34
amd
26.6k21046
I think I I understoond most of it. One last thing, in our pumpkin example., acceleration and velocity are both increasing and both have same sign (negative), then why is that only velocity graph is showing increase in velocity while acceleration graph shows constant ?
– Arnuld
Aug 25 at 4:10
Not at all. Until the pumpkin reaches its maximum height, the velocity and acceleration have different signs—the pumpkin slows down. Once the pumpkin starts dropping, they have the same sign—the pumpkin speeds up. The acceleration (due to gravity) is constant throughout.
– amd
Aug 25 at 4:28
Oh no, they are talking about acceleration due to gravity and I was confusing it with acceleration of the "throw". I was thinking this way: the force with which pumpkin is thrown up makes velocity up and acceleration up. That acceleration is not taken into account I see.
– Arnuld
Aug 25 at 9:08
@Arnuld The initial impetus is instantaneous in this simplified model. If you like, assume that the pumpkin had already been set in motion before you started tracking it. Either way, there is no upward component of acceleration during the time that is being shown.
– amd
Aug 25 at 18:08
understood. Thanks
– Arnuld
Aug 26 at 15:13
 |Â
show 1 more comment
up vote
0
down vote
accepted
up vote
0
down vote
accepted
You’re using the word acceleration to mean two different, though related, things.
In colloquial usage (what you all “real lifeâ€Â), to accelerate means to increase speed, and to decelerate means to decrease speed. However, you’re working with velocities, which are directed—here signed—quantities. In this context, the word “acceleration†refers to any change in velocity, regardless of whether it increases or decreases the object’s speed, which is the velocity’s absolute value. This acceleration is also a signed quantity, but to determine whether it corresponds to an increase or decrease in speed, you also have to consider the sign of the velocity: when the signs are the same, the speed increases—an “acceleration†in the first sense above; when the signs differ, the speed decreases—a “deceleration.â€Â
answered Aug 24 at 21:34
amd
26.6k21046
You’re using the word acceleration to mean two different, though related, things.
In colloquial usage (what you all “real lifeâ€Â), to accelerate means to increase speed, and to decelerate means to decrease speed. However, you’re working with velocities, which are directed—here signed—quantities. In this context, the word “acceleration†refers to any change in velocity, regardless of whether it increases or decreases the object’s speed, which is the velocity’s absolute value. This acceleration is also a signed quantity, but to determine whether it corresponds to an increase or decrease in speed, you also have to consider the sign of the velocity: when the signs are the same, the speed increases—an “acceleration†in the first sense above; when the signs differ, the speed decreases—a “deceleration.â€Â
answered Aug 24 at 21:34
amd
26.6k21046
edited Aug 26 at 15:56
answered Aug 24 at 21:34
amd
26.6k21046
answered Aug 24 at 21:34
amd
26.6k21046
answered Aug 24 at 21:34
amd
26.6k21046
26.6k21046
I think I I understoond most of it. One last thing, in our pumpkin example., acceleration and velocity are both increasing and both have same sign (negative), then why is that only velocity graph is showing increase in velocity while acceleration graph shows constant ?
– Arnuld
Aug 25 at 4:10
Not at all. Until the pumpkin reaches its maximum height, the velocity and acceleration have different signs—the pumpkin slows down. Once the pumpkin starts dropping, they have the same sign—the pumpkin speeds up. The acceleration (due to gravity) is constant throughout.
– amd
Aug 25 at 4:28
Oh no, they are talking about acceleration due to gravity and I was confusing it with acceleration of the "throw". I was thinking this way: the force with which pumpkin is thrown up makes velocity up and acceleration up. That acceleration is not taken into account I see.
– Arnuld
Aug 25 at 9:08
@Arnuld The initial impetus is instantaneous in this simplified model. If you like, assume that the pumpkin had already been set in motion before you started tracking it. Either way, there is no upward component of acceleration during the time that is being shown.
– amd
Aug 25 at 18:08
understood. Thanks
– Arnuld
Aug 26 at 15:13
 |Â
show 1 more comment
I think I I understoond most of it. One last thing, in our pumpkin example., acceleration and velocity are both increasing and both have same sign (negative), then why is that only velocity graph is showing increase in velocity while acceleration graph shows constant ?
– Arnuld
Aug 25 at 4:10
Not at all. Until the pumpkin reaches its maximum height, the velocity and acceleration have different signs—the pumpkin slows down. Once the pumpkin starts dropping, they have the same sign—the pumpkin speeds up. The acceleration (due to gravity) is constant throughout.
– amd
Aug 25 at 4:28
Oh no, they are talking about acceleration due to gravity and I was confusing it with acceleration of the "throw". I was thinking this way: the force with which pumpkin is thrown up makes velocity up and acceleration up. That acceleration is not taken into account I see.
– Arnuld
Aug 25 at 9:08
@Arnuld The initial impetus is instantaneous in this simplified model. If you like, assume that the pumpkin had already been set in motion before you started tracking it. Either way, there is no upward component of acceleration during the time that is being shown.
– amd
Aug 25 at 18:08
understood. Thanks
– Arnuld
Aug 26 at 15:13
I think I I understoond most of it. One last thing, in our pumpkin example., acceleration and velocity are both increasing and both have same sign (negative), then why is that only velocity graph is showing increase in velocity while acceleration graph shows constant ?
– Arnuld
Aug 25 at 4:10
I think I I understoond most of it. One last thing, in our pumpkin example., acceleration and velocity are both increasing and both have same sign (negative), then why is that only velocity graph is showing increase in velocity while acceleration graph shows constant ?
– Arnuld
Aug 25 at 4:10
Not at all. Until the pumpkin reaches its maximum height, the velocity and acceleration have different signs—the pumpkin slows down. Once the pumpkin starts dropping, they have the same sign—the pumpkin speeds up. The acceleration (due to gravity) is constant throughout.
– amd
Aug 25 at 4:28
Not at all. Until the pumpkin reaches its maximum height, the velocity and acceleration have different signs—the pumpkin slows down. Once the pumpkin starts dropping, they have the same sign—the pumpkin speeds up. The acceleration (due to gravity) is constant throughout.
– amd
Aug 25 at 4:28
Oh no, they are talking about acceleration due to gravity and I was confusing it with acceleration of the "throw". I was thinking this way: the force with which pumpkin is thrown up makes velocity up and acceleration up. That acceleration is not taken into account I see.
– Arnuld
Aug 25 at 9:08
Oh no, they are talking about acceleration due to gravity and I was confusing it with acceleration of the "throw". I was thinking this way: the force with which pumpkin is thrown up makes velocity up and acceleration up. That acceleration is not taken into account I see.
– Arnuld
Aug 25 at 9:08
@Arnuld The initial impetus is instantaneous in this simplified model. If you like, assume that the pumpkin had already been set in motion before you started tracking it. Either way, there is no upward component of acceleration during the time that is being shown.
– amd
Aug 25 at 18:08
@Arnuld The initial impetus is instantaneous in this simplified model. If you like, assume that the pumpkin had already been set in motion before you started tracking it. Either way, there is no upward component of acceleration during the time that is being shown.
– amd
Aug 25 at 18:08
understood. Thanks
– Arnuld
Aug 26 at 15:13
understood. Thanks
– Arnuld
Aug 26 at 15:13
 |Â
show 1 more comment
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They are vectors and we usually choose the up direction to be the positive one. That's all.
– tst
Aug 24 at 10:44
Notice that if the number -2 decreases, its absolute value increases.
– tst
Aug 24 at 10:45
@tst Oh.. I need to learn Vectors then. 2nd, how |-2| is related here ?
– Arnuld
Aug 24 at 10:49
De-acceleration is an opinion and it is mostly used when the absolute value of the velocity decreases. You shouldn't think in this terms, there is only acceleration that measures the rate of change of the velocity.
– tst
Aug 24 at 10:52
You mean, while doing Math, I should not think of real-life experience of (de-)acceleration but more of as just a 2nd derivative like any other equation ?
– Arnuld
Aug 24 at 10:59