About velocity and acceleration as derivatives [duplicate]
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This question already has an answer here:
How to prove the derivative of position is velocity and of velocity is acceleration?
3 answers
Why is velocity the derivative of position with respect to time? Why is acceleration the derivative of velocity with respect to position? Is there any way to understand this logically, intuitively? Is there a way to see this graphically? How can I understand the "Why? "?
calculus derivatives functional-calculus
asked Dec 7 '18 at 20:34
Bogdan VladBogdan Vlad
495
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marked as duplicate by David K, amWhy calculus
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Dec 7 '18 at 23:45
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
How to prove the derivative of position is velocity and of velocity is acceleration?
3 answers
Why is velocity the derivative of position with respect to time? Why is acceleration the derivative of velocity with respect to position? Is there any way to understand this logically, intuitively? Is there a way to see this graphically? How can I understand the "Why? "?
calculus derivatives functional-calculus
asked Dec 7 '18 at 20:34
Bogdan VladBogdan Vlad
495
$endgroup$
marked as duplicate by David K, amWhy calculus
Users with the calculus badge can single-handedly close calculus questions as duplicates and reopen them as needed.
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Dec 7 '18 at 23:45
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
3
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Do you know what velocity is? Do you know what acceleration is? Do you understand that these are defined as rates of change?
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– MPW
Dec 7 '18 at 20:40
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The "duplicate" might appear to be a different question, since it asks for "proof" rather than understanding, but "proof" turns out to be a nonsensical thing to ask for and the answers to the other question all necessarily address understanding instead. It ends up being the same question in effect and we might as well post any new answers there, not here.
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– David K
Dec 7 '18 at 20:56
1
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In order to answer this question, you need to specify what is your definition of velocity and acceleration. Not necessarily a formal definition, but more like "the velocity is like a box of chocolates", paraphrasing Forrest Gump. Also note that the acceleration is the derivative of velocity with respect to time, not position.
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– Andrei
Dec 7 '18 at 21:08
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@Andrei From what I understand, velocity is basically speed, but with direction, while acceleration is change of said velocity at one moment.
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– Bogdan Vlad
Dec 7 '18 at 21:22
add a comment |
$begingroup$
This question already has an answer here:
How to prove the derivative of position is velocity and of velocity is acceleration?
3 answers
Why is velocity the derivative of position with respect to time? Why is acceleration the derivative of velocity with respect to position? Is there any way to understand this logically, intuitively? Is there a way to see this graphically? How can I understand the "Why? "?
calculus derivatives functional-calculus
asked Dec 7 '18 at 20:34
Bogdan VladBogdan Vlad
495
$endgroup$
This question already has an answer here:
How to prove the derivative of position is velocity and of velocity is acceleration?
3 answers
Why is velocity the derivative of position with respect to time? Why is acceleration the derivative of velocity with respect to position? Is there any way to understand this logically, intuitively? Is there a way to see this graphically? How can I understand the "Why? "?
This question already has an answer here:
How to prove the derivative of position is velocity and of velocity is acceleration?
3 answers
calculus derivatives functional-calculus
calculus derivatives functional-calculus
asked Dec 7 '18 at 20:34
Bogdan VladBogdan Vlad
495
asked Dec 7 '18 at 20:34
Bogdan VladBogdan Vlad
495
asked Dec 7 '18 at 20:34
Bogdan VladBogdan Vlad
495
asked Dec 7 '18 at 20:34
Bogdan VladBogdan Vlad
495
asked Dec 7 '18 at 20:34
Bogdan VladBogdan Vlad
495
495
marked as duplicate by David K, amWhy calculus
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Dec 7 '18 at 23:45
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by David K, amWhy calculus
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Dec 7 '18 at 23:45
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
3
$begingroup$
Do you know what velocity is? Do you know what acceleration is? Do you understand that these are defined as rates of change?
$endgroup$
– MPW
Dec 7 '18 at 20:40
$begingroup$
The "duplicate" might appear to be a different question, since it asks for "proof" rather than understanding, but "proof" turns out to be a nonsensical thing to ask for and the answers to the other question all necessarily address understanding instead. It ends up being the same question in effect and we might as well post any new answers there, not here.
$endgroup$
– David K
Dec 7 '18 at 20:56
1
$begingroup$
In order to answer this question, you need to specify what is your definition of velocity and acceleration. Not necessarily a formal definition, but more like "the velocity is like a box of chocolates", paraphrasing Forrest Gump. Also note that the acceleration is the derivative of velocity with respect to time, not position.
$endgroup$
– Andrei
Dec 7 '18 at 21:08
$begingroup$
@Andrei From what I understand, velocity is basically speed, but with direction, while acceleration is change of said velocity at one moment.
$endgroup$
– Bogdan Vlad
Dec 7 '18 at 21:22
add a comment |
3
$begingroup$
Do you know what velocity is? Do you know what acceleration is? Do you understand that these are defined as rates of change?
$endgroup$
– MPW
Dec 7 '18 at 20:40
$begingroup$
The "duplicate" might appear to be a different question, since it asks for "proof" rather than understanding, but "proof" turns out to be a nonsensical thing to ask for and the answers to the other question all necessarily address understanding instead. It ends up being the same question in effect and we might as well post any new answers there, not here.
$endgroup$
– David K
Dec 7 '18 at 20:56
1
$begingroup$
In order to answer this question, you need to specify what is your definition of velocity and acceleration. Not necessarily a formal definition, but more like "the velocity is like a box of chocolates", paraphrasing Forrest Gump. Also note that the acceleration is the derivative of velocity with respect to time, not position.
$endgroup$
– Andrei
Dec 7 '18 at 21:08
$begingroup$
@Andrei From what I understand, velocity is basically speed, but with direction, while acceleration is change of said velocity at one moment.
$endgroup$
– Bogdan Vlad
Dec 7 '18 at 21:22
3
3
$begingroup$
Do you know what velocity is? Do you know what acceleration is? Do you understand that these are defined as rates of change?
$endgroup$
– MPW
Dec 7 '18 at 20:40
$begingroup$
Do you know what velocity is? Do you know what acceleration is? Do you understand that these are defined as rates of change?
$endgroup$
– MPW
Dec 7 '18 at 20:40
$begingroup$
The "duplicate" might appear to be a different question, since it asks for "proof" rather than understanding, but "proof" turns out to be a nonsensical thing to ask for and the answers to the other question all necessarily address understanding instead. It ends up being the same question in effect and we might as well post any new answers there, not here.
$endgroup$
– David K
Dec 7 '18 at 20:56
$begingroup$
The "duplicate" might appear to be a different question, since it asks for "proof" rather than understanding, but "proof" turns out to be a nonsensical thing to ask for and the answers to the other question all necessarily address understanding instead. It ends up being the same question in effect and we might as well post any new answers there, not here.
$endgroup$
– David K
Dec 7 '18 at 20:56
1
1
$begingroup$
In order to answer this question, you need to specify what is your definition of velocity and acceleration. Not necessarily a formal definition, but more like "the velocity is like a box of chocolates", paraphrasing Forrest Gump. Also note that the acceleration is the derivative of velocity with respect to time, not position.
$endgroup$
– Andrei
Dec 7 '18 at 21:08
$begingroup$
In order to answer this question, you need to specify what is your definition of velocity and acceleration. Not necessarily a formal definition, but more like "the velocity is like a box of chocolates", paraphrasing Forrest Gump. Also note that the acceleration is the derivative of velocity with respect to time, not position.
$endgroup$
– Andrei
Dec 7 '18 at 21:08
$begingroup$
@Andrei From what I understand, velocity is basically speed, but with direction, while acceleration is change of said velocity at one moment.
$endgroup$
– Bogdan Vlad
Dec 7 '18 at 21:22
$begingroup$
@Andrei From what I understand, velocity is basically speed, but with direction, while acceleration is change of said velocity at one moment.
$endgroup$
– Bogdan Vlad
Dec 7 '18 at 21:22
add a comment |
1 Answer
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The average velocity is defined by the change in position in a time interval, divided by that time interval: $$vec v=frac{Delta vec r}{Delta t}$$
This just says that if I move from $A$ to $B$ in time $Delta t$, my average velocity is a vector pointed from $A$ to $B$, with the magnitude equal to the distance between those points divided by the time it needs to get between those two points. Obviously this definition has some limitations. I could move fast at the beginning, then slow down, or I cam move in a weird pattern between those two points. Also, let's say I move in a circle, and $A=B$. Then the change in the position vector is $0$, so no matter how fast you move, the velocity is $0$. That's why you define the instantaneous velocity. $$vec v=lim_{Delta tto 0}frac{Delta vec r}{Delta t}$$ But this is just the definition of the derivative. So why is this better? Well, for one thing the magnitude of it is the speed that you are used to. The other is that is always tangent to the curve. You can do a similar thought process about velocity and acceleration. But at the end of the day, those are just definitions.
answered Dec 7 '18 at 22:58
AndreiAndrei
11.8k21026
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The average velocity is defined by the change in position in a time interval, divided by that time interval: $$vec v=frac{Delta vec r}{Delta t}$$
This just says that if I move from $A$ to $B$ in time $Delta t$, my average velocity is a vector pointed from $A$ to $B$, with the magnitude equal to the distance between those points divided by the time it needs to get between those two points. Obviously this definition has some limitations. I could move fast at the beginning, then slow down, or I cam move in a weird pattern between those two points. Also, let's say I move in a circle, and $A=B$. Then the change in the position vector is $0$, so no matter how fast you move, the velocity is $0$. That's why you define the instantaneous velocity. $$vec v=lim_{Delta tto 0}frac{Delta vec r}{Delta t}$$ But this is just the definition of the derivative. So why is this better? Well, for one thing the magnitude of it is the speed that you are used to. The other is that is always tangent to the curve. You can do a similar thought process about velocity and acceleration. But at the end of the day, those are just definitions.
answered Dec 7 '18 at 22:58
AndreiAndrei
11.8k21026
$endgroup$
add a comment |
$begingroup$
The average velocity is defined by the change in position in a time interval, divided by that time interval: $$vec v=frac{Delta vec r}{Delta t}$$
This just says that if I move from $A$ to $B$ in time $Delta t$, my average velocity is a vector pointed from $A$ to $B$, with the magnitude equal to the distance between those points divided by the time it needs to get between those two points. Obviously this definition has some limitations. I could move fast at the beginning, then slow down, or I cam move in a weird pattern between those two points. Also, let's say I move in a circle, and $A=B$. Then the change in the position vector is $0$, so no matter how fast you move, the velocity is $0$. That's why you define the instantaneous velocity. $$vec v=lim_{Delta tto 0}frac{Delta vec r}{Delta t}$$ But this is just the definition of the derivative. So why is this better? Well, for one thing the magnitude of it is the speed that you are used to. The other is that is always tangent to the curve. You can do a similar thought process about velocity and acceleration. But at the end of the day, those are just definitions.
answered Dec 7 '18 at 22:58
AndreiAndrei
11.8k21026
$endgroup$
add a comment |
$begingroup$
The average velocity is defined by the change in position in a time interval, divided by that time interval: $$vec v=frac{Delta vec r}{Delta t}$$
This just says that if I move from $A$ to $B$ in time $Delta t$, my average velocity is a vector pointed from $A$ to $B$, with the magnitude equal to the distance between those points divided by the time it needs to get between those two points. Obviously this definition has some limitations. I could move fast at the beginning, then slow down, or I cam move in a weird pattern between those two points. Also, let's say I move in a circle, and $A=B$. Then the change in the position vector is $0$, so no matter how fast you move, the velocity is $0$. That's why you define the instantaneous velocity. $$vec v=lim_{Delta tto 0}frac{Delta vec r}{Delta t}$$ But this is just the definition of the derivative. So why is this better? Well, for one thing the magnitude of it is the speed that you are used to. The other is that is always tangent to the curve. You can do a similar thought process about velocity and acceleration. But at the end of the day, those are just definitions.
answered Dec 7 '18 at 22:58
AndreiAndrei
11.8k21026
$endgroup$
The average velocity is defined by the change in position in a time interval, divided by that time interval: $$vec v=frac{Delta vec r}{Delta t}$$
This just says that if I move from $A$ to $B$ in time $Delta t$, my average velocity is a vector pointed from $A$ to $B$, with the magnitude equal to the distance between those points divided by the time it needs to get between those two points. Obviously this definition has some limitations. I could move fast at the beginning, then slow down, or I cam move in a weird pattern between those two points. Also, let's say I move in a circle, and $A=B$. Then the change in the position vector is $0$, so no matter how fast you move, the velocity is $0$. That's why you define the instantaneous velocity. $$vec v=lim_{Delta tto 0}frac{Delta vec r}{Delta t}$$ But this is just the definition of the derivative. So why is this better? Well, for one thing the magnitude of it is the speed that you are used to. The other is that is always tangent to the curve. You can do a similar thought process about velocity and acceleration. But at the end of the day, those are just definitions.
answered Dec 7 '18 at 22:58
AndreiAndrei
11.8k21026
answered Dec 7 '18 at 22:58
AndreiAndrei
11.8k21026
answered Dec 7 '18 at 22:58
AndreiAndrei
11.8k21026
answered Dec 7 '18 at 22:58
AndreiAndrei
11.8k21026
11.8k21026
add a comment |
add a comment |
3
$begingroup$
Do you know what velocity is? Do you know what acceleration is? Do you understand that these are defined as rates of change?
$endgroup$
– MPW
Dec 7 '18 at 20:40
$begingroup$
The "duplicate" might appear to be a different question, since it asks for "proof" rather than understanding, but "proof" turns out to be a nonsensical thing to ask for and the answers to the other question all necessarily address understanding instead. It ends up being the same question in effect and we might as well post any new answers there, not here.
$endgroup$
– David K
Dec 7 '18 at 20:56
1
$begingroup$
In order to answer this question, you need to specify what is your definition of velocity and acceleration. Not necessarily a formal definition, but more like "the velocity is like a box of chocolates", paraphrasing Forrest Gump. Also note that the acceleration is the derivative of velocity with respect to time, not position.
$endgroup$
– Andrei
Dec 7 '18 at 21:08
$begingroup$
@Andrei From what I understand, velocity is basically speed, but with direction, while acceleration is change of said velocity at one moment.
$endgroup$
– Bogdan Vlad
Dec 7 '18 at 21:22