Why isn't acceleration always zero whenever velocity is zero, such as the moment a ball bounces off a wall? [duplicate] The 2019 Stack Overflow Developer Survey Results Are InHow can an object's instantaneous speed be zero and it's instantaneous acceleration be nonzero?How can I add an acceleration vector to a velocity with a different direction?Simple Acceleration Problem throwing me offNon-Constant Acceleration due to GravityWhy is acceleration constant in this example?Direction of acceleration at highest pointRain drop sliding on car windshieldAcceleration of a ball thrown into the airWhen finding acceleration in a pulley system, why does $ a = (m_1g_1 - m_2g_2)/(m_1 + m_2)$ hold when none of the accelerations is $g$Atwood's machines: acceleration of massless pulleyCoefficient of restitution for bouncing ball
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Why isn't acceleration always zero whenever velocity is zero, such as the moment a ball bounces off a wall? [duplicate]
The 2019 Stack Overflow Developer Survey Results Are InHow can an object's instantaneous speed be zero and it's instantaneous acceleration be nonzero?How can I add an acceleration vector to a velocity with a different direction?Simple Acceleration Problem throwing me offNon-Constant Acceleration due to GravityWhy is acceleration constant in this example?Direction of acceleration at highest pointRain drop sliding on car windshieldAcceleration of a ball thrown into the airWhen finding acceleration in a pulley system, why does $ a = (m_1g_1 - m_2g_2)/(m_1 + m_2)$ hold when none of the accelerations is $g$Atwood's machines: acceleration of massless pulleyCoefficient of restitution for bouncing ball
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This question already has an answer here:
How can an object's instantaneous speed be zero and it's instantaneous acceleration be nonzero?
3 answers
The answer to this homework problem is $left(textDright) :$
I understand why we rule out options $left(textAright)$ and $left(textBright)$ first. However, I don't get why $left(textDright)$ is the answer.
Won't the ball's velocity at one point be $0$ when it comes in contact with the ceiling? (The same way as its velocity momentarily is $0$ when it is thrown downwards and comes in contact with the ground before bouncing back upwards). That should give zero acceleration at that point, but none of the acceleration profiles include $0 .$
homework-and-exercises newtonian-mechanics acceleration velocity collision
edited Mar 30 at 16:47
knzhou
47k11127226
asked Mar 30 at 4:47
Clark KentClark Kent
271
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marked as duplicate by Aaron Stevens, GiorgioP, Chair, Dvij Mankad, John Rennie newtonian-mechanics
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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.
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This question already has an answer here:
How can an object's instantaneous speed be zero and it's instantaneous acceleration be nonzero?
3 answers
The answer to this homework problem is $left(textDright) :$
I understand why we rule out options $left(textAright)$ and $left(textBright)$ first. However, I don't get why $left(textDright)$ is the answer.
Won't the ball's velocity at one point be $0$ when it comes in contact with the ceiling? (The same way as its velocity momentarily is $0$ when it is thrown downwards and comes in contact with the ground before bouncing back upwards). That should give zero acceleration at that point, but none of the acceleration profiles include $0 .$
homework-and-exercises newtonian-mechanics acceleration velocity collision
edited Mar 30 at 16:47
knzhou
47k11127226
asked Mar 30 at 4:47
Clark KentClark Kent
271
$endgroup$
marked as duplicate by Aaron Stevens, GiorgioP, Chair, Dvij Mankad, John Rennie newtonian-mechanics
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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.
7
$begingroup$
Yes, the velocity will be zero momentarily. Why would that mean the acceleration is zero?
$endgroup$
– d_b
Mar 30 at 4:52
$begingroup$
a=v-u/t? From this, acceleration should be zero when velocity is zero.
$endgroup$
– Clark Kent
Mar 30 at 5:37
5
$begingroup$
Acceleration is the rate of change of velocity, $a = Delta v/Delta t$. The rate of change of $v$ can be non-zero even when $v=0$. Think about the simpler case of a ball thrown in the air. At the top of its motion, momentarily $v=0$, but there is still acceleration due to gravity, which is why the ball starts moving back down.
$endgroup$
– d_b
Mar 30 at 5:47
6
$begingroup$
@ClarkKent By your argument the ball could never be thrown: It starts in-hand with zero velocity and thus cannot be accelerated.
$endgroup$
– HABO
Mar 30 at 15:06
1
$begingroup$
It may help to think about velocity vs. acceleration by analogy to position vs. velocity; these have very similar relationships to each other, and somehow human brains are wired to understand the latter two much more intuitively. So: a car is driving uptown, hitting green lights at every intersection and passing side streets at 20mph. In this city, they've numbered the side streets; it's in the south but heading north. Eventually it will pass 1st Street South, then 0th Street, then 1st Street North. At the moment that its position reaches 0th street, does its velocity drop to 0mph, too?
$endgroup$
– Daniel Wagner
Mar 30 at 18:17
|
show 5 more comments
$begingroup$
This question already has an answer here:
How can an object's instantaneous speed be zero and it's instantaneous acceleration be nonzero?
3 answers
The answer to this homework problem is $left(textDright) :$
I understand why we rule out options $left(textAright)$ and $left(textBright)$ first. However, I don't get why $left(textDright)$ is the answer.
Won't the ball's velocity at one point be $0$ when it comes in contact with the ceiling? (The same way as its velocity momentarily is $0$ when it is thrown downwards and comes in contact with the ground before bouncing back upwards). That should give zero acceleration at that point, but none of the acceleration profiles include $0 .$
homework-and-exercises newtonian-mechanics acceleration velocity collision
edited Mar 30 at 16:47
knzhou
47k11127226
asked Mar 30 at 4:47
Clark KentClark Kent
271
$endgroup$
This question already has an answer here:
How can an object's instantaneous speed be zero and it's instantaneous acceleration be nonzero?
3 answers
The answer to this homework problem is $left(textDright) :$
I understand why we rule out options $left(textAright)$ and $left(textBright)$ first. However, I don't get why $left(textDright)$ is the answer.
Won't the ball's velocity at one point be $0$ when it comes in contact with the ceiling? (The same way as its velocity momentarily is $0$ when it is thrown downwards and comes in contact with the ground before bouncing back upwards). That should give zero acceleration at that point, but none of the acceleration profiles include $0 .$
This question already has an answer here:
How can an object's instantaneous speed be zero and it's instantaneous acceleration be nonzero?
3 answers
homework-and-exercises newtonian-mechanics acceleration velocity collision
homework-and-exercises newtonian-mechanics acceleration velocity collision
edited Mar 30 at 16:47
knzhou
47k11127226
asked Mar 30 at 4:47
Clark KentClark Kent
271
edited Mar 30 at 16:47
knzhou
47k11127226
asked Mar 30 at 4:47
Clark KentClark Kent
271
edited Mar 30 at 16:47
knzhou
47k11127226
edited Mar 30 at 16:47
knzhou
47k11127226
edited Mar 30 at 16:47
knzhou
47k11127226
47k11127226
asked Mar 30 at 4:47
Clark KentClark Kent
271
asked Mar 30 at 4:47
Clark KentClark Kent
271
asked Mar 30 at 4:47
Clark KentClark Kent
271
271
marked as duplicate by Aaron Stevens, GiorgioP, Chair, Dvij Mankad, John Rennie newtonian-mechanics
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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.
7
$begingroup$
Yes, the velocity will be zero momentarily. Why would that mean the acceleration is zero?
$endgroup$
– d_b
Mar 30 at 4:52
$begingroup$
a=v-u/t? From this, acceleration should be zero when velocity is zero.
$endgroup$
– Clark Kent
Mar 30 at 5:37
5
$begingroup$
Acceleration is the rate of change of velocity, $a = Delta v/Delta t$. The rate of change of $v$ can be non-zero even when $v=0$. Think about the simpler case of a ball thrown in the air. At the top of its motion, momentarily $v=0$, but there is still acceleration due to gravity, which is why the ball starts moving back down.
$endgroup$
– d_b
Mar 30 at 5:47
6
$begingroup$
@ClarkKent By your argument the ball could never be thrown: It starts in-hand with zero velocity and thus cannot be accelerated.
$endgroup$
– HABO
Mar 30 at 15:06
1
$begingroup$
It may help to think about velocity vs. acceleration by analogy to position vs. velocity; these have very similar relationships to each other, and somehow human brains are wired to understand the latter two much more intuitively. So: a car is driving uptown, hitting green lights at every intersection and passing side streets at 20mph. In this city, they've numbered the side streets; it's in the south but heading north. Eventually it will pass 1st Street South, then 0th Street, then 1st Street North. At the moment that its position reaches 0th street, does its velocity drop to 0mph, too?
$endgroup$
– Daniel Wagner
Mar 30 at 18:17
|
show 5 more comments
7
$begingroup$
Yes, the velocity will be zero momentarily. Why would that mean the acceleration is zero?
$endgroup$
– d_b
Mar 30 at 4:52
$begingroup$
a=v-u/t? From this, acceleration should be zero when velocity is zero.
$endgroup$
– Clark Kent
Mar 30 at 5:37
5
$begingroup$
Acceleration is the rate of change of velocity, $a = Delta v/Delta t$. The rate of change of $v$ can be non-zero even when $v=0$. Think about the simpler case of a ball thrown in the air. At the top of its motion, momentarily $v=0$, but there is still acceleration due to gravity, which is why the ball starts moving back down.
$endgroup$
– d_b
Mar 30 at 5:47
6
$begingroup$
@ClarkKent By your argument the ball could never be thrown: It starts in-hand with zero velocity and thus cannot be accelerated.
$endgroup$
– HABO
Mar 30 at 15:06
1
$begingroup$
It may help to think about velocity vs. acceleration by analogy to position vs. velocity; these have very similar relationships to each other, and somehow human brains are wired to understand the latter two much more intuitively. So: a car is driving uptown, hitting green lights at every intersection and passing side streets at 20mph. In this city, they've numbered the side streets; it's in the south but heading north. Eventually it will pass 1st Street South, then 0th Street, then 1st Street North. At the moment that its position reaches 0th street, does its velocity drop to 0mph, too?
$endgroup$
– Daniel Wagner
Mar 30 at 18:17
7
7
$begingroup$
Yes, the velocity will be zero momentarily. Why would that mean the acceleration is zero?
$endgroup$
– d_b
Mar 30 at 4:52
$begingroup$
Yes, the velocity will be zero momentarily. Why would that mean the acceleration is zero?
$endgroup$
– d_b
Mar 30 at 4:52
$begingroup$
a=v-u/t? From this, acceleration should be zero when velocity is zero.
$endgroup$
– Clark Kent
Mar 30 at 5:37
$begingroup$
a=v-u/t? From this, acceleration should be zero when velocity is zero.
$endgroup$
– Clark Kent
Mar 30 at 5:37
5
5
$begingroup$
Acceleration is the rate of change of velocity, $a = Delta v/Delta t$. The rate of change of $v$ can be non-zero even when $v=0$. Think about the simpler case of a ball thrown in the air. At the top of its motion, momentarily $v=0$, but there is still acceleration due to gravity, which is why the ball starts moving back down.
$endgroup$
– d_b
Mar 30 at 5:47
$begingroup$
Acceleration is the rate of change of velocity, $a = Delta v/Delta t$. The rate of change of $v$ can be non-zero even when $v=0$. Think about the simpler case of a ball thrown in the air. At the top of its motion, momentarily $v=0$, but there is still acceleration due to gravity, which is why the ball starts moving back down.
$endgroup$
– d_b
Mar 30 at 5:47
6
6
$begingroup$
@ClarkKent By your argument the ball could never be thrown: It starts in-hand with zero velocity and thus cannot be accelerated.
$endgroup$
– HABO
Mar 30 at 15:06
$begingroup$
@ClarkKent By your argument the ball could never be thrown: It starts in-hand with zero velocity and thus cannot be accelerated.
$endgroup$
– HABO
Mar 30 at 15:06
1
1
$begingroup$
It may help to think about velocity vs. acceleration by analogy to position vs. velocity; these have very similar relationships to each other, and somehow human brains are wired to understand the latter two much more intuitively. So: a car is driving uptown, hitting green lights at every intersection and passing side streets at 20mph. In this city, they've numbered the side streets; it's in the south but heading north. Eventually it will pass 1st Street South, then 0th Street, then 1st Street North. At the moment that its position reaches 0th street, does its velocity drop to 0mph, too?
$endgroup$
– Daniel Wagner
Mar 30 at 18:17
$begingroup$
It may help to think about velocity vs. acceleration by analogy to position vs. velocity; these have very similar relationships to each other, and somehow human brains are wired to understand the latter two much more intuitively. So: a car is driving uptown, hitting green lights at every intersection and passing side streets at 20mph. In this city, they've numbered the side streets; it's in the south but heading north. Eventually it will pass 1st Street South, then 0th Street, then 1st Street North. At the moment that its position reaches 0th street, does its velocity drop to 0mph, too?
$endgroup$
– Daniel Wagner
Mar 30 at 18:17
|
show 5 more comments
3 Answers
3
active
oldest
votes
$begingroup$
You could consider the question in terms of the forces acting on the ball.
During the entire time a downward force of gravity acts on it. While it is in contact with the ceiling there is also a downwards normal force. At no time are there balanced forces acting on the ball
This results in an acceleration is always downwards with a spike in magnitude while the ball is in contact with the ceiling.
answered Mar 30 at 12:40
M. EnnsM. Enns
4,54921328
$endgroup$
add a comment |
$begingroup$
Won't the ball's velocity at one point be 0 when it comes in contact with the ceiling? (The same way as its velocity momentarily is 0 when it is thrown downwards and comes in contact with the ground before bouncing back upwards).
Yes. Its velocity will momentarily be zero.
That should give zero acceleration at that point[...]
No! Velocity being zero says nothing about the acceleration. Just like your position being zero (i.e. when returning home after a day at work) says nothing about your velocity (you could be running in high speed when arriving at your starting position, or you could reach it slowly, or you could stand still at the starting spot. Different velocities are possible where the position is zero; position and velocity are unrelated, and so are velocity and acceleration).
When throwing a ball up, the ball will also momentarily reach zero speed before falling back down to your hand. But gravity is there all the time causing a non-zero acceleration all the time - also when the speed is zero. So, acceleration can't be understood from the value of velocity, only from the change in the value of velocity.
And why the peak then in answer D? Because,
- while flying upwards, gravity causes a constant downwards acceleration. The ball slows down at a constant rate.
- When hitting the ceiling, the ceiling suddenly slows down the ball instantly. That requires a much larger downwards acceleration in that instant in order to reduce the speed to zero in very short time. Thus the peak on the graph.
answered Mar 30 at 14:11
SteevenSteeven
27.9k866113
$endgroup$
add a comment |
$begingroup$
It is often supposed that because a quantity (velocity in this case) is zero the rate of change of that quantity (acceleration in this case) is also zero.
If that were true a ball which is thrown upwards would stay at its greatest height above the ground when it was not moving and never come back to the ground.
Just suppose that you were correct and at the instant the velocity of the ball was zero the acceleration $( = fractextchange in velocitytexttime)$ was also zero.
This means that at some instant if time the velocity of the ball does not change from being zero ie the ball is not moving and will stay in a position of rest on the ceiling for all time.
answered Mar 30 at 5:47
FarcherFarcher
52.1k340110
$endgroup$
1
$begingroup$
Okay, I now understand why acceleration isnt zero. Can you explain why D is the answer?
$endgroup$
– Clark Kent
Mar 30 at 6:26
1
$begingroup$
@ClarkKent Up is positive and the acceleration of free fall is downwards ie negative. When the ball hits the ceiling a downward force due to the ceiling acts on the ball increasing the ball’s downward acceleration.
$endgroup$
– Farcher
Mar 30 at 6:31
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You could consider the question in terms of the forces acting on the ball.
During the entire time a downward force of gravity acts on it. While it is in contact with the ceiling there is also a downwards normal force. At no time are there balanced forces acting on the ball
This results in an acceleration is always downwards with a spike in magnitude while the ball is in contact with the ceiling.
answered Mar 30 at 12:40
M. EnnsM. Enns
4,54921328
$endgroup$
add a comment |
$begingroup$
You could consider the question in terms of the forces acting on the ball.
During the entire time a downward force of gravity acts on it. While it is in contact with the ceiling there is also a downwards normal force. At no time are there balanced forces acting on the ball
This results in an acceleration is always downwards with a spike in magnitude while the ball is in contact with the ceiling.
answered Mar 30 at 12:40
M. EnnsM. Enns
4,54921328
$endgroup$
add a comment |
$begingroup$
You could consider the question in terms of the forces acting on the ball.
During the entire time a downward force of gravity acts on it. While it is in contact with the ceiling there is also a downwards normal force. At no time are there balanced forces acting on the ball
This results in an acceleration is always downwards with a spike in magnitude while the ball is in contact with the ceiling.
answered Mar 30 at 12:40
M. EnnsM. Enns
4,54921328
$endgroup$
You could consider the question in terms of the forces acting on the ball.
During the entire time a downward force of gravity acts on it. While it is in contact with the ceiling there is also a downwards normal force. At no time are there balanced forces acting on the ball
This results in an acceleration is always downwards with a spike in magnitude while the ball is in contact with the ceiling.
answered Mar 30 at 12:40
M. EnnsM. Enns
4,54921328
answered Mar 30 at 12:40
M. EnnsM. Enns
4,54921328
answered Mar 30 at 12:40
M. EnnsM. Enns
4,54921328
answered Mar 30 at 12:40
M. EnnsM. Enns
4,54921328
4,54921328
add a comment |
add a comment |
$begingroup$
Won't the ball's velocity at one point be 0 when it comes in contact with the ceiling? (The same way as its velocity momentarily is 0 when it is thrown downwards and comes in contact with the ground before bouncing back upwards).
Yes. Its velocity will momentarily be zero.
That should give zero acceleration at that point[...]
No! Velocity being zero says nothing about the acceleration. Just like your position being zero (i.e. when returning home after a day at work) says nothing about your velocity (you could be running in high speed when arriving at your starting position, or you could reach it slowly, or you could stand still at the starting spot. Different velocities are possible where the position is zero; position and velocity are unrelated, and so are velocity and acceleration).
When throwing a ball up, the ball will also momentarily reach zero speed before falling back down to your hand. But gravity is there all the time causing a non-zero acceleration all the time - also when the speed is zero. So, acceleration can't be understood from the value of velocity, only from the change in the value of velocity.
And why the peak then in answer D? Because,
- while flying upwards, gravity causes a constant downwards acceleration. The ball slows down at a constant rate.
- When hitting the ceiling, the ceiling suddenly slows down the ball instantly. That requires a much larger downwards acceleration in that instant in order to reduce the speed to zero in very short time. Thus the peak on the graph.
answered Mar 30 at 14:11
SteevenSteeven
27.9k866113
$endgroup$
add a comment |
$begingroup$
Won't the ball's velocity at one point be 0 when it comes in contact with the ceiling? (The same way as its velocity momentarily is 0 when it is thrown downwards and comes in contact with the ground before bouncing back upwards).
Yes. Its velocity will momentarily be zero.
That should give zero acceleration at that point[...]
No! Velocity being zero says nothing about the acceleration. Just like your position being zero (i.e. when returning home after a day at work) says nothing about your velocity (you could be running in high speed when arriving at your starting position, or you could reach it slowly, or you could stand still at the starting spot. Different velocities are possible where the position is zero; position and velocity are unrelated, and so are velocity and acceleration).
When throwing a ball up, the ball will also momentarily reach zero speed before falling back down to your hand. But gravity is there all the time causing a non-zero acceleration all the time - also when the speed is zero. So, acceleration can't be understood from the value of velocity, only from the change in the value of velocity.
And why the peak then in answer D? Because,
- while flying upwards, gravity causes a constant downwards acceleration. The ball slows down at a constant rate.
- When hitting the ceiling, the ceiling suddenly slows down the ball instantly. That requires a much larger downwards acceleration in that instant in order to reduce the speed to zero in very short time. Thus the peak on the graph.
answered Mar 30 at 14:11
SteevenSteeven
27.9k866113
$endgroup$
add a comment |
$begingroup$
Won't the ball's velocity at one point be 0 when it comes in contact with the ceiling? (The same way as its velocity momentarily is 0 when it is thrown downwards and comes in contact with the ground before bouncing back upwards).
Yes. Its velocity will momentarily be zero.
That should give zero acceleration at that point[...]
No! Velocity being zero says nothing about the acceleration. Just like your position being zero (i.e. when returning home after a day at work) says nothing about your velocity (you could be running in high speed when arriving at your starting position, or you could reach it slowly, or you could stand still at the starting spot. Different velocities are possible where the position is zero; position and velocity are unrelated, and so are velocity and acceleration).
When throwing a ball up, the ball will also momentarily reach zero speed before falling back down to your hand. But gravity is there all the time causing a non-zero acceleration all the time - also when the speed is zero. So, acceleration can't be understood from the value of velocity, only from the change in the value of velocity.
And why the peak then in answer D? Because,
- while flying upwards, gravity causes a constant downwards acceleration. The ball slows down at a constant rate.
- When hitting the ceiling, the ceiling suddenly slows down the ball instantly. That requires a much larger downwards acceleration in that instant in order to reduce the speed to zero in very short time. Thus the peak on the graph.
answered Mar 30 at 14:11
SteevenSteeven
27.9k866113
$endgroup$
Won't the ball's velocity at one point be 0 when it comes in contact with the ceiling? (The same way as its velocity momentarily is 0 when it is thrown downwards and comes in contact with the ground before bouncing back upwards).
Yes. Its velocity will momentarily be zero.
That should give zero acceleration at that point[...]
No! Velocity being zero says nothing about the acceleration. Just like your position being zero (i.e. when returning home after a day at work) says nothing about your velocity (you could be running in high speed when arriving at your starting position, or you could reach it slowly, or you could stand still at the starting spot. Different velocities are possible where the position is zero; position and velocity are unrelated, and so are velocity and acceleration).
When throwing a ball up, the ball will also momentarily reach zero speed before falling back down to your hand. But gravity is there all the time causing a non-zero acceleration all the time - also when the speed is zero. So, acceleration can't be understood from the value of velocity, only from the change in the value of velocity.
And why the peak then in answer D? Because,
- while flying upwards, gravity causes a constant downwards acceleration. The ball slows down at a constant rate.
- When hitting the ceiling, the ceiling suddenly slows down the ball instantly. That requires a much larger downwards acceleration in that instant in order to reduce the speed to zero in very short time. Thus the peak on the graph.
answered Mar 30 at 14:11
SteevenSteeven
27.9k866113
edited Mar 30 at 21:33
answered Mar 30 at 14:11
SteevenSteeven
27.9k866113
answered Mar 30 at 14:11
SteevenSteeven
27.9k866113
answered Mar 30 at 14:11
SteevenSteeven
27.9k866113
27.9k866113
add a comment |
add a comment |
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It is often supposed that because a quantity (velocity in this case) is zero the rate of change of that quantity (acceleration in this case) is also zero.
If that were true a ball which is thrown upwards would stay at its greatest height above the ground when it was not moving and never come back to the ground.
Just suppose that you were correct and at the instant the velocity of the ball was zero the acceleration $( = fractextchange in velocitytexttime)$ was also zero.
This means that at some instant if time the velocity of the ball does not change from being zero ie the ball is not moving and will stay in a position of rest on the ceiling for all time.
answered Mar 30 at 5:47
FarcherFarcher
52.1k340110
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1
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Okay, I now understand why acceleration isnt zero. Can you explain why D is the answer?
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– Clark Kent
Mar 30 at 6:26
1
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@ClarkKent Up is positive and the acceleration of free fall is downwards ie negative. When the ball hits the ceiling a downward force due to the ceiling acts on the ball increasing the ball’s downward acceleration.
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– Farcher
Mar 30 at 6:31
add a comment |
$begingroup$
It is often supposed that because a quantity (velocity in this case) is zero the rate of change of that quantity (acceleration in this case) is also zero.
If that were true a ball which is thrown upwards would stay at its greatest height above the ground when it was not moving and never come back to the ground.
Just suppose that you were correct and at the instant the velocity of the ball was zero the acceleration $( = fractextchange in velocitytexttime)$ was also zero.
This means that at some instant if time the velocity of the ball does not change from being zero ie the ball is not moving and will stay in a position of rest on the ceiling for all time.
answered Mar 30 at 5:47
FarcherFarcher
52.1k340110
$endgroup$
1
$begingroup$
Okay, I now understand why acceleration isnt zero. Can you explain why D is the answer?
$endgroup$
– Clark Kent
Mar 30 at 6:26
1
$begingroup$
@ClarkKent Up is positive and the acceleration of free fall is downwards ie negative. When the ball hits the ceiling a downward force due to the ceiling acts on the ball increasing the ball’s downward acceleration.
$endgroup$
– Farcher
Mar 30 at 6:31
add a comment |
$begingroup$
It is often supposed that because a quantity (velocity in this case) is zero the rate of change of that quantity (acceleration in this case) is also zero.
If that were true a ball which is thrown upwards would stay at its greatest height above the ground when it was not moving and never come back to the ground.
Just suppose that you were correct and at the instant the velocity of the ball was zero the acceleration $( = fractextchange in velocitytexttime)$ was also zero.
This means that at some instant if time the velocity of the ball does not change from being zero ie the ball is not moving and will stay in a position of rest on the ceiling for all time.
answered Mar 30 at 5:47
FarcherFarcher
52.1k340110
$endgroup$
It is often supposed that because a quantity (velocity in this case) is zero the rate of change of that quantity (acceleration in this case) is also zero.
If that were true a ball which is thrown upwards would stay at its greatest height above the ground when it was not moving and never come back to the ground.
Just suppose that you were correct and at the instant the velocity of the ball was zero the acceleration $( = fractextchange in velocitytexttime)$ was also zero.
This means that at some instant if time the velocity of the ball does not change from being zero ie the ball is not moving and will stay in a position of rest on the ceiling for all time.
answered Mar 30 at 5:47
FarcherFarcher
52.1k340110
answered Mar 30 at 5:47
FarcherFarcher
52.1k340110
answered Mar 30 at 5:47
FarcherFarcher
52.1k340110
answered Mar 30 at 5:47
FarcherFarcher
52.1k340110
52.1k340110
1
$begingroup$
Okay, I now understand why acceleration isnt zero. Can you explain why D is the answer?
$endgroup$
– Clark Kent
Mar 30 at 6:26
1
$begingroup$
@ClarkKent Up is positive and the acceleration of free fall is downwards ie negative. When the ball hits the ceiling a downward force due to the ceiling acts on the ball increasing the ball’s downward acceleration.
$endgroup$
– Farcher
Mar 30 at 6:31
add a comment |
1
$begingroup$
Okay, I now understand why acceleration isnt zero. Can you explain why D is the answer?
$endgroup$
– Clark Kent
Mar 30 at 6:26
1
$begingroup$
@ClarkKent Up is positive and the acceleration of free fall is downwards ie negative. When the ball hits the ceiling a downward force due to the ceiling acts on the ball increasing the ball’s downward acceleration.
$endgroup$
– Farcher
Mar 30 at 6:31
1
1
$begingroup$
Okay, I now understand why acceleration isnt zero. Can you explain why D is the answer?
$endgroup$
– Clark Kent
Mar 30 at 6:26
$begingroup$
Okay, I now understand why acceleration isnt zero. Can you explain why D is the answer?
$endgroup$
– Clark Kent
Mar 30 at 6:26
1
1
$begingroup$
@ClarkKent Up is positive and the acceleration of free fall is downwards ie negative. When the ball hits the ceiling a downward force due to the ceiling acts on the ball increasing the ball’s downward acceleration.
$endgroup$
– Farcher
Mar 30 at 6:31
$begingroup$
@ClarkKent Up is positive and the acceleration of free fall is downwards ie negative. When the ball hits the ceiling a downward force due to the ceiling acts on the ball increasing the ball’s downward acceleration.
$endgroup$
– Farcher
Mar 30 at 6:31
add a comment |
7
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Yes, the velocity will be zero momentarily. Why would that mean the acceleration is zero?
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– d_b
Mar 30 at 4:52
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a=v-u/t? From this, acceleration should be zero when velocity is zero.
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– Clark Kent
Mar 30 at 5:37
5
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Acceleration is the rate of change of velocity, $a = Delta v/Delta t$. The rate of change of $v$ can be non-zero even when $v=0$. Think about the simpler case of a ball thrown in the air. At the top of its motion, momentarily $v=0$, but there is still acceleration due to gravity, which is why the ball starts moving back down.
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– d_b
Mar 30 at 5:47
6
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@ClarkKent By your argument the ball could never be thrown: It starts in-hand with zero velocity and thus cannot be accelerated.
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– HABO
Mar 30 at 15:06
1
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It may help to think about velocity vs. acceleration by analogy to position vs. velocity; these have very similar relationships to each other, and somehow human brains are wired to understand the latter two much more intuitively. So: a car is driving uptown, hitting green lights at every intersection and passing side streets at 20mph. In this city, they've numbered the side streets; it's in the south but heading north. Eventually it will pass 1st Street South, then 0th Street, then 1st Street North. At the moment that its position reaches 0th street, does its velocity drop to 0mph, too?
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– Daniel Wagner
Mar 30 at 18:17