Velocity, acceleration, derivatives without function












1












$begingroup$


I'm having trouble with an exercise about position, velocity and acceleration.
Apologies for the formatting but the prompt has a graph.
prompt



So I was able to fill out the table up to a certain point, but I"m having issues understanding how I can find the velocity if I don't have the function. I know that I should use the derivatives, but without having the expression of the function, how can I find it?



Thanks a lot for your help,






integration ordinary-differential-equations derivatives







share|cite|improve this question









$endgroup$












  • $begingroup$
    But who writed the values in the "velocity" column ?
    $endgroup$
    – Mauro ALLEGRANZA
    Dec 8 '18 at 20:24










  • $begingroup$
    The implicity assumption seems constant speed. This means that in the first part : $6$ meters travelled in $2$ sec = $3 { text m / text s}$.
    $endgroup$
    – Mauro ALLEGRANZA
    Dec 8 '18 at 20:25












  • $begingroup$
    I did, I tried to guess them based on the relationship of change in position over time. But I'm not sure about what I did and also, how doest that allow me to get to question d? I can't seem to understand if I can find acceleration without the function
    $endgroup$
    – Minimorum
    Dec 8 '18 at 20:26










  • $begingroup$
    Acceleration is the derivative of velocity (second derivative of position) with respect to time. Constant velocity means zero acceleration.
    $endgroup$
    – KM101
    Dec 8 '18 at 20:26












  • $begingroup$
    Ok, I'm a little bit confused. Because the velocity I found in the column is a constant (no x), but it does change, so does that mean that acceleration will be 0 in every case?
    $endgroup$
    – Minimorum
    Dec 8 '18 at 20:30
















1












$begingroup$


I'm having trouble with an exercise about position, velocity and acceleration.
Apologies for the formatting but the prompt has a graph.
prompt



So I was able to fill out the table up to a certain point, but I"m having issues understanding how I can find the velocity if I don't have the function. I know that I should use the derivatives, but without having the expression of the function, how can I find it?



Thanks a lot for your help,






integration ordinary-differential-equations derivatives







share|cite|improve this question









$endgroup$












  • $begingroup$
    But who writed the values in the "velocity" column ?
    $endgroup$
    – Mauro ALLEGRANZA
    Dec 8 '18 at 20:24










  • $begingroup$
    The implicity assumption seems constant speed. This means that in the first part : $6$ meters travelled in $2$ sec = $3 { text m / text s}$.
    $endgroup$
    – Mauro ALLEGRANZA
    Dec 8 '18 at 20:25












  • $begingroup$
    I did, I tried to guess them based on the relationship of change in position over time. But I'm not sure about what I did and also, how doest that allow me to get to question d? I can't seem to understand if I can find acceleration without the function
    $endgroup$
    – Minimorum
    Dec 8 '18 at 20:26










  • $begingroup$
    Acceleration is the derivative of velocity (second derivative of position) with respect to time. Constant velocity means zero acceleration.
    $endgroup$
    – KM101
    Dec 8 '18 at 20:26












  • $begingroup$
    Ok, I'm a little bit confused. Because the velocity I found in the column is a constant (no x), but it does change, so does that mean that acceleration will be 0 in every case?
    $endgroup$
    – Minimorum
    Dec 8 '18 at 20:30














1












1








1





$begingroup$


I'm having trouble with an exercise about position, velocity and acceleration.
Apologies for the formatting but the prompt has a graph.
prompt



So I was able to fill out the table up to a certain point, but I"m having issues understanding how I can find the velocity if I don't have the function. I know that I should use the derivatives, but without having the expression of the function, how can I find it?



Thanks a lot for your help,






integration ordinary-differential-equations derivatives







share|cite|improve this question









$endgroup$




I'm having trouble with an exercise about position, velocity and acceleration.
Apologies for the formatting but the prompt has a graph.
prompt



So I was able to fill out the table up to a certain point, but I"m having issues understanding how I can find the velocity if I don't have the function. I know that I should use the derivatives, but without having the expression of the function, how can I find it?



Thanks a lot for your help,






integration ordinary-differential-equations derivatives




integration ordinary-differential-equations derivatives






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 8 '18 at 20:19









MinimorumMinimorum

175




175












  • $begingroup$
    But who writed the values in the "velocity" column ?
    $endgroup$
    – Mauro ALLEGRANZA
    Dec 8 '18 at 20:24










  • $begingroup$
    The implicity assumption seems constant speed. This means that in the first part : $6$ meters travelled in $2$ sec = $3 { text m / text s}$.
    $endgroup$
    – Mauro ALLEGRANZA
    Dec 8 '18 at 20:25












  • $begingroup$
    I did, I tried to guess them based on the relationship of change in position over time. But I'm not sure about what I did and also, how doest that allow me to get to question d? I can't seem to understand if I can find acceleration without the function
    $endgroup$
    – Minimorum
    Dec 8 '18 at 20:26










  • $begingroup$
    Acceleration is the derivative of velocity (second derivative of position) with respect to time. Constant velocity means zero acceleration.
    $endgroup$
    – KM101
    Dec 8 '18 at 20:26












  • $begingroup$
    Ok, I'm a little bit confused. Because the velocity I found in the column is a constant (no x), but it does change, so does that mean that acceleration will be 0 in every case?
    $endgroup$
    – Minimorum
    Dec 8 '18 at 20:30


















  • $begingroup$
    But who writed the values in the "velocity" column ?
    $endgroup$
    – Mauro ALLEGRANZA
    Dec 8 '18 at 20:24










  • $begingroup$
    The implicity assumption seems constant speed. This means that in the first part : $6$ meters travelled in $2$ sec = $3 { text m / text s}$.
    $endgroup$
    – Mauro ALLEGRANZA
    Dec 8 '18 at 20:25












  • $begingroup$
    I did, I tried to guess them based on the relationship of change in position over time. But I'm not sure about what I did and also, how doest that allow me to get to question d? I can't seem to understand if I can find acceleration without the function
    $endgroup$
    – Minimorum
    Dec 8 '18 at 20:26










  • $begingroup$
    Acceleration is the derivative of velocity (second derivative of position) with respect to time. Constant velocity means zero acceleration.
    $endgroup$
    – KM101
    Dec 8 '18 at 20:26












  • $begingroup$
    Ok, I'm a little bit confused. Because the velocity I found in the column is a constant (no x), but it does change, so does that mean that acceleration will be 0 in every case?
    $endgroup$
    – Minimorum
    Dec 8 '18 at 20:30
















$begingroup$
But who writed the values in the "velocity" column ?
$endgroup$
– Mauro ALLEGRANZA
Dec 8 '18 at 20:24




$begingroup$
But who writed the values in the "velocity" column ?
$endgroup$
– Mauro ALLEGRANZA
Dec 8 '18 at 20:24












$begingroup$
The implicity assumption seems constant speed. This means that in the first part : $6$ meters travelled in $2$ sec = $3 { text m / text s}$.
$endgroup$
– Mauro ALLEGRANZA
Dec 8 '18 at 20:25






$begingroup$
The implicity assumption seems constant speed. This means that in the first part : $6$ meters travelled in $2$ sec = $3 { text m / text s}$.
$endgroup$
– Mauro ALLEGRANZA
Dec 8 '18 at 20:25














$begingroup$
I did, I tried to guess them based on the relationship of change in position over time. But I'm not sure about what I did and also, how doest that allow me to get to question d? I can't seem to understand if I can find acceleration without the function
$endgroup$
– Minimorum
Dec 8 '18 at 20:26




$begingroup$
I did, I tried to guess them based on the relationship of change in position over time. But I'm not sure about what I did and also, how doest that allow me to get to question d? I can't seem to understand if I can find acceleration without the function
$endgroup$
– Minimorum
Dec 8 '18 at 20:26












$begingroup$
Acceleration is the derivative of velocity (second derivative of position) with respect to time. Constant velocity means zero acceleration.
$endgroup$
– KM101
Dec 8 '18 at 20:26






$begingroup$
Acceleration is the derivative of velocity (second derivative of position) with respect to time. Constant velocity means zero acceleration.
$endgroup$
– KM101
Dec 8 '18 at 20:26














$begingroup$
Ok, I'm a little bit confused. Because the velocity I found in the column is a constant (no x), but it does change, so does that mean that acceleration will be 0 in every case?
$endgroup$
– Minimorum
Dec 8 '18 at 20:30




$begingroup$
Ok, I'm a little bit confused. Because the velocity I found in the column is a constant (no x), but it does change, so does that mean that acceleration will be 0 in every case?
$endgroup$
– Minimorum
Dec 8 '18 at 20:30










2 Answers
2






active

oldest

votes


















2












$begingroup$

If the velocity is constant, which is indicated by the fact that the position-time graph is a straight line, you can just take any two points off the graph and use $v=frac {Delta s}{Delta t}$. Any two points on the segment will do. For example, from $t=2$ to $t=6$ it moves from $+6$ to $-10$, so the velocity is $frac {-10-(+6)}{6-2}=-4$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks, that makes sense. But also I have the column about speed. Am I correct to assume that speed if the same as velocity just positive (since velocity is constant anyway?)
    $endgroup$
    – Minimorum
    Dec 8 '18 at 20:44






  • 2




    $begingroup$
    Speed is scalar. The $+$ or $-$ sign in velocity indicates direction, but scalars don’t have direction (vectors do), so you only take the magnitude (positive value).
    $endgroup$
    – KM101
    Dec 8 '18 at 20:45










  • $begingroup$
    Got it. Amazing, thanks
    $endgroup$
    – Minimorum
    Dec 8 '18 at 20:48



















1












$begingroup$

You just look at the slope of the graph. An upward slope indicates motion to the right of the origin/starting point, or positive velocity, and vice-versa. As an example, in the interval $0 < t < 2$, the object travels $6$ meters in $2$ seconds, hence a velocity of $3 frac{m}{s}$. You continue finding the slope of each “section” of the graph. Just remember that velocity depends on whether the slope is upward or downward, while speed is a scalar, not a vector, so it’s always positive.



Finally, acceleration is the derivative of velocity with respect to time. A linear graph for position means constant velocity, hence acceleration is $0$.



For instance:



$$s(t) = nt implies v(t) = n implies a(t) = 0$$



Notice how you need one higher power for position (something involving $t^2$) so you get a non-zero acceleration, such as



$$s(t) = frac{1}{2}at^2 implies v(t) = at implies a(t) = a$$



for some constant $a$.



Hence, for a linear position-time graph, you just find the slope, which becomes the velocity.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Right, the slope becomes the velocity and is constant, got it. Thanks
    $endgroup$
    – Minimorum
    Dec 8 '18 at 20:43






  • 1




    $begingroup$
    Exactly. When you take the derivative of that constant slope, it becomes $0$. Hence, the acceleration becomes $0$.
    $endgroup$
    – KM101
    Dec 8 '18 at 20:44












  • $begingroup$
    Sorry guys, just a little confused by the end of the problem. I worked through the rest. So the acceleration is 0 at t=9 because of what we said previously. But one question asks for the displacement between 0 and 12. I understand that displacement is the derivative of the velocity function right? So since we have velocity has a constant is it going to be 0 as well here? I'm confused because they are constant on each interval, but are changing on different intervals. I understand they don't have variables though. Thanks
    $endgroup$
    – Minimorum
    Dec 8 '18 at 20:55






  • 1




    $begingroup$
    I’m not sure what work you have to show. You could just look at the first and last points and see how the position changed. The objects displaced $6$ meters rightward. (Zero calculation, just inspection.) If you have to show your work, find the displacement of each “section” individually (be sure to get the signs correctly) and just add them up.
    $endgroup$
    – KM101
    Dec 8 '18 at 20:57












  • $begingroup$
    Oh ok, I was imaging it would be something with derivatives again. But this makes sense. thanks I
    $endgroup$
    – Minimorum
    Dec 8 '18 at 20:58











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

If the velocity is constant, which is indicated by the fact that the position-time graph is a straight line, you can just take any two points off the graph and use $v=frac {Delta s}{Delta t}$. Any two points on the segment will do. For example, from $t=2$ to $t=6$ it moves from $+6$ to $-10$, so the velocity is $frac {-10-(+6)}{6-2}=-4$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks, that makes sense. But also I have the column about speed. Am I correct to assume that speed if the same as velocity just positive (since velocity is constant anyway?)
    $endgroup$
    – Minimorum
    Dec 8 '18 at 20:44






  • 2




    $begingroup$
    Speed is scalar. The $+$ or $-$ sign in velocity indicates direction, but scalars don’t have direction (vectors do), so you only take the magnitude (positive value).
    $endgroup$
    – KM101
    Dec 8 '18 at 20:45










  • $begingroup$
    Got it. Amazing, thanks
    $endgroup$
    – Minimorum
    Dec 8 '18 at 20:48
















2












$begingroup$

If the velocity is constant, which is indicated by the fact that the position-time graph is a straight line, you can just take any two points off the graph and use $v=frac {Delta s}{Delta t}$. Any two points on the segment will do. For example, from $t=2$ to $t=6$ it moves from $+6$ to $-10$, so the velocity is $frac {-10-(+6)}{6-2}=-4$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks, that makes sense. But also I have the column about speed. Am I correct to assume that speed if the same as velocity just positive (since velocity is constant anyway?)
    $endgroup$
    – Minimorum
    Dec 8 '18 at 20:44






  • 2




    $begingroup$
    Speed is scalar. The $+$ or $-$ sign in velocity indicates direction, but scalars don’t have direction (vectors do), so you only take the magnitude (positive value).
    $endgroup$
    – KM101
    Dec 8 '18 at 20:45










  • $begingroup$
    Got it. Amazing, thanks
    $endgroup$
    – Minimorum
    Dec 8 '18 at 20:48














2












2








2





$begingroup$

If the velocity is constant, which is indicated by the fact that the position-time graph is a straight line, you can just take any two points off the graph and use $v=frac {Delta s}{Delta t}$. Any two points on the segment will do. For example, from $t=2$ to $t=6$ it moves from $+6$ to $-10$, so the velocity is $frac {-10-(+6)}{6-2}=-4$






share|cite|improve this answer









$endgroup$



If the velocity is constant, which is indicated by the fact that the position-time graph is a straight line, you can just take any two points off the graph and use $v=frac {Delta s}{Delta t}$. Any two points on the segment will do. For example, from $t=2$ to $t=6$ it moves from $+6$ to $-10$, so the velocity is $frac {-10-(+6)}{6-2}=-4$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 8 '18 at 20:38









Ross MillikanRoss Millikan

294k23198371




294k23198371












  • $begingroup$
    Thanks, that makes sense. But also I have the column about speed. Am I correct to assume that speed if the same as velocity just positive (since velocity is constant anyway?)
    $endgroup$
    – Minimorum
    Dec 8 '18 at 20:44






  • 2




    $begingroup$
    Speed is scalar. The $+$ or $-$ sign in velocity indicates direction, but scalars don’t have direction (vectors do), so you only take the magnitude (positive value).
    $endgroup$
    – KM101
    Dec 8 '18 at 20:45










  • $begingroup$
    Got it. Amazing, thanks
    $endgroup$
    – Minimorum
    Dec 8 '18 at 20:48


















  • $begingroup$
    Thanks, that makes sense. But also I have the column about speed. Am I correct to assume that speed if the same as velocity just positive (since velocity is constant anyway?)
    $endgroup$
    – Minimorum
    Dec 8 '18 at 20:44






  • 2




    $begingroup$
    Speed is scalar. The $+$ or $-$ sign in velocity indicates direction, but scalars don’t have direction (vectors do), so you only take the magnitude (positive value).
    $endgroup$
    – KM101
    Dec 8 '18 at 20:45










  • $begingroup$
    Got it. Amazing, thanks
    $endgroup$
    – Minimorum
    Dec 8 '18 at 20:48
















$begingroup$
Thanks, that makes sense. But also I have the column about speed. Am I correct to assume that speed if the same as velocity just positive (since velocity is constant anyway?)
$endgroup$
– Minimorum
Dec 8 '18 at 20:44




$begingroup$
Thanks, that makes sense. But also I have the column about speed. Am I correct to assume that speed if the same as velocity just positive (since velocity is constant anyway?)
$endgroup$
– Minimorum
Dec 8 '18 at 20:44




2




2




$begingroup$
Speed is scalar. The $+$ or $-$ sign in velocity indicates direction, but scalars don’t have direction (vectors do), so you only take the magnitude (positive value).
$endgroup$
– KM101
Dec 8 '18 at 20:45




$begingroup$
Speed is scalar. The $+$ or $-$ sign in velocity indicates direction, but scalars don’t have direction (vectors do), so you only take the magnitude (positive value).
$endgroup$
– KM101
Dec 8 '18 at 20:45












$begingroup$
Got it. Amazing, thanks
$endgroup$
– Minimorum
Dec 8 '18 at 20:48




$begingroup$
Got it. Amazing, thanks
$endgroup$
– Minimorum
Dec 8 '18 at 20:48











1












$begingroup$

You just look at the slope of the graph. An upward slope indicates motion to the right of the origin/starting point, or positive velocity, and vice-versa. As an example, in the interval $0 < t < 2$, the object travels $6$ meters in $2$ seconds, hence a velocity of $3 frac{m}{s}$. You continue finding the slope of each “section” of the graph. Just remember that velocity depends on whether the slope is upward or downward, while speed is a scalar, not a vector, so it’s always positive.



Finally, acceleration is the derivative of velocity with respect to time. A linear graph for position means constant velocity, hence acceleration is $0$.



For instance:



$$s(t) = nt implies v(t) = n implies a(t) = 0$$



Notice how you need one higher power for position (something involving $t^2$) so you get a non-zero acceleration, such as



$$s(t) = frac{1}{2}at^2 implies v(t) = at implies a(t) = a$$



for some constant $a$.



Hence, for a linear position-time graph, you just find the slope, which becomes the velocity.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Right, the slope becomes the velocity and is constant, got it. Thanks
    $endgroup$
    – Minimorum
    Dec 8 '18 at 20:43






  • 1




    $begingroup$
    Exactly. When you take the derivative of that constant slope, it becomes $0$. Hence, the acceleration becomes $0$.
    $endgroup$
    – KM101
    Dec 8 '18 at 20:44












  • $begingroup$
    Sorry guys, just a little confused by the end of the problem. I worked through the rest. So the acceleration is 0 at t=9 because of what we said previously. But one question asks for the displacement between 0 and 12. I understand that displacement is the derivative of the velocity function right? So since we have velocity has a constant is it going to be 0 as well here? I'm confused because they are constant on each interval, but are changing on different intervals. I understand they don't have variables though. Thanks
    $endgroup$
    – Minimorum
    Dec 8 '18 at 20:55






  • 1




    $begingroup$
    I’m not sure what work you have to show. You could just look at the first and last points and see how the position changed. The objects displaced $6$ meters rightward. (Zero calculation, just inspection.) If you have to show your work, find the displacement of each “section” individually (be sure to get the signs correctly) and just add them up.
    $endgroup$
    – KM101
    Dec 8 '18 at 20:57












  • $begingroup$
    Oh ok, I was imaging it would be something with derivatives again. But this makes sense. thanks I
    $endgroup$
    – Minimorum
    Dec 8 '18 at 20:58
















1












$begingroup$

You just look at the slope of the graph. An upward slope indicates motion to the right of the origin/starting point, or positive velocity, and vice-versa. As an example, in the interval $0 < t < 2$, the object travels $6$ meters in $2$ seconds, hence a velocity of $3 frac{m}{s}$. You continue finding the slope of each “section” of the graph. Just remember that velocity depends on whether the slope is upward or downward, while speed is a scalar, not a vector, so it’s always positive.



Finally, acceleration is the derivative of velocity with respect to time. A linear graph for position means constant velocity, hence acceleration is $0$.



For instance:



$$s(t) = nt implies v(t) = n implies a(t) = 0$$



Notice how you need one higher power for position (something involving $t^2$) so you get a non-zero acceleration, such as



$$s(t) = frac{1}{2}at^2 implies v(t) = at implies a(t) = a$$



for some constant $a$.



Hence, for a linear position-time graph, you just find the slope, which becomes the velocity.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Right, the slope becomes the velocity and is constant, got it. Thanks
    $endgroup$
    – Minimorum
    Dec 8 '18 at 20:43






  • 1




    $begingroup$
    Exactly. When you take the derivative of that constant slope, it becomes $0$. Hence, the acceleration becomes $0$.
    $endgroup$
    – KM101
    Dec 8 '18 at 20:44












  • $begingroup$
    Sorry guys, just a little confused by the end of the problem. I worked through the rest. So the acceleration is 0 at t=9 because of what we said previously. But one question asks for the displacement between 0 and 12. I understand that displacement is the derivative of the velocity function right? So since we have velocity has a constant is it going to be 0 as well here? I'm confused because they are constant on each interval, but are changing on different intervals. I understand they don't have variables though. Thanks
    $endgroup$
    – Minimorum
    Dec 8 '18 at 20:55






  • 1




    $begingroup$
    I’m not sure what work you have to show. You could just look at the first and last points and see how the position changed. The objects displaced $6$ meters rightward. (Zero calculation, just inspection.) If you have to show your work, find the displacement of each “section” individually (be sure to get the signs correctly) and just add them up.
    $endgroup$
    – KM101
    Dec 8 '18 at 20:57












  • $begingroup$
    Oh ok, I was imaging it would be something with derivatives again. But this makes sense. thanks I
    $endgroup$
    – Minimorum
    Dec 8 '18 at 20:58














1












1








1





$begingroup$

You just look at the slope of the graph. An upward slope indicates motion to the right of the origin/starting point, or positive velocity, and vice-versa. As an example, in the interval $0 < t < 2$, the object travels $6$ meters in $2$ seconds, hence a velocity of $3 frac{m}{s}$. You continue finding the slope of each “section” of the graph. Just remember that velocity depends on whether the slope is upward or downward, while speed is a scalar, not a vector, so it’s always positive.



Finally, acceleration is the derivative of velocity with respect to time. A linear graph for position means constant velocity, hence acceleration is $0$.



For instance:



$$s(t) = nt implies v(t) = n implies a(t) = 0$$



Notice how you need one higher power for position (something involving $t^2$) so you get a non-zero acceleration, such as



$$s(t) = frac{1}{2}at^2 implies v(t) = at implies a(t) = a$$



for some constant $a$.



Hence, for a linear position-time graph, you just find the slope, which becomes the velocity.






share|cite|improve this answer











$endgroup$



You just look at the slope of the graph. An upward slope indicates motion to the right of the origin/starting point, or positive velocity, and vice-versa. As an example, in the interval $0 < t < 2$, the object travels $6$ meters in $2$ seconds, hence a velocity of $3 frac{m}{s}$. You continue finding the slope of each “section” of the graph. Just remember that velocity depends on whether the slope is upward or downward, while speed is a scalar, not a vector, so it’s always positive.



Finally, acceleration is the derivative of velocity with respect to time. A linear graph for position means constant velocity, hence acceleration is $0$.



For instance:



$$s(t) = nt implies v(t) = n implies a(t) = 0$$



Notice how you need one higher power for position (something involving $t^2$) so you get a non-zero acceleration, such as



$$s(t) = frac{1}{2}at^2 implies v(t) = at implies a(t) = a$$



for some constant $a$.



Hence, for a linear position-time graph, you just find the slope, which becomes the velocity.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 8 '18 at 20:51

























answered Dec 8 '18 at 20:36









KM101KM101

5,9281523




5,9281523












  • $begingroup$
    Right, the slope becomes the velocity and is constant, got it. Thanks
    $endgroup$
    – Minimorum
    Dec 8 '18 at 20:43






  • 1




    $begingroup$
    Exactly. When you take the derivative of that constant slope, it becomes $0$. Hence, the acceleration becomes $0$.
    $endgroup$
    – KM101
    Dec 8 '18 at 20:44












  • $begingroup$
    Sorry guys, just a little confused by the end of the problem. I worked through the rest. So the acceleration is 0 at t=9 because of what we said previously. But one question asks for the displacement between 0 and 12. I understand that displacement is the derivative of the velocity function right? So since we have velocity has a constant is it going to be 0 as well here? I'm confused because they are constant on each interval, but are changing on different intervals. I understand they don't have variables though. Thanks
    $endgroup$
    – Minimorum
    Dec 8 '18 at 20:55






  • 1




    $begingroup$
    I’m not sure what work you have to show. You could just look at the first and last points and see how the position changed. The objects displaced $6$ meters rightward. (Zero calculation, just inspection.) If you have to show your work, find the displacement of each “section” individually (be sure to get the signs correctly) and just add them up.
    $endgroup$
    – KM101
    Dec 8 '18 at 20:57












  • $begingroup$
    Oh ok, I was imaging it would be something with derivatives again. But this makes sense. thanks I
    $endgroup$
    – Minimorum
    Dec 8 '18 at 20:58


















  • $begingroup$
    Right, the slope becomes the velocity and is constant, got it. Thanks
    $endgroup$
    – Minimorum
    Dec 8 '18 at 20:43






  • 1




    $begingroup$
    Exactly. When you take the derivative of that constant slope, it becomes $0$. Hence, the acceleration becomes $0$.
    $endgroup$
    – KM101
    Dec 8 '18 at 20:44












  • $begingroup$
    Sorry guys, just a little confused by the end of the problem. I worked through the rest. So the acceleration is 0 at t=9 because of what we said previously. But one question asks for the displacement between 0 and 12. I understand that displacement is the derivative of the velocity function right? So since we have velocity has a constant is it going to be 0 as well here? I'm confused because they are constant on each interval, but are changing on different intervals. I understand they don't have variables though. Thanks
    $endgroup$
    – Minimorum
    Dec 8 '18 at 20:55






  • 1




    $begingroup$
    I’m not sure what work you have to show. You could just look at the first and last points and see how the position changed. The objects displaced $6$ meters rightward. (Zero calculation, just inspection.) If you have to show your work, find the displacement of each “section” individually (be sure to get the signs correctly) and just add them up.
    $endgroup$
    – KM101
    Dec 8 '18 at 20:57












  • $begingroup$
    Oh ok, I was imaging it would be something with derivatives again. But this makes sense. thanks I
    $endgroup$
    – Minimorum
    Dec 8 '18 at 20:58
















$begingroup$
Right, the slope becomes the velocity and is constant, got it. Thanks
$endgroup$
– Minimorum
Dec 8 '18 at 20:43




$begingroup$
Right, the slope becomes the velocity and is constant, got it. Thanks
$endgroup$
– Minimorum
Dec 8 '18 at 20:43




1




1




$begingroup$
Exactly. When you take the derivative of that constant slope, it becomes $0$. Hence, the acceleration becomes $0$.
$endgroup$
– KM101
Dec 8 '18 at 20:44






$begingroup$
Exactly. When you take the derivative of that constant slope, it becomes $0$. Hence, the acceleration becomes $0$.
$endgroup$
– KM101
Dec 8 '18 at 20:44














$begingroup$
Sorry guys, just a little confused by the end of the problem. I worked through the rest. So the acceleration is 0 at t=9 because of what we said previously. But one question asks for the displacement between 0 and 12. I understand that displacement is the derivative of the velocity function right? So since we have velocity has a constant is it going to be 0 as well here? I'm confused because they are constant on each interval, but are changing on different intervals. I understand they don't have variables though. Thanks
$endgroup$
– Minimorum
Dec 8 '18 at 20:55




$begingroup$
Sorry guys, just a little confused by the end of the problem. I worked through the rest. So the acceleration is 0 at t=9 because of what we said previously. But one question asks for the displacement between 0 and 12. I understand that displacement is the derivative of the velocity function right? So since we have velocity has a constant is it going to be 0 as well here? I'm confused because they are constant on each interval, but are changing on different intervals. I understand they don't have variables though. Thanks
$endgroup$
– Minimorum
Dec 8 '18 at 20:55




1




1




$begingroup$
I’m not sure what work you have to show. You could just look at the first and last points and see how the position changed. The objects displaced $6$ meters rightward. (Zero calculation, just inspection.) If you have to show your work, find the displacement of each “section” individually (be sure to get the signs correctly) and just add them up.
$endgroup$
– KM101
Dec 8 '18 at 20:57






$begingroup$
I’m not sure what work you have to show. You could just look at the first and last points and see how the position changed. The objects displaced $6$ meters rightward. (Zero calculation, just inspection.) If you have to show your work, find the displacement of each “section” individually (be sure to get the signs correctly) and just add them up.
$endgroup$
– KM101
Dec 8 '18 at 20:57














$begingroup$
Oh ok, I was imaging it would be something with derivatives again. But this makes sense. thanks I
$endgroup$
– Minimorum
Dec 8 '18 at 20:58




$begingroup$
Oh ok, I was imaging it would be something with derivatives again. But this makes sense. thanks I
$endgroup$
– Minimorum
Dec 8 '18 at 20:58


















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