Find distance as function of time.












0












$begingroup$


A car starts from rest and accelerates in $a = frac{2cdot m}{3cdot s^3}t$,



After $3$ seconds, The car will be $27$ metres from beginning.



Find distance as function of time.



I know i have to integral the acceleration,But i don't know how.



I found the Equation is $x_{t} = 24 + frac{1}{3} t^2$.



What do you think ?






physics







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    This makes no sense. If the initial speed is 0 and the acceleration is 2/3 then the distance is $t^2/3$ giving a distance 3 m at 3 s, not 27.
    $endgroup$
    – almagest
    Mar 18 '16 at 14:09










  • $begingroup$
    I completely agree. This doesn't make sense. By the way, units of $a$ seems wrong.
    $endgroup$
    – Win Vineeth
    Mar 18 '16 at 14:10


















0












$begingroup$


A car starts from rest and accelerates in $a = frac{2cdot m}{3cdot s^3}t$,



After $3$ seconds, The car will be $27$ metres from beginning.



Find distance as function of time.



I know i have to integral the acceleration,But i don't know how.



I found the Equation is $x_{t} = 24 + frac{1}{3} t^2$.



What do you think ?






physics







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    This makes no sense. If the initial speed is 0 and the acceleration is 2/3 then the distance is $t^2/3$ giving a distance 3 m at 3 s, not 27.
    $endgroup$
    – almagest
    Mar 18 '16 at 14:09










  • $begingroup$
    I completely agree. This doesn't make sense. By the way, units of $a$ seems wrong.
    $endgroup$
    – Win Vineeth
    Mar 18 '16 at 14:10
















0












0








0





$begingroup$


A car starts from rest and accelerates in $a = frac{2cdot m}{3cdot s^3}t$,



After $3$ seconds, The car will be $27$ metres from beginning.



Find distance as function of time.



I know i have to integral the acceleration,But i don't know how.



I found the Equation is $x_{t} = 24 + frac{1}{3} t^2$.



What do you think ?






physics







share|cite|improve this question











$endgroup$




A car starts from rest and accelerates in $a = frac{2cdot m}{3cdot s^3}t$,



After $3$ seconds, The car will be $27$ metres from beginning.



Find distance as function of time.



I know i have to integral the acceleration,But i don't know how.



I found the Equation is $x_{t} = 24 + frac{1}{3} t^2$.



What do you think ?






physics




physics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 18 '16 at 14:16







Noam

















asked Mar 18 '16 at 13:56









NoamNoam

385113




385113








  • 1




    $begingroup$
    This makes no sense. If the initial speed is 0 and the acceleration is 2/3 then the distance is $t^2/3$ giving a distance 3 m at 3 s, not 27.
    $endgroup$
    – almagest
    Mar 18 '16 at 14:09










  • $begingroup$
    I completely agree. This doesn't make sense. By the way, units of $a$ seems wrong.
    $endgroup$
    – Win Vineeth
    Mar 18 '16 at 14:10
















  • 1




    $begingroup$
    This makes no sense. If the initial speed is 0 and the acceleration is 2/3 then the distance is $t^2/3$ giving a distance 3 m at 3 s, not 27.
    $endgroup$
    – almagest
    Mar 18 '16 at 14:09










  • $begingroup$
    I completely agree. This doesn't make sense. By the way, units of $a$ seems wrong.
    $endgroup$
    – Win Vineeth
    Mar 18 '16 at 14:10










1




1




$begingroup$
This makes no sense. If the initial speed is 0 and the acceleration is 2/3 then the distance is $t^2/3$ giving a distance 3 m at 3 s, not 27.
$endgroup$
– almagest
Mar 18 '16 at 14:09




$begingroup$
This makes no sense. If the initial speed is 0 and the acceleration is 2/3 then the distance is $t^2/3$ giving a distance 3 m at 3 s, not 27.
$endgroup$
– almagest
Mar 18 '16 at 14:09












$begingroup$
I completely agree. This doesn't make sense. By the way, units of $a$ seems wrong.
$endgroup$
– Win Vineeth
Mar 18 '16 at 14:10






$begingroup$
I completely agree. This doesn't make sense. By the way, units of $a$ seems wrong.
$endgroup$
– Win Vineeth
Mar 18 '16 at 14:10












2 Answers
2






active

oldest

votes


















0












$begingroup$

You know that $x=ut+frac{1}{2}at^2$



Since the car starts from rest, $u=0$ and $x(t)=frac{1}{2}at^2$



If $x(3)=27$, then $a=3m/s^2$



The equation is $x(t)=frac 32 t^2$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    $a = frac{2cdot m}{3cdot s^2}$
    $endgroup$
    – Noam
    Mar 18 '16 at 14:06










  • $begingroup$
    That would mean it didn't start at $x=0$. Else, it's an impossible question.
    $endgroup$
    – Win Vineeth
    Mar 18 '16 at 14:09



















0












$begingroup$

$a(t) = v'(t) = x''(t)$; we integrate acceleration to find velocity, than integrate that to find position as a function of time. We're given $a(t) = frac{2}{3}t$ and the initial values $x(0) = 0, v(0) = 0$ (because the car starts from rest) and $x(3) = 27$. Can you figure out $x(t)$ from here?






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I got your point. But should I do integral for what ? $int frac{2}{3}t$ or $int frac{2}{3}t^2$ ? I have some mis-understaing with the context of integration and velocity,acceleration and position.
    $endgroup$
    – Noam
    Mar 18 '16 at 14:07












  • $begingroup$
    Take it one step at a time. $v(t) = int a(t) , dt = int frac{2}{3} , dt = frac{2}{3}t + C_1$. Follow a similar process to find $x(t) = int v(t) , dt$, then use the initial values to find $C_1$ and $C_2$.
    $endgroup$
    – DylanSp
    Mar 18 '16 at 14:07












  • $begingroup$
    What do you mean ?
    $endgroup$
    – Noam
    Mar 18 '16 at 14:08










  • $begingroup$
    Sorry, accidentally submitted before I was finished. See my edited version.
    $endgroup$
    – DylanSp
    Mar 18 '16 at 14:11










  • $begingroup$
    $a(t) = frac{2}{3}t$. $v(t) = int a(t) = frac{1}{3}t^2 + v_{0}$. $x(t) = int v(t) = frac{1}{9} + v_{0}t+x_{0}$. $27 = frac{1}{9}t^{3}+x_{0}$. the final eqation is - $x(t) = frac{1}{9}t^3+24$. What do you think ?
    $endgroup$
    – Noam
    Mar 18 '16 at 14:19













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2 Answers
2






active

oldest

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2 Answers
2






active

oldest

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active

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active

oldest

votes









0












$begingroup$

You know that $x=ut+frac{1}{2}at^2$



Since the car starts from rest, $u=0$ and $x(t)=frac{1}{2}at^2$



If $x(3)=27$, then $a=3m/s^2$



The equation is $x(t)=frac 32 t^2$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    $a = frac{2cdot m}{3cdot s^2}$
    $endgroup$
    – Noam
    Mar 18 '16 at 14:06










  • $begingroup$
    That would mean it didn't start at $x=0$. Else, it's an impossible question.
    $endgroup$
    – Win Vineeth
    Mar 18 '16 at 14:09
















0












$begingroup$

You know that $x=ut+frac{1}{2}at^2$



Since the car starts from rest, $u=0$ and $x(t)=frac{1}{2}at^2$



If $x(3)=27$, then $a=3m/s^2$



The equation is $x(t)=frac 32 t^2$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    $a = frac{2cdot m}{3cdot s^2}$
    $endgroup$
    – Noam
    Mar 18 '16 at 14:06










  • $begingroup$
    That would mean it didn't start at $x=0$. Else, it's an impossible question.
    $endgroup$
    – Win Vineeth
    Mar 18 '16 at 14:09














0












0








0





$begingroup$

You know that $x=ut+frac{1}{2}at^2$



Since the car starts from rest, $u=0$ and $x(t)=frac{1}{2}at^2$



If $x(3)=27$, then $a=3m/s^2$



The equation is $x(t)=frac 32 t^2$






share|cite|improve this answer









$endgroup$



You know that $x=ut+frac{1}{2}at^2$



Since the car starts from rest, $u=0$ and $x(t)=frac{1}{2}at^2$



If $x(3)=27$, then $a=3m/s^2$



The equation is $x(t)=frac 32 t^2$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Mar 18 '16 at 14:04









Win VineethWin Vineeth

3,270527




3,270527












  • $begingroup$
    $a = frac{2cdot m}{3cdot s^2}$
    $endgroup$
    – Noam
    Mar 18 '16 at 14:06










  • $begingroup$
    That would mean it didn't start at $x=0$. Else, it's an impossible question.
    $endgroup$
    – Win Vineeth
    Mar 18 '16 at 14:09


















  • $begingroup$
    $a = frac{2cdot m}{3cdot s^2}$
    $endgroup$
    – Noam
    Mar 18 '16 at 14:06










  • $begingroup$
    That would mean it didn't start at $x=0$. Else, it's an impossible question.
    $endgroup$
    – Win Vineeth
    Mar 18 '16 at 14:09
















$begingroup$
$a = frac{2cdot m}{3cdot s^2}$
$endgroup$
– Noam
Mar 18 '16 at 14:06




$begingroup$
$a = frac{2cdot m}{3cdot s^2}$
$endgroup$
– Noam
Mar 18 '16 at 14:06












$begingroup$
That would mean it didn't start at $x=0$. Else, it's an impossible question.
$endgroup$
– Win Vineeth
Mar 18 '16 at 14:09




$begingroup$
That would mean it didn't start at $x=0$. Else, it's an impossible question.
$endgroup$
– Win Vineeth
Mar 18 '16 at 14:09











0












$begingroup$

$a(t) = v'(t) = x''(t)$; we integrate acceleration to find velocity, than integrate that to find position as a function of time. We're given $a(t) = frac{2}{3}t$ and the initial values $x(0) = 0, v(0) = 0$ (because the car starts from rest) and $x(3) = 27$. Can you figure out $x(t)$ from here?






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I got your point. But should I do integral for what ? $int frac{2}{3}t$ or $int frac{2}{3}t^2$ ? I have some mis-understaing with the context of integration and velocity,acceleration and position.
    $endgroup$
    – Noam
    Mar 18 '16 at 14:07












  • $begingroup$
    Take it one step at a time. $v(t) = int a(t) , dt = int frac{2}{3} , dt = frac{2}{3}t + C_1$. Follow a similar process to find $x(t) = int v(t) , dt$, then use the initial values to find $C_1$ and $C_2$.
    $endgroup$
    – DylanSp
    Mar 18 '16 at 14:07












  • $begingroup$
    What do you mean ?
    $endgroup$
    – Noam
    Mar 18 '16 at 14:08










  • $begingroup$
    Sorry, accidentally submitted before I was finished. See my edited version.
    $endgroup$
    – DylanSp
    Mar 18 '16 at 14:11










  • $begingroup$
    $a(t) = frac{2}{3}t$. $v(t) = int a(t) = frac{1}{3}t^2 + v_{0}$. $x(t) = int v(t) = frac{1}{9} + v_{0}t+x_{0}$. $27 = frac{1}{9}t^{3}+x_{0}$. the final eqation is - $x(t) = frac{1}{9}t^3+24$. What do you think ?
    $endgroup$
    – Noam
    Mar 18 '16 at 14:19


















0












$begingroup$

$a(t) = v'(t) = x''(t)$; we integrate acceleration to find velocity, than integrate that to find position as a function of time. We're given $a(t) = frac{2}{3}t$ and the initial values $x(0) = 0, v(0) = 0$ (because the car starts from rest) and $x(3) = 27$. Can you figure out $x(t)$ from here?






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I got your point. But should I do integral for what ? $int frac{2}{3}t$ or $int frac{2}{3}t^2$ ? I have some mis-understaing with the context of integration and velocity,acceleration and position.
    $endgroup$
    – Noam
    Mar 18 '16 at 14:07












  • $begingroup$
    Take it one step at a time. $v(t) = int a(t) , dt = int frac{2}{3} , dt = frac{2}{3}t + C_1$. Follow a similar process to find $x(t) = int v(t) , dt$, then use the initial values to find $C_1$ and $C_2$.
    $endgroup$
    – DylanSp
    Mar 18 '16 at 14:07












  • $begingroup$
    What do you mean ?
    $endgroup$
    – Noam
    Mar 18 '16 at 14:08










  • $begingroup$
    Sorry, accidentally submitted before I was finished. See my edited version.
    $endgroup$
    – DylanSp
    Mar 18 '16 at 14:11










  • $begingroup$
    $a(t) = frac{2}{3}t$. $v(t) = int a(t) = frac{1}{3}t^2 + v_{0}$. $x(t) = int v(t) = frac{1}{9} + v_{0}t+x_{0}$. $27 = frac{1}{9}t^{3}+x_{0}$. the final eqation is - $x(t) = frac{1}{9}t^3+24$. What do you think ?
    $endgroup$
    – Noam
    Mar 18 '16 at 14:19
















0












0








0





$begingroup$

$a(t) = v'(t) = x''(t)$; we integrate acceleration to find velocity, than integrate that to find position as a function of time. We're given $a(t) = frac{2}{3}t$ and the initial values $x(0) = 0, v(0) = 0$ (because the car starts from rest) and $x(3) = 27$. Can you figure out $x(t)$ from here?






share|cite|improve this answer











$endgroup$



$a(t) = v'(t) = x''(t)$; we integrate acceleration to find velocity, than integrate that to find position as a function of time. We're given $a(t) = frac{2}{3}t$ and the initial values $x(0) = 0, v(0) = 0$ (because the car starts from rest) and $x(3) = 27$. Can you figure out $x(t)$ from here?







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Mar 18 '16 at 16:17

























answered Mar 18 '16 at 14:01









DylanSpDylanSp

1,589714




1,589714












  • $begingroup$
    I got your point. But should I do integral for what ? $int frac{2}{3}t$ or $int frac{2}{3}t^2$ ? I have some mis-understaing with the context of integration and velocity,acceleration and position.
    $endgroup$
    – Noam
    Mar 18 '16 at 14:07












  • $begingroup$
    Take it one step at a time. $v(t) = int a(t) , dt = int frac{2}{3} , dt = frac{2}{3}t + C_1$. Follow a similar process to find $x(t) = int v(t) , dt$, then use the initial values to find $C_1$ and $C_2$.
    $endgroup$
    – DylanSp
    Mar 18 '16 at 14:07












  • $begingroup$
    What do you mean ?
    $endgroup$
    – Noam
    Mar 18 '16 at 14:08










  • $begingroup$
    Sorry, accidentally submitted before I was finished. See my edited version.
    $endgroup$
    – DylanSp
    Mar 18 '16 at 14:11










  • $begingroup$
    $a(t) = frac{2}{3}t$. $v(t) = int a(t) = frac{1}{3}t^2 + v_{0}$. $x(t) = int v(t) = frac{1}{9} + v_{0}t+x_{0}$. $27 = frac{1}{9}t^{3}+x_{0}$. the final eqation is - $x(t) = frac{1}{9}t^3+24$. What do you think ?
    $endgroup$
    – Noam
    Mar 18 '16 at 14:19




















  • $begingroup$
    I got your point. But should I do integral for what ? $int frac{2}{3}t$ or $int frac{2}{3}t^2$ ? I have some mis-understaing with the context of integration and velocity,acceleration and position.
    $endgroup$
    – Noam
    Mar 18 '16 at 14:07












  • $begingroup$
    Take it one step at a time. $v(t) = int a(t) , dt = int frac{2}{3} , dt = frac{2}{3}t + C_1$. Follow a similar process to find $x(t) = int v(t) , dt$, then use the initial values to find $C_1$ and $C_2$.
    $endgroup$
    – DylanSp
    Mar 18 '16 at 14:07












  • $begingroup$
    What do you mean ?
    $endgroup$
    – Noam
    Mar 18 '16 at 14:08










  • $begingroup$
    Sorry, accidentally submitted before I was finished. See my edited version.
    $endgroup$
    – DylanSp
    Mar 18 '16 at 14:11










  • $begingroup$
    $a(t) = frac{2}{3}t$. $v(t) = int a(t) = frac{1}{3}t^2 + v_{0}$. $x(t) = int v(t) = frac{1}{9} + v_{0}t+x_{0}$. $27 = frac{1}{9}t^{3}+x_{0}$. the final eqation is - $x(t) = frac{1}{9}t^3+24$. What do you think ?
    $endgroup$
    – Noam
    Mar 18 '16 at 14:19


















$begingroup$
I got your point. But should I do integral for what ? $int frac{2}{3}t$ or $int frac{2}{3}t^2$ ? I have some mis-understaing with the context of integration and velocity,acceleration and position.
$endgroup$
– Noam
Mar 18 '16 at 14:07






$begingroup$
I got your point. But should I do integral for what ? $int frac{2}{3}t$ or $int frac{2}{3}t^2$ ? I have some mis-understaing with the context of integration and velocity,acceleration and position.
$endgroup$
– Noam
Mar 18 '16 at 14:07














$begingroup$
Take it one step at a time. $v(t) = int a(t) , dt = int frac{2}{3} , dt = frac{2}{3}t + C_1$. Follow a similar process to find $x(t) = int v(t) , dt$, then use the initial values to find $C_1$ and $C_2$.
$endgroup$
– DylanSp
Mar 18 '16 at 14:07






$begingroup$
Take it one step at a time. $v(t) = int a(t) , dt = int frac{2}{3} , dt = frac{2}{3}t + C_1$. Follow a similar process to find $x(t) = int v(t) , dt$, then use the initial values to find $C_1$ and $C_2$.
$endgroup$
– DylanSp
Mar 18 '16 at 14:07














$begingroup$
What do you mean ?
$endgroup$
– Noam
Mar 18 '16 at 14:08




$begingroup$
What do you mean ?
$endgroup$
– Noam
Mar 18 '16 at 14:08












$begingroup$
Sorry, accidentally submitted before I was finished. See my edited version.
$endgroup$
– DylanSp
Mar 18 '16 at 14:11




$begingroup$
Sorry, accidentally submitted before I was finished. See my edited version.
$endgroup$
– DylanSp
Mar 18 '16 at 14:11












$begingroup$
$a(t) = frac{2}{3}t$. $v(t) = int a(t) = frac{1}{3}t^2 + v_{0}$. $x(t) = int v(t) = frac{1}{9} + v_{0}t+x_{0}$. $27 = frac{1}{9}t^{3}+x_{0}$. the final eqation is - $x(t) = frac{1}{9}t^3+24$. What do you think ?
$endgroup$
– Noam
Mar 18 '16 at 14:19






$begingroup$
$a(t) = frac{2}{3}t$. $v(t) = int a(t) = frac{1}{3}t^2 + v_{0}$. $x(t) = int v(t) = frac{1}{9} + v_{0}t+x_{0}$. $27 = frac{1}{9}t^{3}+x_{0}$. the final eqation is - $x(t) = frac{1}{9}t^3+24$. What do you think ?
$endgroup$
– Noam
Mar 18 '16 at 14:19




















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Wiesbaden

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Wappen Deutschlandkarte 50.0825 8.24 117 Koordinaten: 50° 5′  N , 8° 14′  O Basisdaten Bundesland: Hessen Regierungsbezirk: Darmstadt Höhe: 117 m ü. NHN Fläche: 203,93 km 2 Einwohner: 278.654 (31. Dez. 2017) [1] Bevölkerungsdichte: 1366 Einwohner je km 2 Postleitzahlen: 65183–65207, 55246, 55252 Vorlage:Infobox Gemeinde in Deutschland/Wartung/PLZ enthält Text Vorwahlen: 0611, 06122, 06127, 06134 Kfz-Kennzeichen: WI Gemeindeschlüssel: 06 4 14 000 LOCODE: DE WIB NUTS: DE714 Stadtgliederung: 26 Ortsbezirke Adresse der Stadtverwaltung: Schlossplatz 6 65183 Wiesbaden Webpräsenz: www.wiesbaden.de Oberbürgermeister: Sven Gerich (SPD) Lage der Landeshauptstadt Wiesbaden in Hessen und im Regierungsbezirk Darmstadt Offizielles Logo der Landeshauptstadt Wiesbaden Flagge der Landeshauptstadt Wiesbaden Wiesbaden is
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Marschland

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Kilometerweite Marsch in den Vier- und Marschlanden am Elbdeich Die Wedeler Marschlandschaft in Schleswig-Holstein Der Hadelner Kanal zur Entwässerung der Marsch Die tief gelegene Marsch und ein Entwässerungskanal Das Schöpfwerk Otterndorf im Land Hadeln zur Entwässerung der Medem verfügte damals bereits über die größte Kreiselpumpe Europas. Das Schöpfwerk in Neuhaus für die Aue und des Neuhaus-Bülkauer Kanals Als Marsch(land) (v. niederdt., altsächs. mersc ) – auch Masch , Mersch oder Schwemmland genannt – bezeichnet man eine nacheiszeitlich entstandene geomorphologische Landform im Gebiet der nordwestdeutschen Küsten und Flüsse sowie vergleichbare Landformen weltweit. Inhaltsverzeichnis 1 Landschaft 2 Entstehung 3 Marschböden 4 Nutzung 5 Marschgebiete in Deutschland 6 Siehe auch 7 Literatur 8 Weblinks 9 Einzelnachweise Landschaft | Marschen sind generell flache Landstriche ohne natürliche Erhebungen. Si
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Dieringhausen

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Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7
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