Find distance as function of time.
$begingroup$
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
asked Mar 18 '16 at 13:56
NoamNoam
385113
$endgroup$
add a comment |
$begingroup$
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
asked Mar 18 '16 at 13:56
NoamNoam
385113
$endgroup$
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
add a comment |
$begingroup$
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
asked Mar 18 '16 at 13:56
NoamNoam
385113
$endgroup$
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
asked Mar 18 '16 at 13:56
NoamNoam
385113
asked Mar 18 '16 at 13:56
NoamNoam
385113
edited Mar 18 '16 at 14:16
Noam
asked Mar 18 '16 at 13:56
NoamNoam
385113
asked Mar 18 '16 at 13:56
NoamNoam
385113
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
add a comment |
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
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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 $x(t)=frac 32 t^2$
answered Mar 18 '16 at 14:04
Win VineethWin Vineeth
3,270527
$endgroup$
$begingroup$
$a = frac{2cdot m}{3cdot s^2}$
$endgroup$
– 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
add a comment |
$begingroup$
$a(t) = v'(t) = x''(t)$; 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?
answered Mar 18 '16 at 14:01
DylanSpDylanSp
1,589714
$endgroup$
$begingroup$
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
|
show 2 more comments
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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votes
active
oldest
votes
$begingroup$
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 $x(t)=frac 32 t^2$
answered Mar 18 '16 at 14:04
Win VineethWin Vineeth
3,270527
$endgroup$
$begingroup$
$a = frac{2cdot m}{3cdot s^2}$
$endgroup$
– 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
add a comment |
$begingroup$
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 $x(t)=frac 32 t^2$
answered Mar 18 '16 at 14:04
Win VineethWin Vineeth
3,270527
$endgroup$
$begingroup$
$a = frac{2cdot m}{3cdot s^2}$
$endgroup$
– 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
add a comment |
$begingroup$
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 $x(t)=frac 32 t^2$
answered Mar 18 '16 at 14:04
Win VineethWin Vineeth
3,270527
$endgroup$
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 $x(t)=frac 32 t^2$
answered Mar 18 '16 at 14:04
Win VineethWin Vineeth
3,270527
answered Mar 18 '16 at 14:04
Win VineethWin Vineeth
3,270527
answered Mar 18 '16 at 14:04
Win VineethWin Vineeth
3,270527
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
add a comment |
$begingroup$
$a = frac{2cdot m}{3cdot s^2}$
$endgroup$
– 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
add a comment |
$begingroup$
$a(t) = v'(t) = x''(t)$; 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?
answered Mar 18 '16 at 14:01
DylanSpDylanSp
1,589714
$endgroup$
$begingroup$
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
|
show 2 more comments
$begingroup$
$a(t) = v'(t) = x''(t)$; 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?
answered Mar 18 '16 at 14:01
DylanSpDylanSp
1,589714
$endgroup$
$begingroup$
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
|
show 2 more comments
$begingroup$
$a(t) = v'(t) = x''(t)$; 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?
answered Mar 18 '16 at 14:01
DylanSpDylanSp
1,589714
$endgroup$
$a(t) = v'(t) = x''(t)$; 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?
answered Mar 18 '16 at 14:01
DylanSpDylanSp
1,589714
edited Mar 18 '16 at 16:17
answered Mar 18 '16 at 14:01
DylanSpDylanSp
1,589714
answered Mar 18 '16 at 14:01
DylanSpDylanSp
1,589714
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
|
show 2 more comments
$begingroup$
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
|
show 2 more comments
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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