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

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 ?

NM2
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    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. – almagest Mar 18 '16 at 14:09
  • I completely agree. This doesn't make sense. By the way, units of $a$ seems wrong. – Win Vineeth Mar 18 '16 at 14:10

2 Answers2

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$a(t) = v'(t) = x''(t)$; 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?

DylanSp
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  • 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. – NM2 Mar 18 '16 at 14:07
  • 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 $C_2$. – DylanSp Mar 18 '16 at 14:07
  • What do you mean ? – NM2 Mar 18 '16 at 14:08
  • Sorry, accidentally submitted before I was finished. See my edited version. – DylanSp Mar 18 '16 at 14:11
  • $a(t) = \frac{2}{3}t$. $v(t) = \int a(t) = \frac{1}{3}t^2 + v_{0}$. $x(t) = \int v(t) = \frac{1}{9} + v_{0}t+x_{0}$. $27 = \frac{1}{9}t^{3}+x_{0}$. the final eqation is - $x(t) = \frac{1}{9}t^3+24$. What do you think ? – NM2 Mar 18 '16 at 14:19
  • Ah, if the acceleration is nonconstant, that explains everyone's confusion. Editing my answer. – DylanSp Mar 18 '16 at 15:01
  • @Noam Yes, that looks correct. – DylanSp Mar 18 '16 at 16:18
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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 $x(t)=\frac 32 t^2$

Win Vineeth
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