Find distance as function of time.
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 ?
$\endgroup$ 22 Answers
$\begingroup$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 $x(t)=\frac 32 t^2$
$\endgroup$ 2 $\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?
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