Velocity/Displacement question

Question

An object moves in a straight line with an acceleration of $8~\text{m}/\text{s}^2$. If after $1$ second, it passes through point $O$ and after $3$ seconds it is $30$ metres from $O$, find its initial position relative to $O$.

What is your method of thinking through this question? I feel like this should be very simple, but I am not getting the correct answer (which is $3$). I assume that the initial velocity was $0$ and therefore, by doing two integrations, got displacement $= 4t^2$, let $t = 3$ and equated it to $30 + c$, where $c$ is the initial distance. My thinking was that after $3$ seconds, the distance will be the initial distance $+ 30$, and after the same time, the distance will equal $36$ ($4 \cdot 3^2$).

Answer 1

Let's not make any extra assumptions about the velocity. Let's denote the position as a function of time as $x(t)$. We know that $$ x''(t) = 8 \qquad \Rightarrow \qquad x(t) = 4t^2 + v_0 t + x_0 $$ We also know that, at $t=1$, it passes the point $O$, for which we set the coordinate $x_1$ $$ x(1) = x_1 $$ And then, after 3 seconds (I interpret this as $t=3$) it's $30$ m from $x_1$, so $$ x(3) = x_1 + 30 $$ Plugging in these results, we get $$ \left\{ \begin{array}{ccc} 4 + v_0 + x_0 &=& x_1 \\ 36+ 3v_0 + x_0 &=& x_1 +30 \end{array} \right. $$ Can you continue from here?

Answer 2

$$\begin{cases}\dfrac 821^2+v_01+x_0=0,\\\dfrac823^2+v_03+x_0=30\end{cases}$$ so that

$$(3-1)x_0=3\cdot(0-4)-(30-36).$$

Answer 3

Displacement is initial velocity times time, plus one half acceleration times time squared.

$S = ut + \frac{1}{2}at^2$

$30 = 3u + 4\cdot 3^2$

$30 = 3u + 36$

$3u = -6$

$u = -2$

If the velocity after $1$ second was $-2$ m/s, then the velocity at $t_0$ was $-10$ m/s

Displacement from $t_0$ to $t_1$ is the distance from O to the start

$S = ut + \frac{1}{2}at^2$

$S = -10\cdot 1 + 4\cdot 1^2$

$S = -10 + 4$

$S = -6$ m

So if right is positive displacement, then the starting position is $6$ m to the right of O.

BTW, the equation for the motion will be:

$x = \frac{1}{2}at^2 + v_0t + x_0$

$x = 4t^2 - 10t + 6$