Kinematics (Calculus Form)

**Question: **

A particle travels along a straight line which passes through $\displaystyle A$ and $\displaystyle B$. During the motion the velocity of the particle is $\displaystyle 6t+t^2$ metres per second, where $\displaystyle t$ seconds is the time measured from a certain instant. The particle passes through $\displaystyle A$ when $\displaystyle t=2$ and through $\displaystyle B$ when $\displaystyle t=5$. Find, in terms of $\displaystyle t$, the acceleration of the particle and its distance from $\displaystyle A$ at any instant during the motion. Calculate, also, the distance of $\displaystyle B$ from $\displaystyle A$.

**My Answers:**

$\displaystyle a = \frac{\mathrm{d}v}{\mathrm{d}t} = \frac{\mathrm{d}(6t+t^2)}{\mathrm{d}t} = 6 + 2t$

$\displaystyle v = (6t+t^2) = \frac{\mathrm{d}x}{\mathrm{d}t}$

$\displaystyle \therefore x = 3t^2 + \frac13 t^3 + c$

$\displaystyle x_{\vec{AB}} = \left[3t^2 + \frac13 t^3\right]^5_2 = 102$

**My Problem:**

I don't know how to do this part (don't even understand what it is saying (Worried)): 'Find...its distance from $\displaystyle A$ at any instant during the motion'*. Can someone help? Thanks in advance.

*The answer is $\displaystyle x = 3t^2+ \frac13 t^3 - \frac{44}{3}$.