Finding distance when the force is a function of time I'm having trouble with this homework question

A mysterious rocket-propelled object of mass 49.0 kg is initially at rest in the middle of the horizontal, frictionless surface of an ice-covered lake. Then a force directed east and with magnitude $F(t) = (16.3\text{ N/s})t$ is applied.
How far does the object travel in the first 5.50s after the force is applied?

For some reason I'm not getting the correct answer. I think maybe I'm not understanding how to use the magnitude of the force they are giving me. I know how to use a constant force, but is this different because the force is a function of time?
I tried starting it like this:
$$\begin{align}F &= ma\\
16.3(t) &= (49)(a) \\
16.3(5.5) &= (49)(a) \\
89.65 &= (49)(a) \\
a &= 1.82959\ \mathrm{m/s^2}\end{align}$$
So now we know that: 
$$\begin{align}t &= 5.5\text{ s} \\
a &= 1.82959 m/s^2 \\
V_o &= 0\end{align}$$
So I plug it into my equation:
$$\begin{align}\Delta X &= V_o t + 1/2 a t^2\\
\Delta X &= (0)(5.5) + (1/2)(1.82959)(5.5)^2\\
\Delta X &= 27.7\text{ m}\end{align}$$
But that's not the right answer.
 A: Your mistake here is that you're using the three equations of motion ($v=u+at$ et al), even though they are only applicable for constant acceleration.
The correct way is this (i'm writing 16.3 as $k$) :$$F=ma=kt$$
$$\therefore a=kt/m$$
$$\therefore \frac{dv}{dt}=kt/m$$
$$\therefore dv=\frac{ktdt}{m}$$
$$\therefore \int\limits_0^vdv=\int\limits_0^t\frac{ktdt}{m}$$
$$\therefore v=\frac{kt^2}{2m}$$
$$\therefore \frac{dx}{dt}=\frac{kt^2}{2m}$$
$$\therefore dx=\frac{kt^2dt}{2m}$$
$$\therefore \int\limits_0^xdx=\int\limits_0^t\frac{kt^2dt}{2m}$$
$$\therefore x=\frac{kt^3}{6m}$$
I had to use a bit of calculus here. Any problem involving rate of change requires calculus. Those three equations of motion are derived from "acceleration is rate of change of velocity" and "velocity is rate of change of displacement" at constant acceleration. If $a$ is not constant, you have to use these equations: $$v=\frac{dx}{dt}$$,$$a=\frac{dv}{dt}=\frac{d^2x}{dt^2}$$. A shortcut formula for when acceleration is given in terms of displacement is $$a=v\frac{dv}{dx}$$.
A: If the force is constant with time, then the distance is a polynomial with time of order 2. If force varies linearly with time, then the distance is a polynomial of order 3. 
$$ x(t) = C_0 + C_1\, t + C_2\, t^2 + C_3\, t^3 $$
So there are four (4) unknown coefficients to the expression for distance. Two of them are given from the initial conditions ($x=0$ and $v=0$ at $t=0$), leaving only the coefficients for $t^2$ and $t^3$ to be determined.
The other two are found from the equations of motion
$$ F(t) = m \frac{{\rm d}^2 x(t)}{{\rm d}t^2} $$
given that $F(t)=k\,t$ and $x(t)=\ldots+C_2\,t^2+C_3\,t^3$
I hope you have had calculus and knows how to do a 2nd order derivative of a polynomial. This will give you the coefficients $C_2$ and $C_3$ since the above needs to be solved for ALL values of time $t$.
In the end plug $k=16.3$ and $t=5.5$ and whatever values the initial conditions give you and you will get your answer.
