Variable acceleration, time, distance

[Markisen]

Homework Statement

A particle moves along a straight line. At t=0 the acceleration is k(t+2), initial velocity = 0 and x=5m.
a) Write down the position of the particle as a function of time.
b) At what point in time does the particle reach the velocity 16m/s, and where is the particle at this time?

Homework Equations

$$a=k(t+2)=kt+2k$$
$$v=\int a\textit{dt}=(kt^2)/2+2kt+C_1$$
$$x=\int v\textit{dt}=(kt^3)/6+kt^2+C_2$$

The Attempt at a Solution

I think I got part a) With the above equations that I calculated and with the initial values inserted I get: $$x=(kt^3)/6+kt^2+5$$

for part b I'm more unsure that I'm correct. Entering the velocity as 16m/s I get the equation: $$16=(kt^2)/2+2kt$$ which when calculated gets the time. $$t=-2+4\sqrt{1+{64/k}^2}$$ and then this time is inserted into $$x=(kt^3)/6+kt^2+5$$ to calculate the distance. This gives me $$1/6\,k \left( -2+4<br /> \,\sqrt {1+64\,{k}^{-2}} \right) ^{3}+k \left( -2+4\,\sqrt {1+64\,{k}^{-2}} \right) ^{2}+5$$ Seems a bit to complicated to be right? Or is it right and I'm just confused

Thanks for the help!

 

[CompuChip]

Note that
$$x=\int v\textit{dt}=(kt^3)/6+kt^2+C_1 t + C_2,$$
you forgot to integrate the C1 term.

Finally, you can probably simplify your answer a bit but your way of calculating it looks correct. I am afraid that you are just confused and it is just an ugly answer