Pretty sure I have every part of this problem solved but my final answer concerns me as it seems a little high.
$\displaystyle a(t)=(3-t)42-9.8m/s^2$ for 0<=t<=3
and $\displaystyle a(t)=-9.8m/s^2$ for t>3
I should find the maximum altitude and time in which it gets there. My answer is 56s and 18565.18meters.
**additional info**
initial velocity ($\displaystyle v_i$ is 366.4m/s)
initial position ($\displaystyle x_i$) is 0
$\displaystyle \int (3-t)42-9.8 dx=v(t_i)=126t-42t-9.8t+v_i$
after taking integral of v(t) $\displaystyle x(t_i)=-7t^3+63t^2+366.4t+x_i$
**
$\displaystyle x(t_i)$= position fxn until 3 seconds.
$\displaystyle x(t_i)=-7t^3+63t^2+366.4t=1477.2m$ after 3 seconds of burn
$\displaystyle x(t_m)$=position fxn after 3s initial burn.
$\displaystyle x(t_m)=\frac{9.88t^2}{2}+555.4t+1477.2$
$\displaystyle x(t_m)=\frac{9.88t^2}{2}+555.4t+1477.2=0$ at t=56.2145s
$\displaystyle x'(t_m)$where t=56.2145s
$\displaystyle x'(56.2145)=17087.98m$
17087.98+1477.2=18565.18meters
Is it me or is that pretty dang high?