# Math Help - 3 second burn rocket problem

1. ## 3 second burn rocket problem

Pretty sure I have every part of this problem solved but my final answer concerns me as it seems a little high.

$a(t)=(3-t)42-9.8m/s^2$ for 0<=t<=3
and $a(t)=-9.8m/s^2$ for t>3

I should find the maximum altitude and time in which it gets there. My answer is 56s and 18565.18meters.

initial velocity ( $v_i$ is 366.4m/s)
initial position ( $x_i$) is 0
$\int (3-t)42-9.8 dx=v(t_i)=126t-42t-9.8t+v_i$
after taking integral of v(t) $x(t_i)=-7t^3+63t^2+366.4t+x_i$
**
$x(t_i)$= position fxn until 3 seconds.
$x(t_i)=-7t^3+63t^2+366.4t=1477.2m$ after 3 seconds of burn

$x(t_m)$=position fxn after 3s initial burn.
$x(t_m)=\frac{9.88t^2}{2}+555.4t+1477.2$
$x(t_m)=\frac{9.88t^2}{2}+555.4t+1477.2=0$ at t=56.2145s

$x'(t_m)$where t=56.2145s
$x'(56.2145)=17087.98m$

17087.98+1477.2=18565.18meters

Is it me or is that pretty dang high?

2. ## Re: 3 second burn rocket problem

I guess I don't need to add the two x positions since the second one already has it in there. So 17087.98meters.

3. ## Re: 3 second burn rocket problem

$a(t)=(3-t)42-9.8m/s^2$ for 0<=t<=3
and $a(t)=-9.8m/s^2$ for t>3
So,
initial velocity ( $v_i$ is 366.4m/s)
initial position ( $x_i$) is 0
$\int (3-t)42-9.8 dx=v(t_i)=-21t^2=116.2t+v_i$
after taking integral of v(t) $x(t_i)=-7t^3+58.1t^2+366.4t+x_i$

Then, at 3 seconds:
$x(3)=-7(3)^3+58.1(3)^2+366.4(3)=1433.1=x_m$
and
$v(3)=-21(3)^2=116.2(3)+366.4=526=v_m$

so after 3 seconds
$a(t)=-9.8m/s^2$
$\int a(t)=-9.8dx$
$=-9.8t+v_m=-9.8t+526$
$=v(t)=-9.8t+526$
$\int v(t)=-9.8t+526dx = x(t)=\frac{9.88t^2}{2}+526t+1433.1$

Finally:
$=v(t)=-9.8t+526=0?$ at t=53.23seconds
then
$x(53.23)=\frac{9.88t^2}{2}+526t+1433.1=15479.02$

So, it reaches a height of 15479.02m after flying for 53.23s. Still seems high to me...

4. ## Re: 3 second burn rocket problem

For some reason I switched from 9.8 to 9.88 as grav. constant. I adjusted them on my paper but don't have the time to do it again here. The values are very nearly the same anyway.