My solution attempt is at the bottom. I believe that there is no solution from the given choices.
This is a one-dimensional problem. No external forces are applied (gravity is already included in the given acceleration graph). All friction forces are negligible. In essence, it is purely an integration problem.
A rocket is being launched vertically from a platform $6.4$ meters high at $t = 0$, where $v_0 = 2.1 m/s$ towards the heavens. Below is a graph of its vertical acceleration as a function of time:
What will be the height of the rocket (i.e. its distance from the ground) when $6 < t < 8$? Choose from:
The chart above shows us the rate of change of the velocity per unit of time, so it can be used to deduce the graph of the velocity as a function of time by summing the area under it in every uniform segment:
From here, I calculated the area under the graph by summing the areas of the two leftmost trapezoids and the one rectangle. This gives us the total distance the rocket travelled up until $t = 6$. We then add $6.4m$ to that distance to get the rocket's distance from the ground:
$$x(6) = 6.4 + \frac{2(2.1 + 22.1)}{2} + \frac{2(22.1 + 30.1)}{2} + 30.1 \cdot 2 = 143 $$
Now, I have the following information for $t \geq 6$: $$\begin{cases} v(6) = 30.1m/s\\ a = -4m/s^2\\ h(6) = 143m \end{cases}$$ Plugging the values above in the kinematics formula of $h(t) = h_0 + v_0\cdot t + \frac{at^2}{2}$ yields:
$$\boxed{h(t) = 143 + 30.1t -2t^2}$$
However, this doesn't fit any of the answers. I am hoping that I missed something, and that I am not wasting everyone's time with a flawed problem.
You were doing fine until you plugged the computed values of altitude, velocity and acceleration at $t=6$ into the kinematic equation. You didn’t take into account that the segment starts at $t=6$ instead of $t=0$, so the correct equation is $h_0+v_0(t-6)+\frac12a(t-6)^2$. Substituting your values and simplifying produces $-2t^2+54.1t-109.6$, which is the fourth option.