Answer
a. $680\ ft$
b. See graph.
Work Step by Step
a. Step 1. Given the velocity function $v(t)$ as shown in the graph of the exercise, we have $\frac{ds}{dt}=v(t)$ and $s(8)=\int_0^8 v(t)dt$ which means that the total distance (height) is the area under the velocity curve in $0\leq t\leq 8$.
Step 2. From $t=2$ to $t=8$ the velocity can be model as a line passing $(2,190)$ and $(8,0)$, which gives $v=-\frac{95}{3}(t-8)$. The area under this part is
$A_2=\int_2^8(-\frac{95}{3}(t-8))dt=-\frac{95}{6}t^2|_2^8+\frac{760}{3}t|_2^8=570\ ft$
Step 3. From $t=0$ to $t=2$, we need to count the number of small squares under the curve. Each small square represents
$\Delta s=0.4(sec)\times 10(ft/sec)=4\ ft$.
We can use approximations (left-end, right-end, midpoint, trapezoid) for each interval and take the sum. Or just count the number of small squares layer by layer as
$4+3.5+2.9+2.7+2.4+1.9+1.8+1.6+1.4+1.2+0.9+0.8+0.7+0.6+0.4+0.3+0.2+0.1=27.5$
which gives the area (height) for this part as
$A_1=27.5\times4=110\ ft$
Step 4. The total height is then $H=A_1+A_2=680\ ft$
b. From $t=0$ to $t=2\ sec$, the velocity curve resembles a parabola. Its integral (which represents the height) can be modeled as a cubic function, concave up. From $t=2$ to $t=8\ sec$, the velocity is a line and its integral (which represents the height) can be modeled as a parabola concave down. We can sketch the $s(t)$ function as shown in the figure.
