8- below figure is graph acceleration versus time for a particle moving along an x axis. the acceleration…

Explanation:

Step1: Recall the relationship between acceleration and velocity

The change in velocity $\Delta v$ is given by the integral of acceleration $a$ with respect to time, $\Delta v=\int_{t_1}^{t_2}a\mathrm{d}t$. Geometrically, it is the area under the $a - t$ graph between $t_1$ and $t_2$. Here, $t_1=- 2.0\mathrm{s}$, $t_2 = 5.0\mathrm{s}$, and $v(t_1)=10.0\mathrm{m/s}$.

Step2: Divide the time - interval and calculate the area under the graph

The acceleration - time graph is a straight - line. We can divide the time interval from $t=-2.0\mathrm{s}$ to $t = 5.0\mathrm{s}$ into parts. The equation of the line is $a=a_0+bt$. From the graph, when $t = 0$, $a = 8\mathrm{m/s}^2$, and the slope $b=\frac{-8}{4}=-2\mathrm{m/s}^3$. So $a = 8-2t$. We calculate the area under the graph in two parts:

1. From $t=-2\mathrm{s}$ to $t = 4\mathrm{s}$: The area of the trapezoid from $t=-2\mathrm{s}$ to $t = 4\mathrm{s}$: The formula for the area of a trapezoid is $A_1=\frac{(a_1 + a_2)h}{2}$, where $a_1$ and $a_2$ are the parallel sides and $h$ is the height. Here, $a_1$ (at $t=-2\mathrm{s}$) is $12\mathrm{m/s}^2$, $a_2$ (at $t = 4\mathrm{s}$) is $0\mathrm{m/s}^2$, and $h=(4-( - 2))=6\mathrm{s}$. So $A_1=\frac{(12 + 0)\times6}{2}=36\mathrm{m/s}$.
2. From $t = 4\mathrm{s}$ to $t = 5\mathrm{s}$: The area of the triangle from $t = 4\mathrm{s}$ to $t = 5\mathrm{s}$: The formula for the area of a triangle is $A_2=\frac{1}{2}bh$, where $b$ is the base and $h$ is the height. Here, $b = 1\mathrm{s}$ and $h=-2\mathrm{m/s}^2$. So $A_2=\frac{1}{2}\times1\times(-2)=-1\mathrm{m/s}$. The total area $A=A_1+A_2=36-1 = 35\mathrm{m/s}$.

Step3: Calculate the final velocity

We know that $v(t_2)=v(t_1)+\int_{t_1}^{t_2}a\mathrm{d}t$. Substituting $v(t_1)=10\mathrm{m/s}$ and $\int_{t_1}^{t_2}a\mathrm{d}t = 35\mathrm{m/s}$, we get $v(5.0\mathrm{s})=10+35=45\mathrm{m/s}$.

Answer:

$45\mathrm{m/s}$