given the velocity and initial position of a body moving along a coordinate line at time t, find the bodys…

Explanation:

Step1: Recall the relationship between velocity and position

The velocity $v(t)$ is the derivative of the position function $s(t)$, i.e., $v(t)=s^\prime(t)$. To find $s(t)$ from $v(t)$, we integrate $v(t)$ with respect to $t$. Given $v(t)=- 13t + 6$, we use the power - rule of integration $\int x^n dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$). [s(t)=\int(-13t + 6)dt=-13\int tdt+6\int dt] [s(t)=-13\times\frac{t^{2}}{2}+6t + C=-\frac{13}{2}t^{2}+6t + C]

Step2: Use the initial condition to find the constant $C$

We know that $s(0) = 11$. Substitute $t = 0$ into $s(t)=-\frac{13}{2}t^{2}+6t + C$. [s(0)=-\frac{13}{2}(0)^{2}+6(0)+C] Since $s(0) = 11$, then $C = 11$. So the position function is $s(t)=-\frac{13}{2}t^{2}+6t + 11$.

Answer:

D. $s =-\frac{13}{2}t^{2}+6t + 11$