a body moves on a coordinate line such that it has a position (s = f(t)=\frac{16}{t^{2}}-\frac{4}{t}) on the…

a body moves on a coordinate line such that it has a position (s = f(t)=\frac{16}{t^{2}}-\frac{4}{t}) on the interval (1leq tleq4), with (s) in meters and (t) in seconds.\na. find the bodys displacement and average velocity for the given time interval.\nb. find the bodys speed and acceleration at the endpoints of the interval.\nc. when, if ever, during the interval does the body change direction?\nthe bodys displacement for the given time interval is - 12 m.\n(type an integer or a simplified fraction.)\nthe bodys average velocity for the given time interval is (square) m/s.\n(type an integer or a simplified fraction.)

a body moves on a coordinate line such that it has a position (s = f(t)=\frac{16}{t^{2}}-\frac{4}{t}) on the interval (1leq tleq4), with (s) in meters and (t) in seconds.\na. find the bodys displacement and average velocity for the given time interval.\nb. find the bodys speed and acceleration at the endpoints of the interval.\nc. when, if ever, during the interval does the body change direction?\nthe bodys displacement for the given time interval is - 12 m.\n(type an integer or a simplified fraction.)\nthe bodys average velocity for the given time interval is (square) m/s.\n(type an integer or a simplified fraction.)

Answer

Explanation:

Step1: Recall displacement formula

Displacement $\Delta s=s(4)-s(1)$. First, find $s(4)$ and $s(1)$. For $s(t)=\frac{16}{t^{2}}-\frac{4}{t}$, when $t = 4$, $s(4)=\frac{16}{4^{2}}-\frac{4}{4}=1 - 1=0$. When $t = 1$, $s(1)=\frac{16}{1^{2}}-\frac{4}{1}=16 - 4 = 12$. So, $\Delta s=0 - 12=- 12$ m.

Step2: Recall average - velocity formula

Average velocity $v_{avg}=\frac{\Delta s}{\Delta t}$. Here, $\Delta t=4 - 1 = 3$ s and $\Delta s=-12$ m. So, $v_{avg}=\frac{-12}{3}=-4$ m/s.

Answer:

-4