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

Explanation:

Step1: Recall position - velocity - acceleration relations

The velocity $v(t)$ is the derivative of the position function $s(t)=\frac{16}{t^{2}}-\frac{4}{t}=16t^{- 2}-4t^{-1}$, and the acceleration $a(t)$ is the derivative of the velocity function. $v(t)=s^\prime(t)=-32t^{-3}+4t^{-2}=-\frac{32}{t^{3}}+\frac{4}{t^{2}}$ $a(t)=v^\prime(t)=96t^{-4}-8t^{-3}=\frac{96}{t^{4}}-\frac{8}{t^{3}}$

Step2: Calculate displacement and average velocity

Displacement $\Delta s=s(4)-s(1)$. $s(4)=\frac{16}{4^{2}}-\frac{4}{4}=1 - 1=0$ $s(1)=\frac{16}{1^{2}}-\frac{4}{1}=16 - 4 = 12$ $\Delta s=0 - 12=-12$ m Average velocity $v_{avg}=\frac{s(4)-s(1)}{4 - 1}=\frac{-12}{3}=-4$ m/s

Step3: Calculate speed and acceleration at endpoints

For speed (magnitude of velocity): At $t = 1$: $v(1)=-\frac{32}{1^{3}}+\frac{4}{1^{2}}=-32 + 4=-28$, speed $|v(1)| = 28$ m/s $a(1)=\frac{96}{1^{4}}-\frac{8}{1^{3}}=96 - 8 = 88$ m/s² At $t = 4$: $v(4)=-\frac{32}{4^{3}}+\frac{4}{4^{2}}=-\frac{32}{64}+\frac{4}{16}=-\frac{1}{2}+\frac{1}{4}=-\frac{1}{4}$, speed $|v(4)|=\frac{1}{4}$ m/s $a(4)=\frac{96}{4^{4}}-\frac{8}{4^{3}}=\frac{96}{256}-\frac{8}{64}=\frac{3}{8}-\frac{1}{8}=\frac{1}{4}$ m/s²

Step4: Find when the body changes direction

The body changes direction when $v(t)=0$. $-\frac{32}{t^{3}}+\frac{4}{t^{2}} = 0$ Multiply through by $t^{3}$: $-32 + 4t=0$ $4t=32$, so $t = 8$. But $8\notin[1,4]$. So the body does not change direction during the interval $[1,4]$.

Answer:

a. Displacement: $- 12$ m; Average velocity: $-4$ m/s b. Speeds at endpoints: $28$ m/s, $\frac{1}{4}$ m/s; Accelerations at endpoints: $88$ m/s², $\frac{1}{4}$ m/s² c. B. The body does not change direction during the interval