Question

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 1 ≤ t ≤ 4, with s in meters and t in seconds.
a. find the bodys displacement and average velocity for the given time interval.
b. find the bodys speed and acceleration at the endpoints of the interval.
c. when, if ever, during the interval does the body change direction?
(type an integer or a simplified fraction.)
the bodys average velocity for the given time interval is - 4 m/s.
(type an integer or a simplified fraction.)
the bodys speeds at the left and right endpoints of the interval are 28 m/s and \frac{1}{4} m/s, respectively.
(type integers or simplified fractions.)
the bodys accelerations at the left and right endpoints of the interval are 88 m/s² and \frac{1}{4} m/s², respectively.
(type integers or simplified fractions.)
when, if ever, during the interval does the body change direction? select the correct choice below and fill in any answer boxes within your choice.
a. the body changes direction at t = s.
(type an integer or a simplified fraction.)
b. the body does not change direction during the interval

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 $8
otin[1,4]$. 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