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?
the bodys displacement for the given time interval is - 12 m
(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 m/s and m/s, respectively
(type integers or simplified fractions.)

Explanation:

Step1: Find the velocity function

The velocity $v(t)$ is the derivative of the position function $s(t)=\frac{16}{t^{2}}-\frac{4}{t}=16t^{- 2}-4t^{-1}$. Using the power - rule $\frac{d}{dt}(t^{n})=nt^{n - 1}$, we have $v(t)=s^\prime(t)=-32t^{-3}+4t^{-2}=-\frac{32}{t^{3}}+\frac{4}{t^{2}}$.

Step2: Evaluate velocity at endpoints

For $t = 1$:
$v(1)=-\frac{32}{1^{3}}+\frac{4}{1^{2}}=-32 + 4=-28$. The speed at $t = 1$ is $|v(1)| = 28$ m/s.
For $t = 4$:
$v(4)=-\frac{32}{4^{3}}+\frac{4}{4^{2}}=-\frac{32}{64}+\frac{4}{16}=-\frac{1}{2}+\frac{1}{4}=-\frac{1}{4}$. The speed at $t = 4$ is $|v(4)|=\frac{1}{4}$ m/s.

Answer:

28, $\frac{1}{4}$