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 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 m/s² and m/s², respectively. (type integers or simplified fractions.)

Explanation:

Step1: Recall position - velocity - acceleration relationships

Velocity $v(t)=s^\prime(t)$ and acceleration $a(t)=v^\prime(t)=s^{\prime\prime}(t)$. First, find the derivative of $s(t)=\frac{16}{t^{2}}-\frac{4}{t}=16t^{- 2}-4t^{-1}$.
Using the power - rule $\frac{d}{dt}(t^{n})=nt^{n - 1}$, we have $v(t)=s^\prime(t)=-32t^{-3}+4t^{-2}=-\frac{32}{t^{3}}+\frac{4}{t^{2}}$.
Then find the second - derivative $a(t)=v^\prime(t)=96t^{-4}-8t^{-3}=\frac{96}{t^{4}}-\frac{8}{t^{3}}$.

Step2: Calculate acceleration at endpoints

For $t = 1$:
Substitute $t = 1$ into the acceleration formula $a(t)=\frac{96}{t^{4}}-\frac{8}{t^{3}}$.
$a(1)=\frac{96}{1^{4}}-\frac{8}{1^{3}}=96 - 8=88$ m/s².
For $t = 4$:
Substitute $t = 4$ into the acceleration formula $a(t)=\frac{96}{t^{4}}-\frac{8}{t^{3}}$.
$a(4)=\frac{96}{4^{4}}-\frac{8}{4^{3}}=\frac{96}{256}-\frac{8}{64}=\frac{96}{256}-\frac{32}{256}=\frac{96 - 32}{256}=\frac{64}{256}=\frac{1}{4}$ m/s².

Step3: Determine when the body changes direction

The body changes direction when $v(t)=0$.
Set $v(t)=-\frac{32}{t^{3}}+\frac{4}{t^{2}} = 0$.
Multiply through by $t^{3}$ to get $-32 + 4t=0$.
Solve for $t$: $4t=32$, so $t = 8$. But $8
otin[1,4]$, so the body does not change direction in the interval $1\leq t\leq4$.

Answer:

The body's accelerations at the left and right endpoints of the interval are $88$ m/s² and $\frac{1}{4}$ m/s², respectively.
The body does not change direction during the interval $1\leq t\leq4$.