Question

1. if $s(t)$ denotes the position of an object in feet at time $t$ seconds, $s(t)$ is the acceleration of the object in ft/s² at time $t$ seconds. if $s(t)=t^{4/3}-7t + 4$ ft, find the acceleration at time $t = 8$ s.

Explanation:

Step1: Find the first - derivative

The power rule for differentiation is $\frac{d}{dt}(t^n)=nt^{n - 1}$.
For $s(t)=t^{\frac{4}{3}}-7t + 4$, we have $s'(t)=\frac{d}{dt}(t^{\frac{4}{3}})-\frac{d}{dt}(7t)+\frac{d}{dt}(4)$.
$s'(t)=\frac{4}{3}t^{\frac{4}{3}-1}-7+0=\frac{4}{3}t^{\frac{1}{3}}-7$.

Step2: Find the second - derivative

Differentiate $s'(t)$ with respect to $t$.
$s''(t)=\frac{d}{dt}(\frac{4}{3}t^{\frac{1}{3}})-\frac{d}{dt}(7)$.
Using the power rule again, $s''(t)=\frac{4}{3}\times\frac{1}{3}t^{\frac{1}{3}-1}-0=\frac{4}{9}t^{-\frac{2}{3}}$.

Step3: Evaluate at $t = 8$

Substitute $t = 8$ into $s''(t)$.
$s''(8)=\frac{4}{9}\times8^{-\frac{2}{3}}$.
Since $8^{-\frac{2}{3}}=(8^{\frac{1}{3}})^{-2}=2^{-2}=\frac{1}{4}$, then $s''(8)=\frac{4}{9}\times\frac{1}{4}=\frac{1}{9}$ ft/s².

Answer:

$\frac{1}{9}$ ft/s²