Question

the position s of a point (in feet) is given as a function of time t (in seconds). s = \frac{2}{t}-\frac{2}{t^{2}}; t = 4. find the acceleration of the point at the given t value. leave your answer in fraction form.

Explanation:

Step1: Rewrite the position - function

Rewrite $s(t)=\frac{2}{t}-\frac{2}{t^{2}} = 2t^{-1}-2t^{-2}$.

Step2: Find the first - derivative (velocity function)

Using the power rule $\frac{d}{dt}(t^{n})=nt^{n - 1}$, we have $v(t)=s^\prime(t)=2(-1)t^{-2}-2(-2)t^{-3}=-2t^{-2}+4t^{-3}$.

Step3: Find the second - derivative (acceleration function)

Differentiate $v(t)$ with respect to $t$ again. $a(t)=v^\prime(t)=-2(-2)t^{-3}+4(-3)t^{-4}=4t^{-3}-12t^{-4}$.

Step4: Evaluate the acceleration at $t = 4$

Substitute $t = 4$ into $a(t)$. $a(4)=4\times4^{-3}-12\times4^{-4}=\frac{4}{4^{3}}-\frac{12}{4^{4}}=\frac{4}{64}-\frac{12}{256}=\frac{16 - 12}{256}=\frac{4}{256}=\frac{1}{64}$.

Answer:

$\frac{1}{64}$