Solve the problem.
-A particle is moving on a horizontal line. Its distance from the origin at time $t$ seconds is $f(t)=\frac{18}{t+1}+\frac{13}{t^{2}+1}$ feet. What is the instantaneous velocity at 2 seconds?
A) $-\frac{14}{5} \mathrm{ft} / \mathrm{sec}$
B) $\frac{43}{5} \mathrm{ft} / \mathrm{sec}$
C) $-\frac{70}{9} \mathrm{ft} / \mathrm{sec}$
D) $-\frac{102}{25} \mathrm{ft} / \mathrm{sec}$

Question

Quizplus · Accepted Answer

To find the instantaneous velocity, we need to compute the derivative $f'(t)$ and evaluate it at $t=2$. The derivative $f'(t)$ is calculated using the quotient rule for each term. After finding $f'(t)$, substitute $t=2$ to get the instantaneous velocity, which results in $-\frac{102}{25} \mathrm{ft} / \mathrm{sec}$.

Solve the Problem ttt Seconds Is f(t)=18t+1+13t2+1f(t)=\frac{18}{t+1}+\frac{13}{t^{2}+1}f(t)=t+118​+t2+113​ Feet What Is the Instantaneous Velocity at 2 Seconds? A)

Solve the Problem $t$ Seconds Is $f(t)=\frac{18}{t+1}+\frac{13}{t^{2}+1}$ Feet What Is the Instantaneous Velocity at 2 Seconds?
A)