Single Variable Calculus, Chapter 3, 3.1, Section 3.1, Problem 38

Suppose that a particle moves along a straight line with equation of motion
$s = f(t) = t^{-1} -t$ , where $s$ is measured in meters and $t$
in seconds. Determine the velocity and the speed when $t=5$.



Based from the definition of instantaneous velocity,



$
\quad
\begin{equation}
\begin{aligned}
\nu (a) &= \lim\limits_{h \to 0} \frac{f(a+h) - f(t)}{h}\\
f(t) &= t^{-1} - t = \frac{1}{t} -t
\end{aligned}
\end{equation}
$



$
\quad
\begin{equation}
\begin{aligned}
\nu(t) &= \lim\limits_{h \to 0 } \frac{\left[ \frac{1}{t+h} - (t+h) \right] - \left[ \frac{1}{t} - t \right]}{h}\\
\nu(t) &= \lim\limits_{h \to 0 } \frac{\frac{1-(t+h)^2}{t+h} - \frac{(1-t^2)}{t}}{h}\\
\nu(t) &= \lim\limits_{h \to 0 } \frac{t-t(t+h)^2 - (1-t^2)(t+h)}{t(t+h)h}\\
\nu(t) &= \lim\limits_{h \to 0 } \frac{ \cancel{t} - \cancel{t^3} - 2t^2h - th - \cancel{t} + \cancel{t^3} - h + t^2h}{t(t+h)h}\\
\nu(t) &= \lim\limits_{h \to 0 } \frac{-t^2h-th^2-h}{t(t+h)h}\\
\nu(t) &= \lim\limits_{h \to 0 } \frac{\cancel{h}(-t^2 - th - 1)}{t(t+h)\cancel{h}}\\
\nu(t) &= \lim\limits_{h \to 0 } \frac{-t^2-th-1}{t^2+th}\\
\nu(t) &= \frac{-t^2-t(0)-1}{t^2+t(0)}\\
\nu(t) &= \frac{-1-t^2}{t^2}\\
\nu(t) &= \frac{-1}{t^2} - \frac{t^2}{t^2}\\
\nu(t) &= \frac{-1}{t^2} - 1
\end{aligned}
\end{equation}
$



The velocity after $5s$ is $\displaystyle \nu(5) = \frac{-1}{5^2} - 1 = \frac{-26}{25} \frac{m}{s}$



A negative velocity means that the particle moves in opposite direction the way it should be.
Assuming that the function is moving to the east.



Therefore, the speed and velocity of the particle are $\displaystyle \frac{26}{25} \frac{m}{s}$ and
$\displaystyle \frac{26}{25} \frac{m}{s}$ west respectively.