Single Variable Calculus, Chapter 3, 3.1, Section 3.1, Problem 15

Suppose that a particle moving in a straight line has an equation of motion $\displaystyle s = \frac{1}{t^2}$
, where $s$ and $t$ are measured in meters and seconds respectively. Find the velocity of the particle at time $t=a$, $t=1$, $t=2$ and $t = 3$.
From the definition of instantaneous velocity,

$
\displaystyle
\nu(a)= \lim\limits_{h \to 0} \frac{f(a+h)-f(a)}{h}\\
$


$
\begin{equation}
\begin{aligned}
s = f(t)
& = \frac{1}{t^2}\\
\nu(a)
& = \lim \limits_{h \to 0} \frac{\frac{1}{(a+h)^2}-\left(\frac{1}{a^2}\right)}{h}\\
\nu(a)
& = \lim \limits_{h \to 0} \frac{a^2 - (a + h)^2}{a^2 ( a + h )^2 h}\\
\nu(a)
& = \lim \limits_{h \to 0} \frac{\cancel{a^2} - \cancel{a^2}-2ah-h^2}{h(a^2)(a^2+2ah+h^2)}\\
\nu(a)
& = \lim \limits_{h \to 0} \frac{\cancel{h}(-2a-h)}{\cancel{h}(a^2)(a^2+2ah+h^2)}\\
\nu(a)
& = \lim \limits_{h \to 0} \frac{-2a-h}{(a^2)(a^2+2ah+h^2)}\\
\nu(a)
& = \frac{-2a-0}{(a^2)[a^2+2a(0)+(0)^2]}\\
\nu(a)
& = \frac{-2a}{(a^2)(a^2)} = \frac{-2\cancel{a}}{a^{\cancel{4}}} = \frac{-2}{a^3}\\
\nu(a)
& = -\frac{2}{a^3}
\end{aligned}
\end{equation}
$

The velocity of the particle at $t = a $ is $\displaystyle \nu(a) = \frac{-2}{a^3} \frac{m}{s}$
The velocity of the particle at $t = 1 $ is $\displaystyle \nu(1) = \frac{-2}{1^3} = -2 \frac{m}{s}$
The velocity of the particle at $t = 2 $ is $\displaystyle \nu(2) = \frac{-2}{2^3} = \frac{-1}{4} \frac{m}{s}$
The velocity of the particle at $t = 3 $ is $\displaystyle \nu(3) = \frac{-2}{3^3} = \frac{-2}{27}\frac{m}{s}$