Find the velocity $v$ of the falling rock from Example 1 at:

Solution

(a) Use the definition of instantaneous velocity with $f( t) =16t^{2}$ and $t_{0}=1$. $$\begin{eqnarray*} v&=&\lim\limits_{\Delta t\rightarrow 0}\dfrac{\Delta s}{\Delta t} =\lim\limits_{t\rightarrow 1}\dfrac{f( t) -f(1) }{t-1} =\lim\limits_{t\rightarrow 1}\dfrac{16t^{2}-16}{t-1}=\lim\limits_{t \rightarrow 1}\dfrac{16( t^{2}-1) }{t-1}\\[6pt] &=&\lim\limits_{t \rightarrow 1}\dfrac{16( t-1) ( t+1) }{t-1} =\lim\limits_{t\rightarrow 1}[ 16( t+1) ] =32 \end{eqnarray*}$$

After 1 second, the velocity of the rock is $32$ ft/s.

(b) For $t_{0}=7.4$ s, $$\begin{eqnarray*} v &=&\lim\limits_{\Delta t\rightarrow 0}\dfrac{\Delta s}{\Delta t} =\lim\limits_{t\rightarrow 7.4}\dfrac{f( t) -f ( 7.4 ) }{ t-7.4}=\lim\limits_{t\rightarrow 7.4}\dfrac{16t^{2}-16\cdot ( 7.4 ) ^{2}}{t-7.4}\\[8pt] &=&\lim\limits_{t\rightarrow 7.4}\dfrac{16 [ t^{2}- ( 7.4 ) ^{2} ] }{t-7.4} =\lim\limits_{t\rightarrow 7.4}\dfrac{16 ( t-7.4 ) ( t+7.4 ) }{t-7.4}\\[8pt] &=&\lim\limits_{t\rightarrow 7.4} [ 16 ( t+7.4 ) ] =16 ( 14.8 ) =236.8 \end{eqnarray*}$$

After 7.4 seconds, the velocity of the rock is 236.8 ft/s.

(c) $$\begin{eqnarray*} v & = & \lim\limits_{t\rightarrow t_{0}}\dfrac{f( t) -f\left( t_{0}\right) }{t-t_{0}}=\lim\limits_{t\rightarrow t_{0}}\dfrac{ 16t^{2}-16t_{0}{}^{2}}{t-t_{0}}=\lim\limits_{t\rightarrow t_{0}}\dfrac{ 16\left( t-t_{0}{}\right) \left( t+t_{0}{}\right) }{t-t_{0}}\\[6pt] & = & 16\lim\limits_{t\rightarrow t_{0}}\left( t+t_{0}{}\right) =32 t_{0} \end{eqnarray*}$$

At $t_{0}$ seconds, the velocity of the rock is $32t_0$ ft/s.