## Precalculus (6th Edition) Blitzer

a) The instantaneous velocity of the debris at $\frac{1}{2}$ seconds after the explosion is $56\text{ feet per second}$ and the instantaneous velocity $4$ seconds after the explosion is $-56\text{ feet per second}$. b) The instantaneous velocity just before the debris hits the ground is $-72\text{ feet per second}$.
(a) Consider that the function $s\left( t \right)=-16{{t}^{2}}+72t$ describes the height of the debris above the ground in feet, t seconds after the explosion with an initial velocity of $72\text{ feet per second}$. Compute the derivative of $s\left( t \right)=-16{{t}^{2}}+72t$ using the formula $s'\left( t \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{s\left( t+h \right)-s\left( t \right)}{h}$ as follows: To compute $s\left( t+h \right)$, substitute $t=t+h$ in the function $s\left( t \right)=-16{{t}^{2}}+72t$. So, $s\left( t+h \right)=-16{{\left( t+h \right)}^{2}}+72\left( t+h \right)$ Put value of $s\left( t+h \right)$ in $s'\left( t \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{s\left( t+h \right)-s\left( t \right)}{h}$ , \begin{align} & s'\left( t \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{s\left( t+h \right)-s\left( t \right)}{h} \\ & =\underset{h\to 0}{\mathop{\lim }}\,\frac{\left( -16{{\left( t+h \right)}^{2}}+72\left( t+h \right) \right)-\left( -16{{t}^{2}}+72t \right)}{h} \end{align} Now, simplify ${{\left( t+h \right)}^{2}}$ by using the property ${{\left( A+B \right)}^{2}}={{A}^{2}}+2AB+{{B}^{2}}$ \begin{align} & s'\left( t \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{\left( -16\left( {{t}^{2}}+{{h}^{2}}+2th \right)+72\left( t+h \right) \right)-\left( -16{{t}^{2}}+72t \right)}{h} \\ & =\underset{h\to 0}{\mathop{\lim }}\,\frac{-16{{t}^{2}}-16{{h}^{2}}-32th+72t+72h+16{{t}^{2}}-72t}{h} \end{align} Combine the like terms in the numerator, then divide the numerator and denominator by h; this gives, $s'\left( t \right)=\underset{h\to 0}{\mathop{\lim }}\,\left( -32t-16h+72 \right)$ Apply the limits, $s'\left( t \right)=\left( -32t+72 \right)$ Now, substitute $t=\frac{1}{2}$ in $s'\left( t \right)$ to compute the instantaneous velocity at $\frac{1}{2}$ second after the explosion. \begin{align} & s'\left( \frac{1}{2} \right)=\left( -32\left( \frac{1}{2} \right)+72 \right) \\ & =-16+72 \\ & =56 \end{align} Thus, the instantaneous velocity $\frac{1}{2}$ second after the explosion is $56\text{ feet per second}$. Now, substitute $t=4$ in $s'\left( t \right)$ to compute the instantaneous velocity at $4$ second after the explosion. \begin{align} & s'\left( 4 \right)=\left( -32\left( 4 \right)+72 \right) \\ & =-128+72 \\ & =-56 \end{align} Thus, the instantaneous velocity $4$ seconds after the explosion is $-56\text{ feet per second}$. (b) Consider that the function $s\left( t \right)=-16{{t}^{2}}+72t$ describes the height of the debris above the ground in feet, t seconds after the explosion with an initial velocity of $64\text{ feet per second}$ From part (a), $s'\left( t \right)=\left( -32t+72 \right)$ The instantaneous velocity has to be calculated just before the debris hits the ground. So, set $s\left( t \right)=0$ \begin{align} & -16{{t}^{2}}+72t=0 \\ & 8t\left( 2t-9 \right)=0 \end{align} Now, put the factors $8t,\left( 2t-9 \right)$ equal to zero and solve for t: \begin{align} & 8t=0 \\ & t=0 \end{align} Or, \begin{align} & \left( 2t-9 \right)=0 \\ & 2t=9 \\ & t=\frac{9}{2} \end{align} Now, substitute $t=\frac{9}{2}$ in $s'\left( t \right)$ to compute the instantaneous velocity at $\frac{9}{2}$ second after the explosion. \begin{align} & s'\left( \frac{9}{2} \right)=-32\left( \frac{9}{2} \right)+72 \\ & =-144+72 \\ & =-72 \end{align} Thus, the instantaneous velocity just before the debris hits the ground is $-72\text{ feet per second}$.