## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 11 - Section 11.4 - Introduction to Derrivatives - Exercise Set - Page 1175: 43

#### Answer

a) The instantaneous velocity $1$ second after the explosion is $32\text{ feet per second}$ and the instantaneous velocity $3$ seconds after the explosion is $-32\text{ feet per second}$. b) The instantaneous velocity just before the debris hits the ground is $-64\text{ feet per second}$.

#### Work Step by Step

(a) Consider that the function $s\left( t \right)=-16{{t}^{2}}+64t$ describes the height of the debris above the ground in feet, t seconds after the explosion with an initial velocity of $64\text{ feet per second}$. Compute the derivative of $s\left( t \right)=-16{{t}^{2}}+64t$ 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 $s\left( t \right)=-16{{t}^{2}}+64t$. $s\left( t+h \right)=-16{{\left( t+h \right)}^{2}}+64\left( t+h \right)$ 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}}+64\left( t+h \right) \right)-\left( -16{{t}^{2}}+64t \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)+64\left( t+h \right) \right)-\left( -16{{t}^{2}}+64t \right)}{h} \\ & =\underset{h\to 0}{\mathop{\lim }}\,\frac{-16{{t}^{2}}-16{{h}^{2}}-32th+64t+64h+16{{t}^{2}}-64t}{h} \end{align} Combine the like terms in the numerator, then divide the numerator and denominator by h; this gives, $s'\left( t \right)=\underset{h\to 0}{\mathop{\lim }}\,\left( -32t-16h+64 \right)$ Apply the limits, $s'\left( t \right)=-32t+64$ Now, substitute $t=1$ in $s'\left( t \right)$ to compute the instantaneous velocity at $1$ second after the explosion. \begin{align} & s'\left( 1 \right)=-32\left( 1 \right)+64 \\ & =-32+64 \\ & =32 \end{align} Thus, the instantaneous velocity $1$ second after the explosion is $32\text{ feet per second}$. Now, substitute $t=3$ in $s'\left( t \right)$ to compute the instantaneous velocity at $3$ second after the explosion. \begin{align} & s'\left( 3 \right)=-32\left( 3 \right)+64 \\ & =-96+64 \\ & =-32 \end{align} Thus, the instantaneous velocity $3$ seconds after the explosion is $-32\text{ feet per second}$. (b) Consider that the function $s\left( t \right)=-16{{t}^{2}}+64t$ 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), $s'\left( t \right)=\left( -32t+64 \right)$ 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}}+64t=0 \\ & {{t}^{2}}-4t=0 \\ & t\left( t-4 \right)=0 \end{align} Now, put the factors $t,\left( t-4 \right)$ equal to zero and solve for t: $t=0$ Or \begin{align} & \left( t-4 \right)=0 \\ & t=4 \end{align} 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)=-32\left( 4 \right)+64 \\ & =-128+64 \\ & =-64 \end{align} Thus, the instantaneous velocity just before the debris hits the ground is $-64\text{ feet per second}$.

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