\hfill \hfill } %} \newtheorem{theorem}{Theorem} \newtheorem{acknowledgement}[theorem]{Acknowledgement} \newtheorem{algorithm}[theorem]{Algorithm} \newtheorem{axiom}[theorem]{Axiom} \newtheorem{case}[theorem]{Case} \newtheorem{claim}[theorem]{Claim} \newtheorem{conclusion}[theorem]{Conclusion} \newtheorem{condition}[theorem]{Condition} \newtheorem{conjecture}[theorem]{Conjecture} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{criterion}[theorem]{Criterion} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{exercise}{Exercise} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{notation}[theorem]{Notation} \newtheorem{problem}[theorem]{Problem} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{remark}[theorem]{Remark} \newtheorem{solution}[theorem]{Solution} \newtheorem{summary}[theorem]{Summary} \newenvironment{proof}[1][Proof]{\noindent\textbf{#1.} }{\ \rule{0.5em}{0.5em}} \input{tcilatex} \geometry{left=0.5in,right=0.5in,top=0.5in,bottom=0.4in} \pagestyle{fancy} \lhead{\color{blue} \Large Lecture notes} \chead{\color{black} \LARGE Average Velocity - Part 1} \rhead{\large page \ \thepage} \cfoot{} \lfoot{\small \copyright $\;$ Hidegkuti, 2015} \rfoot{\small Last revised: September 8, 2016} \textwidth 7.5in \textheight 9.6in \setlength{\headheight}{29pt} \setlength{\parindent}{0in} \begin{document} There is a significant difference between \textbf{speed} and \textbf{velocity% }. \ Speed refers to the 'fastness' of an object's motion without any concern about its direction. \ For example, if car A is moving to East and car B is moving South but both cars travel exactly $45$ miles in one hour, then cars A and B have the same speed, namely $45$ $\dfrac{\unit{mi}}{\unit{h% }}$ (miles per hour). \ These two cars however will have different velocities as the concept of velocity also includes the direction of movement. \ Two objects have the same velocity if they move with the same speed, in the same direction, along parallel lines or on the same line.\bigskip Speed and velocity usually denoted by $v$, distance traveled by $s$, and of course time by $t$.\bigskip To find the average velocity of an object, we need to divide the \textbf{% displacement} by the time. \ (The displacement is the directed distance between the starting point and the ending point). \ To find the average speed of an object, we need to divide the \textbf{distance traveled} by the time.\bigskip For instance, if an object traveled $10$ meters to the right and then $10$ meters to the left during $50$ seconds, then the displacement is $0$ meters and the distance traveled is $20$ meters.% \begin{equation*} v_{\text{av}}=\dfrac{0\unit{m}}{50\unit{s}}=0\dfrac{\unit{m}}{\unit{s}}\text{ \ \ \ and \ \ }sp_{\text{av}}=\dfrac{20\unit{m}}{50\unit{s}}=0.4\dfrac{\unit{% m}}{\unit{s}} \end{equation*} \bigskip \bigskip \begin{center} {\Large Sample Problems\bigskip } \end{center} \begin{enumerate} \item A car traveled North for $8$ hours. \ After $8$ hours, it has traveled a distance of $400$ miles. \ What is the average velocity and average speed of the car? \item Jesse is driving across the country. On Wednesday, he drove for $8$ hours and traveled $384$miles. On Thursday, he drove for $3$ hours and traveled $123$ miles. What was his average speed for the two days? Give both the exact value and an approximation (accurate up to three or more decimal places) of the answer. \item Erica is driving across the country. On Tuesday, she drove for $3$ hours with an average velocity of $59$ miles per hour. On Wednesday, she drove in the same direction, for $11$ hours with an average velocity of $51$ miles per hour. What was her average speed for the two days? Give both the exact value and an approximation (accurate up to three or more decimal places) of the answer. \item We drove for two days. On the first day, we drove for $4$ hours and had an average speed of $50$ miles per hour. On the second day, we drove for $6$ hours. Our average speed for the two days was $53$ miles per hour. What was our average speed on the second day? \item First we traveled towards North. \ We covered $200$ miles in four hours. \ Then we traveled toward West for five hours and covered $150$ miles. a) \ What was our average speed for the first part of the trip? b) \ What was our average speed for the second part of the trip? c) \ What was our average speed for the entire trip? d) \ What was our average velocity for the first part of the trip? e) \ What was average velocity for the second part of the trip? f) \ What was our average velocity for the entire trip? \item A bus travels between cities A and B. \ The distance between these cities is $60$ miles. It takes the bus $2$ hours to get from A to B. On its way back, the traveling time was only $1.5$ hours. Find the average speed of the bus for a) \ the trip from A to B b) \ the trip from B to A c) \ for the roundtrip. \item A bus travels between cities A and B. \ From A to B, the bus has an average speed of $v_{1}$. On its way back, the average speed is $v_{2.}$ \ Express the average speed of the bus in terms of $v_{1}$ \ and $v_{2}$. \item[8*.] (Enrichment) \ A long train is moving slowly, with a constant speed. \ We walk next to the train, with a constant speed, higher than that of the train. \ When we walk from the end of the train to the font of it, it takes $200$ steps. \ Then we turn around and walk from the front of the train to the end, and count $120$ steps. \ How many steps long is the train? \end{enumerate} \bigskip \begin{center} {\Large Sample Problems- Solutions\bigskip }\bigskip \end{center} \begin{enumerate} \item A car traveled North for $8$ hours. \ After $8$ hours, it has traveled a distance of $400$ miles. \ What is the average velocity and average speed of the car? Solution: \ Both average velocity and average speed is \begin{equation*} v_{\text{av}}=\dfrac{400\unit{mi}}{8\unit{h}}=50\dfrac{\unit{mi}}{\unit{h}} \end{equation*} \item Jesse is driving across the country. On Wednesday, he drove for $8$ hours and traveled $384$miles. On Thursday, he drove for $3$ hours and traveled $123$ miles. What was his average speed for the two days? Give both the exact value and an approximation (accurate up to three or more decimal places) of the answer. Solution: \ \begin{equation*} v_{\text{av}}=\dfrac{\text{distance traveled}}{\text{time}}=\dfrac{% s_{1}+s_{2}}{t_{1}+t_{2}}=\dfrac{384\unit{mi}+123\unit{mi}}{8\unit{h}+3\unit{% h}}=\dfrac{507\unit{mi}}{11\unit{h}}=\dfrac{507}{11}\dfrac{\unit{mi}}{\unit{h% }} \end{equation*}% Exact value: \ $\dfrac{507}{11}\dfrac{\unit{mi}}{\unit{h}}$ \ \ \ \ Approximate value: \ $46.\,\allowbreak 09091\dfrac{\unit{mi}}{\unit{h}}$ \item Erica is driving across the country. On Tuesday, she drove for $3$ hours with an average velocity of $59$ miles per hour. On Wednesday, she drove in the same direction, for $11$ hours with an average velocity of $51$ miles per hour. What was her average speed for the two days? Give both the exact value and an approximation (accurate up to three or more decimal places) of the answer. Solution:\ \begin{equation*} v_{\text{av}}=\dfrac{\text{distance traveled}}{\text{time}}=\dfrac{% s_{1}+s_{2}}{t_{1}+t_{2}} \end{equation*}% But we do not know $s_{1}$ and $t_{1}$. \ However, $v=\dfrac{s}{t}$ implies that $vt=s$. \ So,% \begin{equation*} s_{1}=v_{1}t_{1}=59\dfrac{\unit{mi}}{\unit{h}}\cdot 3\unit{h}=177\unit{mi}% \text{ \ \ and \ }s_{2}=v_{2}t_{2}=51\dfrac{\unit{mi}}{\unit{h}}\cdot 11% \unit{h}=561\unit{mi} \end{equation*}% \begin{equation*} v_{\text{av}}=\dfrac{\text{distance traveled}}{\text{time}}=\dfrac{% s_{1}+s_{2}}{t_{1}+t_{2}}=\dfrac{177\unit{mi}+561\unit{mi}}{3\unit{h}+11% \unit{h}}=\dfrac{738\unit{mi}}{14\unit{h}}=\dfrac{369}{7}\dfrac{\unit{mi}}{% \unit{h}} \end{equation*}% Exact value: \ $\dfrac{369}{7}\dfrac{\unit{mi}}{\unit{h}}$ \ \ \ \ Approximate value: \ $52.\,\allowbreak 714\,3\dfrac{\unit{mi}}{\unit{h}}$ \item We drove for two days. On the first day, we drove for $4$ hours and had an average speed of $50$ miles per hour. On the second day, we drove for $6$ hours. Our average speed for the two days was $53$ miles per hour. What was our average speed on the second day? Solution: \ \ \ Recall that $v=\dfrac{s}{t}$ and \ $s=vt$ \ \ \begin{eqnarray*} v_{\text{av}} &=&\dfrac{\text{distance traveled}}{\text{time}}=\dfrac{% s_{1}+s_{2}}{t_{1}+t_{2}}=\dfrac{v_{1}t_{1}+v_{2}t_{2}}{t_{1}+t_{2}} \\ v_{\text{av}} &=&\dfrac{v_{1}t_{1}+v_{2}t_{2}}{t_{1}+t_{2}} \\ 53\dfrac{\unit{mi}}{\unit{h}} &=&\dfrac{50\dfrac{\unit{mi}}{\unit{h}}\cdot 4% \unit{h}+v_{2}\cdot 6\unit{h}}{4\unit{h}+6\unit{h}} \\ 53 &=&\dfrac{200+6v_{2}}{10}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ multiply by }9 \\ 530 &=&200+6v_{2}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ subtract }150 \\ 330 &=&6v_{2}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ divide by }6 \\ 55 &=&v_{2} \end{eqnarray*}% So the average speed for the second day must have been $55\dfrac{\unit{mi}}{% \unit{h}}$. \item First we traveled towards North. \ We covered $200$ miles in four hours. \ Then we traveled toward West for five hours and covered $150$ miles. a) \ What was our average speed for the first part of the trip? Solution: \ \begin{equation*} v_{\text{av}}=\dfrac{\text{distance traveled}}{\text{time}}=\dfrac{s}{t}=% \dfrac{200\unit{mi}}{4\unit{h}}=50\dfrac{\unit{mi}}{\unit{h}} \end{equation*}% b) \ What was our average speed for the second part of the trip? Solution: \ \begin{equation*} v_{\text{av}}=\dfrac{\text{distance traveled}}{\text{time}}=\dfrac{s}{t}=% \dfrac{150\unit{mi}}{5\unit{h}}=30\dfrac{\unit{mi}}{\unit{h}} \end{equation*}% c) \ What was our average speed for the entire trip? Solution:% \begin{equation*} v_{\text{av}}=\dfrac{\text{distance traveled}}{\text{time}}=\dfrac{% s_{1}+s_{2}}{t_{1}+t_{2}}=\dfrac{v_{1}t_{1}+v_{2}t_{2}}{t_{1}+t_{2}}=\dfrac{% 200\unit{mi}+150\unit{mi}}{4\unit{h}+5\unit{h}}=\dfrac{350\unit{mi}}{9\unit{h% }}=38.\overline{8}\dfrac{\unit{mi}}{\unit{h}} \end{equation*} d) \ What was our average velocity for the first part of the trip? Solution:% \begin{equation*} v_{\text{av}}=\dfrac{\text{distance traveled}}{\text{time}}=\dfrac{s}{t}=% \dfrac{200\unit{mi}}{4\unit{h}}=50\dfrac{\unit{mi}}{\unit{h}} \end{equation*}% So the average velocity was $50\dfrac{\unit{mi}}{\unit{h}}$ to North. e) \ What was average velocity for the second part of the trip? Solution: \ \begin{equation*} v_{\text{av}}=\dfrac{\text{distance traveled}}{\text{time}}=\dfrac{s}{t}=% \dfrac{150\unit{mi}}{5\unit{h}}=30\dfrac{\unit{mi}}{\unit{h}} \end{equation*}% So the average velocity was $30\dfrac{\unit{mi}}{\unit{h}}$ to West. \pagebreak f) \ What was our average velocity for the entire trip? Solution: \ The displacement involves the distance between the starting and ending point of the trip.\FRAME{dtbpF}{2.2719in}{2.3229in}{0pt}{}{}{pic1.bmp% }{\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 2.2719in;height 2.3229in;depth 0pt;original-width 0.0865in;original-height 0.0891in;cropleft "0.0144";croptop "1";cropright "0.9856";cropbottom "0";filename 'pic1.bmp';file-properties "XNPEU";}}The distance between $A$ and $B$ can be found using the Pythagorean Theorem. \begin{eqnarray*} \left( AB\right) ^{2} &=&\left( 200\unit{mi}\right) ^{2}+\left( 150\unit{mi}% \right) ^{2} \\ \left( AB\right) ^{2} &=&40\,000\unit{mi}^{2}+22\,500\unit{mi}^{2} \\ \left( AB\right) ^{2} &=&62\,500\unit{mi}^{2} \\ AB &=&\pm 250\unit{mi}~~~\Longrightarrow ~~AB=250\unit{mi} \end{eqnarray*}% The angle at $A$ can also be found, using trigonometry% \begin{equation*} \tan \alpha =\dfrac{150}{200}=\dfrac{3}{4}~~~~~\alpha =\tan ^{-1}\left( \dfrac{3}{4}\right) \approx 36.\,\allowbreak 869\,9^{\circ } \end{equation*}% So the average velocity was% \begin{equation*} v=\dfrac{s}{t}=\dfrac{250\unit{mi}}{9\unit{h}}\approx 27.\overline{7}\dfrac{% \unit{mi}}{\unit{h}} \end{equation*}% So our average velocity was $27.\overline{7}\dfrac{\unit{mi}}{\unit{h}},$ $% 36.8699^{\circ }$ West from due North. \item A bus travels between cities A and B. \ The distance between these cities is $60$ miles. It takes the bus $2$ hours to get from A to B. On its way back, the traveling time was only $1.5$ hours. Find the average speed of the bus for a) \ the trip from A to B Solution:% \begin{equation*} v=\dfrac{s}{t}=\dfrac{60\unit{mi}}{2\unit{h}}=30\dfrac{\unit{mi}}{\unit{h}} \end{equation*}% b) \ the trip from B to A Solution:% \begin{equation*} v=\dfrac{s}{t}=\dfrac{60\unit{mi}}{1.5\unit{h}}=40\dfrac{\unit{mi}}{\unit{h}} \end{equation*}% c) \ for the roundtrip. Solution: \ \begin{equation*} v_{\text{av}}=\dfrac{\text{distance traveled}}{\text{time}}=\dfrac{% s_{1}+s_{2}}{t_{1}+t_{2}}=\dfrac{60\unit{mi}+60\unit{mi}}{2\unit{h}+1.5\unit{% h}}=\dfrac{120}{3.5}\dfrac{\unit{mi}}{\unit{h}}\approx 34.\,\allowbreak 28571% \dfrac{\unit{mi}}{\unit{h}} \end{equation*} \pagebreak \item A bus travels between cities A and B. \ From A to B, the bus has an average speed of $v_{1}$. On its way back, the average speed is $v_{2.}$ \ Express the average speed of the bus in terms of $v_{1}$ \ and $v_{2}$. \ Solution: \ Recall that $v=\dfrac{s}{t}$ and so $t=\dfrac{s}{v}$ \begin{eqnarray*} v_{\text{av}} &=&\dfrac{\text{distance traveled}}{\text{time}}=\dfrac{% s_{1}+s_{2}}{t_{1}+t_{2}}=\dfrac{s+s}{\dfrac{s}{v_{1}}+\dfrac{s}{v_{2}}}=% \dfrac{2s}{\dfrac{sv_{2}}{v_{1}v_{2}}+\dfrac{sv_{1}}{v_{1}v_{2}}}=\dfrac{2s}{% ~\dfrac{sv_{2}+sv_{1}}{v_{1}v_{2}}~} \\ &=&2s\cdot \dfrac{v_{1}v_{2}}{s\left( v_{2}+v_{1}\right) }=\dfrac{2\NEG% {s}v_{1}v_{2}}{\NEG{s}\left( v_{2}+v_{1}\right) }=\dfrac{2v_{1}v_{2}}{% v_{2}+v_{1}} \end{eqnarray*} \end{enumerate} \vspace{4in} \vspace{3in} \bigskip \bigskip \bigskip {\small %TCIMACRO{\TeXButton{\small}{\small}}% %BeginExpansion \small% %EndExpansion } \href{https://teaching.martahidegkuti.com/shared/lnotes/lecturenotes.html}{% For more documents like this, visit our page at\ https://teaching.martahidegkuti.com and click on Lecture Notes. \ E-mail questions or comments to mhidegkuti@ccc.edu.} \end{document}