Q. 54 - Calculus

Question

$$\begin{gathered} \mathrm{Suppose} \mathrm{you} \mathrm{drive} \mathrm{on} \mathrm{a} \mathrm{racetrack} \mathrm{for} 10 \mathrm{minutes} \mathrm{with} \mathrm{velocity} \mathrm{as} \mathrm{shown} \mathrm{in} \mathrm{the} \mathrm{graph} \mathrm{at} \mathrm{the} \mathrm{right}. \\ \left(\mathrm{a}\right) \mathrm{Describe} \mathrm{in} \mathrm{words} \mathrm{the} \mathrm{behavior} \mathrm{of} \mathrm{your} \mathrm{race} \mathrm{car} \mathrm{over} \mathrm{the} 10 \mathrm{minutes} \mathrm{as} \mathrm{shown} \mathrm{in} \mathrm{the} \mathrm{graph}. \\ \left(\mathrm{b}\right) \mathrm{Find} \mathrm{a} \mathrm{piecewise}-\mathrm{defined} \mathrm{formula} \mathrm{for} \mathrm{your} \mathrm{velocity} \mathrm{v}\left(\mathrm{t}\right), \mathrm{in} \mathrm{miles} \mathrm{per} \mathrm{hour}, \mathrm{t} \mathrm{hours} \mathrm{after} \mathrm{you} \mathrm{start} \mathrm{from} \mathrm{rest}. \\ (\mathrm{Note} \mathrm{that} 1 \mathrm{minute} \mathrm{is} \frac{1}{60} \mathrm{of} \mathrm{an} \mathrm{hour}.) \\ \left(\mathrm{c}\right) \mathrm{Approximate} \mathrm{the} \mathrm{distance} \mathrm{you} \mathrm{travelled} \mathrm{over} \mathrm{the} 10 \mathrm{minutes} \mathrm{by} \mathrm{using} 10 \mathrm{subintervals} \mathrm{of} 1 \mathrm{minute} \mathrm{over} \mathrm{which} \mathrm{you} \mathrm{assume} \mathrm{a} \mathrm{constant} \mathrm{velocity}. \\ \mathrm{Illustrate} \mathrm{this} \mathrm{approximation} \mathrm{by} \mathrm{showing} \mathrm{rectangles} \mathrm{on} \mathrm{the} \mathrm{graph} \mathrm{of} \mathrm{v}\left(\mathrm{t}\right). \\ \left(\mathrm{d}\right) \mathrm{Given} \mathrm{that} \mathrm{distance} \mathrm{travelled} \mathrm{is} \mathrm{the} \mathrm{area} \mathrm{under} \mathrm{the} \mathrm{velocity} \mathrm{graph}, \mathrm{use} \mathrm{triangles} \mathrm{and} \mathrm{squares} \mathrm{to} \mathrm{calculate} \mathrm{the} \mathrm{exact} \mathrm{distance} \mathrm{travelled}. \end{gathered}$$

Step-by-Step Solution

Answer

$$\begin{gathered} \left(\mathrm{a}\right). \mathrm{The} \mathrm{velocity} \mathrm{of} \mathrm{car} \mathrm{increases} \mathrm{during} 0-2 \mathrm{minutes}, \mathrm{remains} \mathrm{constant} \mathrm{during} 2-9 \mathrm{minutes} \mathrm{and} \mathrm{starts} \mathrm{decreasing} \mathrm{during} 9-10 \mathrm{minutes}. \\ \mathrm{The} \mathrm{car} \mathrm{finally} \mathrm{stops} \mathrm{after} {10}^{\mathrm{th}} \mathrm{minute}. \\ \left(\mathrm{b}\right). \mathrm{The} \mathrm{piecewise} \mathrm{defined} \mathrm{formula} \mathrm{for} \mathrm{v}\left(\mathrm{t}\right) = \left\{\begin{array}{l}(0-\frac{100}{88}=\frac{25}{22})\mathrm{mph}, 0\le \mathrm{t}<\frac{1}{30}\\ \left(\frac{25}{22}\right)\mathrm{mph}, \frac{1}{30}\le \mathrm{t}<\frac{3}{20}\\ (\frac{25}{22} - 0)\mathrm{mph}, \frac{3}{20}\le \mathrm{t}<\frac{1}{6}\end{array}\right. \\ \left(\mathrm{c}\right). \mathrm{The} \mathrm{total} \mathrm{distance} \mathrm{travelled} \mathrm{by} \mathrm{racing} \mathrm{car} \mathrm{in} 10 \mathrm{minutes} \mathrm{is} 850\mathrm{ft}. \\ \left(\mathrm{d}\right). \mathrm{The} \mathrm{total} \mathrm{distance} \mathrm{travelled} \mathrm{by} \mathrm{racing} \mathrm{car} \mathrm{in} 10 \mathrm{minutes} \mathrm{is} 850\mathrm{ft}. \end{gathered}$$

Step 1. Given Information

time = 10 minutes

Step 2. Solution (a) : Behavior of race car over 10 minutes.

$$\begin{gathered} \mathrm{We} \mathrm{know} \mathrm{that}, \mathrm{slope} \mathrm{of} \mathrm{velocity}-\mathrm{time} \mathrm{graph} \mathrm{represents} \mathrm{acceleration} \mathrm{of} \mathrm{an} \mathrm{object} \\ \mathrm{Therefore}, \\ \mathrm{For} 0\le t<2 : \mathrm{a} = \frac{(100-0) \mathrm{ft}/\min }{(2-0) \min }=+50 \mathrm{ft}/{\min }^2. \\ \mathrm{as} \mathrm{acceleration} \mathrm{is} \mathrm{positive}, \mathrm{therefore} \mathrm{velocity} \mathrm{of} \mathrm{car} \mathrm{increases} \mathrm{in} \mathrm{time} \mathrm{interval} 0-2 \min . \\ \mathrm{For} 2\le \mathrm{t}<9 : \mathrm{a} = \frac{(100-100) \mathrm{ft}/\min }{(9-2) \min }=0 \mathrm{ft}/{\min }^2. \\ \mathrm{as} \mathrm{acceleration} \mathrm{is} \mathrm{zero}, \mathrm{therefore} \mathrm{velocity} \mathrm{remains} \mathrm{constant} \mathrm{in} \mathrm{time} \mathrm{interval} 2-9 \min . \\ \mathrm{For} 9\le \mathrm{t}<10 : \mathrm{a} = \frac{(0-100) \mathrm{ft}/\min }{(10-9) \min }=-100 \mathrm{ft}/{\min }^2. \\ \mathrm{as} \mathrm{acceleration} \mathrm{is} \mathrm{negative}, \mathrm{therefore} \mathrm{velcity} \mathrm{of} \mathrm{car} \mathrm{decreases} \mathrm{in} \mathrm{time} \mathrm{interval} 9-10 \min \mathrm{and} \\ \mathrm{car} \mathrm{finally} \mathrm{stops} \mathrm{after} {10}^{\mathrm{th}} \min . \end{gathered}$$

Step 3. Solution (b) : Piecewise-defined formula for velocity v(t) in miles per hour starting from rest

$$\begin{gathered} \mathrm{v}\left(\mathrm{t}\right) (\mathrm{ft}/\min ) = \mathrm{v}\left(\mathrm{t}\right) \times \frac{1}{88}\mathrm{miles}/\mathrm{hour} \mathrm{as} 1\mathrm{mph} = 88 \mathrm{ft}./\min \\ \mathrm{Continuous} \mathrm{time} \mathrm{intervals} \mathrm{used} \mathrm{are} : 0\le \mathrm{t}<2, 2\le \mathrm{t}<9, 9\le \mathrm{t}<10. \\ \mathrm{Hence}, \mathrm{the} \mathrm{piecewise} \mathrm{defined} \mathrm{formula} \mathrm{for} \mathrm{v}\left(\mathrm{t}\right) = \left\{\begin{array}{l}(0-\frac{100}{88}=\frac{25}{22})\mathrm{mph}, 0\le \mathrm{t}<\frac{1}{30}\\ \left(\frac{25}{22}\right)\mathrm{mph}, \frac{1}{30}\le \mathrm{t}<\frac{3}{20}\\ (\frac{25}{22} - 0)\mathrm{mph}, \frac{3}{20}\le \mathrm{t}<\frac{1}{6}\end{array}\right. \end{gathered}$$

Step 4. Solution (c) : Finding appx. distance in 10 minutes by using 10 subintervals in 1 minute over which we assume velocity constant.

$$\begin{gathered} \mathrm{Formula} \mathrm{used} : \mathrm{distance} = \mathrm{velocity}\times \mathrm{time}. \\ \mathrm{For} \mathrm{time} \mathrm{interval} 0 \mathrm{to} 2 \mathrm{minutes} \mathrm{the} \mathrm{constant} \mathrm{velocity} \mathrm{is} \mathrm{equal} \mathrm{to} \mathrm{the} \mathrm{average} \mathrm{velocity} \\ {\mathrm{v}}_{(0-2)}=\frac{(0+100) \mathrm{ft}/\min }{2}= 50\mathrm{ft}/\min . \\ \mathrm{Therefore}, \mathrm{the} \mathrm{distance} \mathrm{travelled} \mathrm{for} \mathrm{first} \mathrm{two} \mathrm{minutes} \mathrm{of} \mathrm{journey} \mathrm{is} \\ {\mathrm{d}}_{(0-2)}=50\frac{\mathrm{ft}}{\min } \mathrm{x} 2 \min =100\mathrm{ft} ....... \left(1\right) \\ \mathrm{For} \mathrm{time} \mathrm{interval} 2 \mathrm{to} 9 \mathrm{minutes} \mathrm{the} \mathrm{constant} \mathrm{velocity} \mathrm{is} \mathrm{v}(2-9) 100\mathrm{ft}/\min . \\ \mathrm{Therefore}, \mathrm{the} \mathrm{distance} \mathrm{travelled} \mathrm{in} 2 \mathrm{to} 9 \mathrm{minutes} \mathrm{of} \mathrm{journey} \mathrm{minutes} \mathrm{is} \\ {\mathrm{d}}_{(2-9)} =100\frac{\mathrm{ft}}{\min }\times (9-2) \min =700\mathrm{ft} ...... \left(2\right) \\ \mathrm{For} \mathrm{time} \mathrm{interval} 9 \mathrm{to} 10 \mathrm{minutes} \mathrm{the} \mathrm{constant} \mathrm{velocity} \mathrm{is} \mathrm{equal} \mathrm{to} \mathrm{the} \mathrm{average} \mathrm{velocity} \\ {\mathrm{v}}_{(9-10)}=\frac{(0+100)\mathrm{ft}/\min }{2}= 50\mathrm{ft}/\min . \\ \mathrm{Therefore}, \mathrm{the} \mathrm{distance} \mathrm{travelled} \mathrm{in} \mathrm{last} 1\mathrm{minute} \mathrm{of} \mathrm{journey} \mathrm{is} \\ {\mathrm{d}}_{(9-10)}= 50\frac{\mathrm{ft}}{\min } \mathrm{x} 1 \min = 50\mathrm{ft}..... \left(3\right) \\ \mathrm{The} \mathrm{total} \mathrm{distance} \mathrm{travelled} \mathrm{by} \mathrm{racing} \mathrm{car} \mathrm{is} \\ \mathrm{d} = {\mathrm{d}}_{(0-2)}+{\mathrm{d}}_{(2-9)}+{\mathrm{d}}_{(9-10)}= (100+700 + 50) \mathrm{ft}=850\mathrm{ft}. \\ \mathrm{The} \mathrm{division} \mathrm{of} \mathrm{time} 10 \mathrm{minutes} \mathrm{in} 10 \mathrm{subinervals} \mathrm{is} \mathrm{shown} \mathrm{in} \mathrm{figure} \mathrm{here}. \end{gathered}$$

Step 5. Solution (d) : Distance travelled as the area of velocity-time graph

$$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{a} \mathrm{triangle} △=\frac{1}{2}\times \mathrm{base}\times \mathrm{height}, \mathrm{and} \mathrm{the} \mathrm{area} \mathrm{of} \mathrm{a} \mathrm{rectangle} ▭=\mathrm{Length}\times \mathrm{breadth}. \\ \mathrm{Calculation}: \\ \mathrm{From} \mathrm{the} \mathrm{velocity}-\mathrm{time} \mathrm{graph} \mathrm{given} \mathrm{in} \mathrm{part} \left(\mathrm{a}\right) \mathrm{of} \mathrm{racing} \mathrm{car}, \mathrm{there} \mathrm{are} \mathrm{two} \mathrm{triangles} \mathrm{and} \mathrm{one} \mathrm{rectangle}. \\ \mathrm{The} \mathrm{total} \mathrm{distance} \mathrm{travelled} \mathrm{by} \mathrm{car} \mathrm{is} \mathrm{the} \mathrm{sum} \mathrm{of} \mathrm{the} \mathrm{area} \mathrm{of} \mathrm{two} \mathrm{triangles} \mathrm{and} \mathrm{one} \mathrm{rectangle}. \\ \mathrm{Now}, \mathrm{the} \mathrm{distance} \mathrm{travelled} \mathrm{for} \mathrm{first} \mathrm{two} \mathrm{minutes} \mathrm{is} \mathrm{equal} \mathrm{to} \mathrm{the} \mathrm{area} △\mathrm{OA}2, \mathrm{i}.\mathrm{e}. \\ {\mathrm{d}}_{(0-2)}=△\mathrm{OA}2=\frac{1}{2}\times (2-0)\times 100=100\mathrm{ft}. \\ \mathrm{The} \mathrm{distance} \mathrm{travelled} \mathrm{in} 2 \mathrm{to} 9 \mathrm{minutes} \mathrm{of} \mathrm{journey} \mathrm{minutes} \mathrm{is} \mathrm{area} \mathrm{of} \mathrm{rectangle} ▭2\mathrm{AB}92 \\ {\mathrm{d}}_{(2-9)}=▭2\mathrm{AB}92 = 100 \mathrm{x} (9- 2) = 700\mathrm{ft}. \\ \mathrm{The} \mathrm{distance} \mathrm{travelled} \mathrm{last} \mathrm{minute} \mathrm{of} \mathrm{joumey} \mathrm{is} \mathrm{equal} \mathrm{to} \mathrm{the} \mathrm{area} △9\mathrm{B}10, \mathrm{ie}. \\ {\mathrm{d}}_{(9-10)}= △9\mathrm{B}10=\frac{1}{2}\times (10-9)\times 100=50\mathrm{ft}. \\ \mathrm{Hence}, \mathrm{the} \mathrm{total} \mathrm{distance} \mathrm{travelled} \mathrm{by} \mathrm{racing} \mathrm{car} \mathrm{is} {\mathrm{d}}_{(0-2)} + {\mathrm{d}}_{(2-9)} + {\mathrm{d}}_{(9-10)}= 850\mathrm{ft}. \\ \mathrm{Therefore}, \mathrm{the} \mathrm{total} \mathrm{distance} \mathrm{travelled} \mathrm{by} \mathrm{racing} \mathrm{car} \mathrm{is} 850\mathrm{ft}. \end{gathered}$$