From: W. Trevor King Date: Fri, 14 Jan 2011 17:12:36 +0000 (-0500) Subject: Add week 3 recitation solutions. X-Git-Url: http://git.tremily.us/?p=course.git;a=commitdiff_plain;h=3b3b1efc8443b0b1f8530e23a4ccf9d89aec6429 Add week 3 recitation solutions. --- diff --git a/latex/problems/Serway_and_Jewett_8/problem02.05.tex b/latex/problems/Serway_and_Jewett_8/problem02.05.tex new file mode 100644 index 0000000..16528df --- /dev/null +++ b/latex/problems/Serway_and_Jewett_8/problem02.05.tex @@ -0,0 +1,65 @@ +\begin{problem*}{2.5} +A position-time graph for a particle moving along the $x$ axis is +shown in Figure P2.5. \Part{a} Find the average velocity in the time +interval $t=1.50\U{s}$ to $t=4.00\U{s}$. \Part{b} Determine the +instantaneous velocity at $t=2.00\U{s}$ by measuring the slope of the +tangent to the graph. \Part{c} At what value of $t$ is the velocity +zero? +\begin{center} +\begin{asy} +import graph; + +size(4cm, IgnoreAspect); + +/* Parameter computation from the back-of-the-book answers: + * 2*a*2 + b = -3.8 (1) (answer to part b) + * 2*a*4 + b = 0 (2) (answer to part c) + * 2*a*2 = 3.8 (3) ((2)-(1)) + * a = 3.8/4 = 0.95 ~= 1.0 (4) (solve (3), rounding to likely value) + * b = -8*a (5) (rearrange (2)) + * b = -8*1 = -8 (6) (plug (4) into (5)) + * a*4^2 + b*4 + c = 2 (7) (location of minimum from graph) + * c = 2 - 16a - 4b (8) (rearrange (7)) + * c = 2 - 16*1 + 4*8 = 18 (9) (plug (4) and (6) into (8)) + */ +real a=1, b=-8, c=18; +real tanj_time = 2; + +real parab(real t) { + return a*t*t + b*t + c; +} + +real tanj(real t) { + real slope = 2*a*tanj_time + b; + return slope * (t - tanj_time) + parab(tanj_time); +} + +draw(graph(tanj, 0, 3.5), green); +draw(graph(parab, 1, 6.5), red); + +xaxis("$t\U{s}$", BottomTop, LeftTicks(extend=true, ptick=gray+thin())); +yaxis("$x\U{m}$", LeftRight, RightTicks(extend=true, ptick=gray+thin())); +\end{asy} +\end{center} +\end{problem*} + +\begin{solution} +\Part{a} +The average velocity is the total displacement over the elapsed time, so +\begin{equation} + v_\text{avg} = \frac{x(4.00\U{s}) - x(1.50\U{s})}{4.00\U{s}-1.50\U{s}} + = \frac{2.0\U{m} - 8.0\U{m}}{2.5\U{s}} + = \ans{-2.4\U{m/s}} +\end{equation} + +\Part{b} +Using rise-over-run to determine the tangent slope +\begin{equation} + v(t=2.00\U{s}) = \frac{0\U{m} - 11.5\U{m}}{3.5\U{s} - 0.5\U{s}} + = \ans{-3.8\U{m/s}} +\end{equation} + +\Part{c} +Looking for the minimum of $x(t)$ (where the tangent curve is flat), +we see that $v(\ans{4.0\U{s}})=0$. +\end{solution} diff --git a/latex/problems/Serway_and_Jewett_8/problem02.11.tex b/latex/problems/Serway_and_Jewett_8/problem02.11.tex new file mode 100644 index 0000000..e097bd5 --- /dev/null +++ b/latex/problems/Serway_and_Jewett_8/problem02.11.tex @@ -0,0 +1,88 @@ +\begin{problem*}{2.11} +A hare and a tortoise compete in a race over a straight course +$1.00\U{km}$ long. The tortoise crawls at a speed of $0.200\U{m/s}$ +toward the finish line. The hare runs at a speed of $8.00\U{m/s}$ +toward the finish line for $0.800\U{km}$ and then stops to tease the +slow-moving tortoise as the tortoise eventually passes by. The hare +waits for a while after the tortoise passes by and then runs toward +the finish line again at $8.00\U{m/s}$. Both the hare and the tortise +cross the finish line at exactly the same instant. Assume both +animals, when moving, move steadily at their respective +speeds. \Part{a} How far is the tortoise from the finish line when +the hare resumes the race? \Part{b} For how long in time was the hare +stationary? +\end{problem*} + +\begin{solution} +Sometimes it is useful to draw a graph to get a feel for what's going +on. +\begin{center} +\begin{asy} +import graph; +import Mechanics; + +real u = 0.005cm; +real v_scale = 10; // increase v_scale to decrease time spread + +real L = 1e3*u; // length of race +real p = 0.8e3u; // hare pause location +real vt = 0.2 * v_scale; // tortoise velocity +real vh = 8 * v_scale; // hare velocity +real T = L / vt; // the tortoise walks the whole time +pair f = (L,T); // finish +pair h1 = (p, p/vh); // hare pause event +pair h2 = (p, T-(L-p)/vh); // hare restart event + +dot((0,0)); // start +dot(f); // finish + +Vector T1 = Vector((0,0), mag=length(f), dir=degrees(f), "tortoise"); +T1.draw(rotateLabel=true, labelOffset=-f/2); + +Vector H1 = Vector((0,0), mag=length(h1), dir=degrees(h1), "hare"); +H1.draw(rotateLabel=true, labelOffset=-h1/2); +Vector H2 = Vector(h1, mag=length(h2-h1), dir=degrees(h2-h1), "pause"); +H2.draw(rotateLabel=true, labelOffset=-(h2-h1)*2/3); +Vector H3 = Vector(h2, mag=length(f-h2), dir=degrees(f-h2), "hare"); +H3.draw(rotateLabel=true, labelOffset=-(f-h2)/2); + +xaxis("$x$"); +yaxis("$t$"); +\end{asy} +\end{center} + +\Part{a} +The final distance run by the hare is +\begin{equation} + x_\text{h,2} = L - \Delta x_\text{h,1} +\end{equation} +which takes the hare +\begin{equation} + t_\text{h,2} = \frac{L - x_\text{h,1}}{v_\text{h}} +\end{equation} +In this time, the tortoise covers +\begin{align} + x_\text{t,2} &= v_\text{t} \cdot t_\text{h,2} + = v_\text{t}\frac{L - x_\text{h,1}}{v_\text{h}} + = \frac{v_\text{t}}{v_\text{h}}(L - x_\text{h,1}) \\ + &= \frac{0.200\U{m/s}}{8.00\U{m/s}}(1.00\U{km} - 0.800\U{km}) + = \ans{5.00\U{m}} +\end{align} + +\Part{b} +The hare is running for +\begin{equation} + t_\text{h,run} = \frac{L}{v_\text{h}} +\end{equation} +The tortoise is running for +\begin{equation} + t_\text{t,run} = \frac{L}{v_\text{h}} +\end{equation} +and the tortoise is running for the whole race, so the hare pauses for +\begin{align} + t_\text{h,pause} &= t_\text{t,run} - t_\text{h,run} + = \frac{L}{v_\text{h}} - \frac{L}{v_\text{h}} + = \frac{1.00\U{km}}{8.00\U{m/s}} - \frac{1.00\U{km}}{0.200\U{m/s}} \\ + &= \ans{4.88\U{ks}} = 1.35\U{hours} +\end{align} +\end{solution} diff --git a/latex/problems/Serway_and_Jewett_8/problem02.29.tex b/latex/problems/Serway_and_Jewett_8/problem02.29.tex new file mode 100644 index 0000000..f58e21b --- /dev/null +++ b/latex/problems/Serway_and_Jewett_8/problem02.29.tex @@ -0,0 +1,32 @@ +\begin{problem*}{2.29} +The driver of a car slams on the brakes when he sees a tree blocking +the road. The car slows uniformly with an acceleration of +$-5.60\U{m/s$^2$}$ for $4.20\U{s}$, making straight skid marks +$62.4\U{m}$ long, all the way to the tree. With what speed does the +car then strike the tree? +\end{problem*} + +\begin{solution} +The speed of the car is given by +\begin{equation} + v(t) = at + v_0 +\end{equation} +and its position is given by +\begin{equation} + x(t) = \frac{1}{2} a t^2 + v_0 t + x_0 +\end{equation} + +We know $a$, $t$, and $x-x_0$, so we can use the second equation to +find $v_0$. +\begin{equation} + v_0 = \frac{x-x_0-\frac{1}{2}at^2}{t} +\end{equation} +We can plug this into the first equation to find the final velocity. +\begin{align} + v &= at + \frac{x-x_0-\frac{1}{2}at^2}{t} \\ + &= -5.60\U{m/s$^2$} \cdot 4.20\U{s} + +\frac{62.4\U{m} - \frac{1}{2}\cdot(-5.60\U{m/s$^2$})\cdot(4.20\U{s})^2} + {4.20\U{s}} \\ + &= \ans{3.10\U{m/s}} +\end{align} +\end{solution} diff --git a/latex/problems/Serway_and_Jewett_8/problem02.33.tex b/latex/problems/Serway_and_Jewett_8/problem02.33.tex new file mode 100644 index 0000000..b879d78 --- /dev/null +++ b/latex/problems/Serway_and_Jewett_8/problem02.33.tex @@ -0,0 +1,71 @@ +\begin{problem*}{2.33} +An object moves with constant acceleration $4.00\U{m/s$^2$}$ and over +a time interval reaches a final velocity of $12.0\U{m/s}$. \Part{a} +If its initial velocity is $6.00\U{m/s}$, what is its displacement +during the time interval? \Part{b} What is the distance it travels +during the time interval? \Part{c} If its initial velocity is +$-6.00\U{m/s}$, what is its displacement during the time +interval? \Part{d} What is the total distance it travels during the +interval in \Part{c}? +\end{problem*} + +\begin{solution} +\Part{a} +The position of an object undergoing constant acceleration is +\begin{equation} + x(t) = \frac{1}{2} a t^2 + v_0 t + x_0 +\end{equation} +Solving for time using the quadratic formula +\begin{equation} + t = \frac{-v_0 \pm \sqrt{v_0^2 - 4(a/2)\cdot(x_0-x)}}{2(a/2)} + = \frac{-v_0 \pm \sqrt{v_0^2 - 2a(x_0-x)}}{a} +\end{equation} +Plugging this into the velocity formula +\begin{align} + v &= a t + v_0 + = a \cdot \frac{-v_0 \pm \sqrt{v_0^2 - 2a(x_0-x)}}{a} + v_0 \\ + &= -v_0 \pm \sqrt{v_0^2 - 2a(x_0-x)} + v_0 + = \pm \sqrt{v_0^2 + 2a(x-x_0)} +\end{align} +The $\pm\sqrt{ }$ is annoying, so square both sides +\begin{equation} + v^2 = v_0^2 + 2a(x-x_0) +\end{equation} +Solving for displacement +\begin{equation} + x-x_0 = \frac{v^2-v_0^2}{2a} + = \frac{12.0^2 - 6.00^2}{2\cdot4.00}\U{m} + = \ans{13.5\U{m}} +\end{equation} + +\Part{b} +With positive accerleration and a positive initial velocity, the +objects velocity will always be increasingly positive, so the +distance traveled is the same as the displacement. +\begin{equation} + D_b = \ans{13.5\U{m}} +\end{equation} + +\Part{c} +A negative initial velocity has no effect on the answer to \Part{a}, +because $v$ only shows up as a square: +\begin{equation} + x-x_0 = \frac{v^2-v_0^2}{2a} + = \frac{12.0^2 - (-6.00)^2}{2\cdot4.00}\U{m} + = \ans{13.5\U{m}} +\end{equation} + +\Part{d} +With positive accerleration and a \emph{negative} initial velocity, +the objects velocity will drop to zero, after which it will be +increasingly positive. The total distance traveled is therefore the +distance traveled in the negative direction while the velocity is +decreasingly negative, plus the distance traveled in the positive +direction while velocity is increasingly positive. +\begin{equation} + D_d = \p|{\frac{0-v_0^2}{2a}}| + \p|{\frac{v^2-0}{2a}}| + = \frac{v^2+v_0^2}{2|a|} + = \frac{12.0^2 + (-6.00)^2}{2\cdot4.00}\U{m} + = \ans{22.5\U{m}} +\end{equation} +\end{solution} diff --git a/latex/problems/Serway_and_Jewett_8/problem02.61.tex b/latex/problems/Serway_and_Jewett_8/problem02.61.tex new file mode 100644 index 0000000..13c3c3e --- /dev/null +++ b/latex/problems/Serway_and_Jewett_8/problem02.61.tex @@ -0,0 +1,96 @@ +\begin{problem*}{2.61} +Kathy tests her new sports car by racing with Stan, an experienced +racer. Both start from rest, but Kathy leaves the starting line +$1.00\U{s}$ after Stan does. Stan moves with a constant acceleration +of $3.50\U{m/s$^2$}$, while Kathy maintains an acceleration of +$4.90\U{m/s$^2$}$. Find \Part{a} the time at which Kathy overtakes +Stan, \Part{b} the distance she travels before she catches him, +and \Part{c} the speeds of both cars at the instant Kathy overtakes +Stan. +\end{problem*} + +\begin{solution} +\begin{center} +\begin{asy} +import graph; + +size(6cm, IgnoreAspect); + +real dt = 1; // Kathy's delay +real ak = 4.9; // Kathy's acceleration +real as = 3.5; // Stan's acceleration + +real k(real t) { + if (t < dt) + return 0; + t -= dt; + return 0.5*ak*t*t; +} + +real s(real t) { + return 0.5*as*t*t; +} + +real A = ak/as - 1; // quadratic terms for tint +real B = -2 * dt * ak/as; +real C = dt*dt * ak/as; +real tint = (-B + sqrt(B*B-4*A*C))/(2*A); // intersection time +real xint = k(tint); // intersection location +pair int = (tint, xint); + +dot((0,0)); // start +dot(int); // intersection + +draw(graph(k, 0, 1.1 tint), green); +draw(graph(s, 0, 1.1 tint), red); + +xaxis("$t\U{s}$", 0, LeftTicks); +yaxis("$x\U{m}$", 0, LeftTicks); +\end{asy} +\end{center} + +\Part{a} +Kathy overtakes Stan when their positions match. +\begin{align} + x_k = \frac{1}{2}a_k (t-\Delta_t)^2 &= x_s = \frac{1}{2}a_s t^2 \\ + \frac{a_k}{a_s}(t^2 - 2\Delta_t t + \Delta_t^2) &= t^2 \\ + \p({\frac{a_k}{a_s} - 1})t^2 + - 2\Delta_t\frac{a_k}{a_s} t + + \Delta_t^2\frac{a_k}{a_s} &= 0 \\ + a t^2 + b t + c &= 0 \\ + a &= \p({\frac{a_k}{a_s} - 1}) = \p({\frac{4.90}{3.50} - 1}) = 0.400 \\ + b &= - 2\Delta_t\frac{a_k}{a_s} = - 2\cdot(1.00\U{s})\frac{4.90}{3.50} + = -2.80\U{s} \\ + c &= \Delta_t^2\frac{a_k}{a_s} = (1.00\U{s})^2\frac{4.90}{3.50} + = 1.40\U{s$^2$} +\end{align} +Solving with the quadratic equation +\begin{align} + t &= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} + = (6.46, 0.542)\U{s} +\end{align} + +The smaller time occurs before Kathy starts, where the quadratic +equation for $x_k(t)$ does not hold ($x_k(t < \Delta(t))=0$, as Kathy +is sitting at the starting line). Therefore, the Kathy passes Stan at +the larger time $\ans{t_p=6.46\U{s}}$ (measured since Stan left the +starting line). + +\Part{b} +Kathy travels the same distance as Stan +\begin{equation} + x_s(t_p) = \frac{1}{2}a_s t_p^2 + = \frac{1}{2} (3.50\U{m/s$^2$}) \cdot (6.46\U{s})^2 + = \ans{73.0\U{m}} +\end{equation} + +\Part{c} +The velocities at the passing point are +\begin{align} + v_s(t_p) &= a_s t_p = (3.50\U{m/s$^2$}) \cdot (6.46\U{s}) + = \ans{22.6\U{m/s}} \\ + v_k(t_p) &= a_k (t_p-\Delta_t) + = (4.90\U{m/s$^2$}) \cdot (6.46\U{s}-1.00\U{s}) + = \ans{26.7\U{m/s}} +\end{align} +\end{solution}