--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
projects / course.git / commitdiff