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concepts/mechanics/u1/m1-2-position-displacement.tex

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+\subsection{Position, Displacement, Distance, and Reference Frames}
+
+\dfn{Reference frame, position, displacement, and distance}{A \emph{reference frame} is a choice of origin $O$, coordinate axes $x$, $y$, and $z$, and a clock for measuring time $t$. In AP kinematics, we describe motion in an inertial reference frame.
+
+The \emph{position vector} of an object at time $t$ is
+\[
+\vec{r}(t)=x(t)\,\hat{\imath}+y(t)\,\hat{\jmath}+z(t)\,\hat{k},
+\]
+where $x(t)$, $y(t)$, and $z(t)$ are the object's coordinates in the chosen frame.
+
+If the object is at initial position $\vec{r}_i$ and later at final position $\vec{r}_f$, the \emph{displacement} over that time interval is the vector
+\[
+\Delta \vec{r}=\vec{r}_f-\vec{r}_i.
+\]
+
+The \emph{distance traveled}, denoted by $d$, is the total length of the path actually followed. Distance is a scalar, while displacement is a vector.}
+
+\nt{The sign of a coordinate such as $x$ depends on the chosen axis direction, and the value of a position such as $\vec{r}(t)$ depends on the chosen origin. Thus position is frame-dependent. Displacement $\Delta \vec{r}$ compares two positions in the same frame, so changing the origin alone changes $\vec{r}_i$ and $\vec{r}_f$ but not their difference. Distance $d$ is not the same as $\lvert \Delta \vec{r} \rvert$ in general: $d$ is the total path length, while $\lvert \Delta \vec{r} \rvert$ is the straight-line separation between the initial and final positions. If an object turns around or follows a curved path, then $d>\lvert \Delta \vec{r} \rvert$.}
+
+\mprop{Component formulas and the straight-line special case}{If the initial position is
+\[
+\vec{r}_i=x_i\,\hat{\imath}+y_i\,\hat{\jmath}+z_i\,\hat{k}
+\]
+and the final position is
+\[
+\vec{r}_f=x_f\,\hat{\imath}+y_f\,\hat{\jmath}+z_f\,\hat{k},
+\]
+then the displacement is
+\[
+\Delta \vec{r}=(x_f-x_i)\,\hat{\imath}+(y_f-y_i)\,\hat{\jmath}+(z_f-z_i)\,\hat{k}.
+\]
+Its magnitude is
+\[
+\lvert \Delta \vec{r} \rvert=\sqrt{(x_f-x_i)^2+(y_f-y_i)^2+(z_f-z_i)^2}.
+\]
+If the object moves along a straight path without changing direction, then the distance traveled equals the magnitude of the displacement:
+\[
+d=\lvert \Delta \vec{r} \rvert.
+\]
+For any other path, the distance satisfies
+\[
+d\ge \lvert \Delta \vec{r} \rvert.
+\]}
+
+\qs{Worked example}{In a laboratory reference frame, the origin is marked on the floor, the $x$-axis points east, and the $y$-axis points north. A robot starts at the position
+\[
+\vec{r}_i=(2\,\mathrm{m})\,\hat{\imath}+(1\,\mathrm{m})\,\hat{\jmath}.
+\]
+It then moves $5\,\mathrm{m}$ east, then $3\,\mathrm{m}$ south, and then $2\,\mathrm{m}$ west. Find the final position $\vec{r}_f$, the displacement $\Delta \vec{r}$, the magnitude $\lvert \Delta \vec{r} \rvert$, and the total distance traveled $d$.}
+
+\sol The initial position is
+\[
+\vec{r}_i=(2\,\mathrm{m})\,\hat{\imath}+(1\,\mathrm{m})\,\hat{\jmath}.
+\]
+After the first motion, the robot moves $5\,\mathrm{m}$ east, so its position becomes
+\[
+(7\,\mathrm{m})\,\hat{\imath}+(1\,\mathrm{m})\,\hat{\jmath}.
+\]
+After the second motion, the robot moves $3\,\mathrm{m}$ south, so its position becomes
+\[
+(7\,\mathrm{m})\,\hat{\imath}+(-2\,\mathrm{m})\,\hat{\jmath}.
+\]
+After the third motion, the robot moves $2\,\mathrm{m}$ west, so the final position is
+\[
+\vec{r}_f=(5\,\mathrm{m})\,\hat{\imath}+(-2\,\mathrm{m})\,\hat{\jmath}.
+\]
+Therefore the displacement is
+\[
+\Delta \vec{r}=\vec{r}_f-\vec{r}_i=\bigl[(5-2)\,\mathrm{m}\bigr]\hat{\imath}+\bigl[(-2)-1\bigr]\mathrm{m}\,\hat{\jmath}.
+\]
+So
+\[
+\Delta \vec{r}=(3\,\mathrm{m})\,\hat{\imath}+(-3\,\mathrm{m})\,\hat{\jmath}.
+\]
+Its magnitude is
+\[
+\lvert \Delta \vec{r} \rvert=\sqrt{(3\,\mathrm{m})^2+(-3\,\mathrm{m})^2}=\sqrt{18}\,\mathrm{m}=3\sqrt{2}\,\mathrm{m}.
+\]
+The total distance traveled is the sum of the three path segments:
+\[
+d=5\,\mathrm{m}+3\,\mathrm{m}+2\,\mathrm{m}=10\,\mathrm{m}.
+\]
+Thus,
+\[
+\vec{r}_f=(5\,\mathrm{m})\,\hat{\imath}-(2\,\mathrm{m})\,\hat{\jmath},
+\qquad
+\Delta \vec{r}=(3\,\mathrm{m})\,\hat{\imath}-(3\,\mathrm{m})\,\hat{\jmath},
+\]
+\[
+\lvert \Delta \vec{r} \rvert=3\sqrt{2}\,\mathrm{m},
+\qquad
+d=10\,\mathrm{m}.
+\]