checkpoint 1

concepts/mechanics/u1/m1-1-scalars-vectors.tex

@@ -0,0 +1,113 @@
+\subsection{Scalars, Vectors, and Components}
+
+This subsection introduces the language used throughout Unit 1 for quantities that have magnitude only and for quantities that also have direction.
+
+\dfn{Scalars, vectors, and component form}{A \textbf{scalar} quantity is specified completely by a magnitude. Examples include mass, time, temperature, energy, and speed.
+
+A \textbf{vector} quantity is specified by both a magnitude and a direction. Examples include displacement, velocity, acceleration, and force.
+
+In a chosen Cartesian coordinate system, let $x$, $y$, and $z$ denote coordinates along the $x$-, $y$-, and $z$-axes, and let $\hat{\imath}$, $\hat{\jmath}$, and $\hat{k}$ denote unit vectors along those axes. If $\vec{v}$ is a vector with scalar components $v_x$, $v_y$, and $v_z$, then its unit-vector decomposition is
+\[
+\vec{v} = v_x\hat{\imath} + v_y\hat{\jmath} + v_z\hat{k}.
+\]
+In two dimensions,
+\[
+\vec{v} = v_x\hat{\imath} + v_y\hat{\jmath}.
+\]
+The numbers $v_x$, $v_y$, and $v_z$ are \textbf{components} of $\vec{v}$; they are scalars and can be positive, negative, or zero. The magnitude of $\vec{v}$ is written $|\vec{v}|$.}
+
+\nt{Speed is a scalar, but velocity is a vector. A component such as $v_x$ is a scalar, not a vector by itself. Also, the magnitude $|\vec{v}|$ is not the same thing as a component: $|\vec{v}|\geq 0$, while a component can be negative if the vector points partly in a negative coordinate direction. Likewise, distance is a scalar, while displacement $\Delta \vec{r}$ is a vector. Later in this unit, $\Delta \vec{r}$, $\vec{v}$, and $\vec{a}$ will all be handled with the same component ideas.}
+
+\mprop{Operational rules in components}{Let
+\[
+\vec{u} = u_x\hat{\imath} + u_y\hat{\jmath} + u_z\hat{k}
+\qquad\text{and}\qquad
+\vec{v} = v_x\hat{\imath} + v_y\hat{\jmath} + v_z\hat{k}
+\]
+be vectors written in the same Cartesian coordinate system.
+
+\begin{enumerate}[label=\bfseries\tiny\protect\circled{\small\arabic*}]
+\item Vector addition and subtraction are done component-by-component:
+\[
+\vec{u} + \vec{v} = (u_x+v_x)\hat{\imath} + (u_y+v_y)\hat{\jmath} + (u_z+v_z)\hat{k},
+\]
+\[
+\vec{u} - \vec{v} = (u_x-v_x)\hat{\imath} + (u_y-v_y)\hat{\jmath} + (u_z-v_z)\hat{k}.
+\]
+
+\item In Cartesian coordinates, the magnitude of a vector comes from its components:
+\[
+|\vec{v}| = \sqrt{v_x^2+v_y^2}
+\qquad\text{in 2D,}
+\]
+\[
+|\vec{v}| = \sqrt{v_x^2+v_y^2+v_z^2}
+\qquad\text{in 3D.}
+\]
+
+\item In two dimensions, if $\theta$ is the direction angle of $\vec{v}$ measured from the positive $x$-axis, then
+\[
+\tan\theta = \frac{v_y}{v_x},
+\]
+provided $v_x\neq 0$. The signs of $v_x$ and $v_y$ determine the correct quadrant for $\theta$. If $v_x=0$, then the vector points straight along the positive or negative $y$-axis.
+\end{enumerate}}
+
+\qs{Worked example}{In an $xy$ coordinate system, the positive $x$-axis points east and the positive $y$-axis points north. A student first moves with displacement
+\[
+\Delta \vec{r}_1=(3.0\hat{\imath}+4.0\hat{\jmath})\,\text{m}
+\]
+and then moves with displacement
+\[
+\Delta \vec{r}_2=(-1.0\hat{\imath}+2.0\hat{\jmath})\,\text{m}.
+\]
+Let $\Delta \vec{r}_{\text{tot}}$ denote the total displacement, and let $d$ denote the total distance traveled.
+
+\begin{enumerate}[label=(\alph*)]
+\item Identify whether $\Delta \vec{r}_{\text{tot}}$, the component $(\Delta \vec{r}_2)_x=-1.0\,\text{m}$, and $d$ are scalars or vectors.
+\item Find $\Delta \vec{r}_{\text{tot}}$ in component form.
+\item Find $|\Delta \vec{r}_{\text{tot}}|$.
+\item Let $\theta$ be the direction of $\Delta \vec{r}_{\text{tot}}$ measured counterclockwise from the positive $x$-axis. Find $\theta$.
+\end{enumerate}}
+\sol For part (a), $\Delta \vec{r}_{\text{tot}}$ is a vector because it has both magnitude and direction. The quantity $(\Delta \vec{r}_2)_x=-1.0\,\text{m}$ is a scalar because it is one component of a vector. The quantity $d$ is also a scalar because distance gives only a path length.
+
+For part (b), add the two displacements by components:
+\[
+\Delta \vec{r}_{\text{tot}}=\Delta \vec{r}_1+\Delta \vec{r}_2.
+\]
+Therefore,
+\[
+\Delta \vec{r}_{\text{tot}}=(3.0-1.0)\hat{\imath}+(4.0+2.0)\hat{\jmath}.
+\]
+So,
+\[
+\Delta \vec{r}_{\text{tot}}=(2.0\hat{\imath}+6.0\hat{\jmath})\,\text{m}.
+\]
+
+For part (c), use the magnitude formula in two dimensions:
+\[
+|\Delta \vec{r}_{\text{tot}}|=\sqrt{(2.0\,\text{m})^2+(6.0\,\text{m})^2}.
+\]
+Thus,
+\[
+|\Delta \vec{r}_{\text{tot}}|=\sqrt{40}\,\text{m}=6.32\,\text{m}.
+\]
+
+For part (d), use the component ratio. Since both components of $\Delta \vec{r}_{\text{tot}}$ are positive, the vector lies in the first quadrant. Then
+\[
+\tan\theta=\frac{6.0}{2.0}=3.0.
+\]
+So,
+\[
+\theta=\tan^{-1}(3.0)=71.6^\circ.
+\]
+Therefore, the total displacement is
+\[
+\Delta \vec{r}_{\text{tot}}=(2.0\hat{\imath}+6.0\hat{\jmath})\,\text{m},
+\]
+with magnitude $6.32\,\text{m}$ and direction $71.6^\circ$ counterclockwise from the positive $x$-axis.
+
+As a useful comparison, the total distance traveled is
+\[
+d=|\Delta \vec{r}_1|+|\Delta \vec{r}_2|=5.0\,\text{m}+\sqrt{5}\,\text{m}=7.24\,\text{m},
+\]
+which is a scalar and is not equal to the magnitude of the total displacement.

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,
+\[
+\vec{r}_f=(5\,\mathrm{m})\,\hat{\imath}-(2\,\mathrm{m})\,\hat{\jmath},
+\qquad
+\Delta \vec{r}=(3\,\mathrm{m})\,\hat{\imath}-(3\,\mathrm{m})\,\hat{\jmath},
+\]
+\[
+\lvert \Delta \vec{r} \rvert=3\sqrt{2}\,\mathrm{m},
+\qquad
+d=10\,\mathrm{m}.
+\]

concepts/mechanics/u1/m1-3-velocity-acceleration.tex

@@ -0,0 +1,133 @@
+\subsection{Velocity and Acceleration as Derivatives}
+
+This subsection describes motion locally: starting from the position vector $\vec{r}(t)$, velocity and acceleration are defined by derivatives at an instant. Later sections will reverse these local definitions with definite integrals to recover displacement and changes in velocity over a time interval.
+
+\dfn{Average and instantaneous velocity and acceleration}{Let $t$ denote time, let $\Delta t$ denote a nonzero time interval, and let $\vec{r}(t)$ denote the position vector of a particle.
+
+The \emph{average velocity} from time $t$ to time $t+\Delta t$ is
+\[
+\vec{v}_{\mathrm{avg}}=\frac{\vec{r}(t+\Delta t)-\vec{r}(t)}{\Delta t}.
+\]
+If the limit exists, the \emph{instantaneous velocity} at time $t$ is the vector
+\[
+\vec{v}(t)=\lim_{\Delta t\to 0}\frac{\vec{r}(t+\Delta t)-\vec{r}(t)}{\Delta t}=\frac{d\vec{r}}{dt}.
+\]
+
+Let $\vec{v}(t)$ denote the instantaneous velocity. Then the \emph{average acceleration} from time $t$ to time $t+\Delta t$ is
+\[
+\vec{a}_{\mathrm{avg}}=\frac{\vec{v}(t+\Delta t)-\vec{v}(t)}{\Delta t}.
+\]
+If the limit exists, the \emph{instantaneous acceleration} at time $t$ is the vector
+\[
+\vec{a}(t)=\lim_{\Delta t\to 0}\frac{\vec{v}(t+\Delta t)-\vec{v}(t)}{\Delta t}=\frac{d\vec{v}}{dt}.
+\]
+The \emph{speed} at time $t$ is the scalar magnitude $|\vec{v}(t)|$.}
+
+\thm{Derivative relations in vector and component form}{Let
+\[
+\vec{r}(t)=x(t)\,\hat{\imath}+y(t)\,\hat{\jmath}+z(t)\,\hat{k},
+\]
+where $x(t)$, $y(t)$, and $z(t)$ are coordinate functions.
+
+Then the velocity vector is
+\[
+\vec{v}(t)=\frac{d\vec{r}}{dt}=\frac{dx}{dt}\,\hat{\imath}+\frac{dy}{dt}\,\hat{\jmath}+\frac{dz}{dt}\,\hat{k}.
+\]
+If $v_x(t)$, $v_y(t)$, and $v_z(t)$ denote the components of $\vec{v}(t)$, then
+\[
+v_x=\frac{dx}{dt},\qquad v_y=\frac{dy}{dt},\qquad v_z=\frac{dz}{dt}.
+\]
+The acceleration vector is
+\[
+\vec{a}(t)=\frac{d\vec{v}}{dt}=\frac{d^2\vec{r}}{dt^2}=\frac{dv_x}{dt}\,\hat{\imath}+\frac{dv_y}{dt}\,\hat{\jmath}+\frac{dv_z}{dt}\,\hat{k}.
+\]
+If $a_x(t)$, $a_y(t)$, and $a_z(t)$ denote the components of $\vec{a}(t)$, then
+\[
+a_x=\frac{dv_x}{dt}=\frac{d^2x}{dt^2},\qquad a_y=\frac{dv_y}{dt}=\frac{d^2y}{dt^2},\qquad a_z=\frac{dv_z}{dt}=\frac{d^2z}{dt^2}.
+\]
+In two-dimensional motion, the same formulas hold with the $z$-terms omitted.}
+
+\pf{Why these formulas are true}{By definition,
+\[
+\vec{v}(t)=\lim_{\Delta t\to 0}\frac{\vec{r}(t+\Delta t)-\vec{r}(t)}{\Delta t}=\frac{d\vec{r}}{dt}.
+\]
+If
+\[
+\vec{r}(t)=x(t)\,\hat{\imath}+y(t)\,\hat{\jmath}+z(t)\,\hat{k},
+\]
+then differentiating component-by-component gives
+\[
+\frac{d\vec{r}}{dt}=\frac{dx}{dt}\,\hat{\imath}+\frac{dy}{dt}\,\hat{\jmath}+\frac{dz}{dt}\,\hat{k}.
+\]
+Likewise,
+\[
+\vec{a}(t)=\lim_{\Delta t\to 0}\frac{\vec{v}(t+\Delta t)-\vec{v}(t)}{\Delta t}=\frac{d\vec{v}}{dt}.
+\]
+Differentiating the velocity components gives
+\[
+a_x=\frac{dv_x}{dt},\qquad a_y=\frac{dv_y}{dt},\qquad a_z=\frac{dv_z}{dt},
+\]
+and substituting $v_x=dx/dt$, $v_y=dy/dt$, and $v_z=dz/dt$ gives
+\[
+a_x=\frac{d^2x}{dt^2},\qquad a_y=\frac{d^2y}{dt^2},\qquad a_z=\frac{d^2z}{dt^2}.
+\]}
+
+\cor{One-dimensional motion and a speed caution}{If motion is confined to the $x$-axis, so that
+\[
+\vec{r}(t)=x(t)\,\hat{\imath},
+\]
+then
+\[
+\vec{v}(t)=v_x(t)\,\hat{\imath}\qquad\text{and}\qquad \vec{a}(t)=a_x(t)\,\hat{\imath},
+\]
+with
+\[
+v_x=\frac{dx}{dt},\qquad a_x=\frac{dv_x}{dt}=\frac{d^2x}{dt^2}.
+\]
+However, in two or three dimensions, constant speed $|\vec{v}|$ does not by itself imply zero acceleration, because the direction of $\vec{v}$ can change even when its magnitude stays the same.}
+
+\qs{Worked example}{Let $t$ denote time in seconds. A particle moves in the $xy$-plane with position vector
+\[
+\vec{r}(t)=\bigl(t^2-4t\bigr)\hat{\imath}+\bigl(3t-t^2\bigr)\hat{\jmath}\;\mathrm{m}.
+\]
+Let $x(t)=t^2-4t$ and let $y(t)=3t-t^2$, where $x(t)$ and $y(t)$ are measured in meters.
+
+Find $\vec{v}(t)$ and $\vec{a}(t)$. Then find $\vec{v}(1.0\,\mathrm{s})$, $\vec{a}(1.0\,\mathrm{s})$, and the speed at $t=1.0\,\mathrm{s}$. Interpret the signs of the velocity components at $t=1.0\,\mathrm{s}$.}
+
+\sol From
+\[
+x(t)=t^2-4t\qquad\text{and}\qquad y(t)=3t-t^2,
+\]
+the component formulas for velocity give
+\[
+v_x=\frac{dx}{dt}=2t-4\qquad\text{and}\qquad v_y=\frac{dy}{dt}=3-2t.
+\]
+Therefore,
+\[
+\vec{v}(t)=\bigl(2t-4\bigr)\hat{\imath}+\bigl(3-2t\bigr)\hat{\jmath}\;\mathrm{m/s}.
+\]
+
+Differentiate again to find the acceleration components:
+\[
+a_x=\frac{dv_x}{dt}=2\qquad\text{and}\qquad a_y=\frac{dv_y}{dt}=-2.
+\]
+So,
+\[
+\vec{a}(t)=2\hat{\imath}-2\hat{\jmath}\;\mathrm{m/s^2}.
+\]
+
+At $t=1.0\,\mathrm{s}$,
+\[
+\vec{v}(1.0\,\mathrm{s})=\bigl(2(1.0)-4\bigr)\hat{\imath}+\bigl(3-2(1.0)\bigr)\hat{\jmath}=-2\hat{\imath}+\hat{\jmath}\;\mathrm{m/s},
+\]
+and
+\[
+\vec{a}(1.0\,\mathrm{s})=2\hat{\imath}-2\hat{\jmath}\;\mathrm{m/s^2}.
+\]
+
+The speed at $t=1.0\,\mathrm{s}$ is the magnitude of the velocity vector:
+\[
+|\vec{v}(1.0\,\mathrm{s})|=\sqrt{(-2)^2+(1)^2}\;\mathrm{m/s}=\sqrt{5}\;\mathrm{m/s}.
+\]
+
+Because $v_x(1.0\,\mathrm{s})=-2\,\mathrm{m/s}$, the particle is moving in the negative $x$-direction at that instant. Because $v_y(1.0\,\mathrm{s})=1\,\mathrm{m/s}$, the particle is moving in the positive $y$-direction at that instant. So at $t=1.0\,\mathrm{s}$ the particle is moving left and upward, with speed $\sqrt{5}\,\mathrm{m/s}$. 

concepts/mechanics/u1/m1-4-motion-graphs.tex

@@ -0,0 +1,137 @@
+\subsection{Motion Graphs, Slopes, and Signed Areas}
+
+This subsection connects the local language of derivatives to the global language of accumulation. In one-dimensional motion, or when motion is analyzed along the $x$-axis, AP problems often ask you to translate among a verbal description, a graph such as $x(t)$, $v_x(t)$, or $a_x(t)$, and equations relating slope and area.
+
+\dfn{Common motion graphs and the meaning of slope}{Let $t$ denote time. Let $x(t)$ denote the position coordinate along the $x$-axis, let $v_x(t)$ denote the $x$-component of velocity, and let $a_x(t)$ denote the $x$-component of acceleration.
+
+An $x(t)$ graph shows position as a function of time. A $v_x(t)$ graph shows velocity as a function of time. An $a_x(t)$ graph shows acceleration as a function of time.
+
+If $t_1$ and $t_2$ are two times with $t_2>t_1$, then the \emph{average slope} of a graph of a quantity $q(t)$ over the interval from $t_1$ to $t_2$ is
+\[
+\frac{q(t_2)-q(t_1)}{t_2-t_1}.
+\]
+This is the slope of the secant line through the two points on the graph.
+
+The \emph{slope at a point} is the slope of the tangent line at that time. For motion graphs, the slope at a point gives an instantaneous rate of change, while the average slope over an interval gives an average rate of change over that interval.}
+
+\nt{Do not confuse a graph's \emph{value} with its \emph{slope}. On an $x(t)$ graph, a point high above the axis means large position, not large velocity. On a $v_x(t)$ graph, a point at $v_x=0$ means zero velocity at that instant, while a horizontal tangent means zero acceleration at that instant. Likewise, zero slope is not the same as zero value. Also, signed area under a $v_x(t)$ graph gives displacement, not total distance traveled. If velocity changes sign, distance in one dimension is found from $\int |v_x|\,dt$, so areas below the axis must be counted with positive magnitude when finding distance.}
+
+\mprop{Operational graph rules for one-dimensional motion}{Let $t_1$ and $t_2$ be times with $t_2>t_1$. Let $\Delta x=x(t_2)-x(t_1)$ and let $\Delta v_x=v_x(t_2)-v_x(t_1)$.
+
+For position and velocity graphs,
+\[
+\text{slope of }x(t)\text{ at time }t=v_x(t),
+\]
+so the average slope of the position graph over $[t_1,t_2]$ is the average velocity:
+\[
+\frac{x(t_2)-x(t_1)}{t_2-t_1}=\frac{\Delta x}{t_2-t_1}.
+\]
+
+For velocity and acceleration graphs,
+\[
+\text{slope of }v_x(t)\text{ at time }t=a_x(t),
+\]
+so the average slope of the velocity graph over $[t_1,t_2]$ is the average acceleration:
+\[
+\frac{v_x(t_2)-v_x(t_1)}{t_2-t_1}=\frac{\Delta v_x}{t_2-t_1}.
+\]
+
+The signed area under the velocity graph from $t_1$ to $t_2$ gives displacement:
+\[
+\Delta x=\int_{t_1}^{t_2} v_x(t)\,dt.
+\]
+Area above the time axis contributes positively, and area below the time axis contributes negatively.
+
+The signed area under the acceleration graph from $t_1$ to $t_2$ gives change in velocity:
+\[
+\Delta v_x=\int_{t_1}^{t_2} a_x(t)\,dt.
+\]
+
+In one dimension, the total distance traveled from $t_1$ to $t_2$ is
+\[
+\text{distance}=\int_{t_1}^{t_2} |v_x(t)|\,dt,
+\]
+which equals the total unsigned area between the $v_x(t)$ graph and the time axis.}
+
+\qs{Worked example}{A particle moves along the $x$-axis. Its velocity graph $v_x(t)$ is described as follows.
+
+From $t=0$ to $t=2.0\,\mathrm{s}$, the velocity increases linearly from $0$ to $4.0\,\mathrm{m/s}$. From $t=2.0\,\mathrm{s}$ to $t=5.0\,\mathrm{s}$, the velocity is constant at $4.0\,\mathrm{m/s}$. From $t=5.0\,\mathrm{s}$ to $t=7.0\,\mathrm{s}$, the velocity decreases linearly from $4.0\,\mathrm{m/s}$ to $-2.0\,\mathrm{m/s}$.
+
+Find the acceleration on each time interval, the displacement from $t=0$ to $t=7.0\,\mathrm{s}$, the total distance traveled from $t=0$ to $t=7.0\,\mathrm{s}$, and the average velocity over the full $7.0\,\mathrm{s}$ interval. State when the particle moves in the negative $x$-direction.}
+
+\sol Let $a_x$ denote the slope of the velocity graph.
+
+From $t=0$ to $t=2.0\,\mathrm{s}$,
+\[
+a_x=\frac{4.0-0}{2.0-0}=2.0\,\mathrm{m/s^2}.
+\]
+
+From $t=2.0\,\mathrm{s}$ to $t=5.0\,\mathrm{s}$, the graph is horizontal, so
+\[
+a_x=0.
+\]
+
+From $t=5.0\,\mathrm{s}$ to $t=7.0\,\mathrm{s}$,
+\[
+a_x=\frac{-2.0-4.0}{7.0-5.0}=-3.0\,\mathrm{m/s^2}.
+\]
+
+Now find displacement from the signed area under the $v_x(t)$ graph.
+
+From $t=0$ to $t=2.0\,\mathrm{s}$, the area is a triangle with base $2.0\,\mathrm{s}$ and height $4.0\,\mathrm{m/s}$:
+\[
+\Delta x_1=\frac{1}{2}(2.0)(4.0)=4.0\,\mathrm{m}.
+\]
+
+From $t=2.0\,\mathrm{s}$ to $t=5.0\,\mathrm{s}$, the area is a rectangle:
+\[
+\Delta x_2=(3.0)(4.0)=12.0\,\mathrm{m}.
+\]
+
+From $t=5.0\,\mathrm{s}$ to $t=7.0\,\mathrm{s}$, the signed area is a trapezoid:
+\[
+\Delta x_3=\frac{4.0+(-2.0)}{2}(2.0)=2.0\,\mathrm{m}.
+\]
+
+Therefore the total displacement is
+\[
+\Delta x=\Delta x_1+\Delta x_2+\Delta x_3=4.0+12.0+2.0=18.0\,\mathrm{m}.
+\]
+
+For total distance, any part of the velocity graph below the axis must be counted positively. The velocity becomes zero during the last interval, so first find that time. Starting from $v_x=4.0\,\mathrm{m/s}$ at $t=5.0\,\mathrm{s}$ with slope $-3.0\,\mathrm{m/s^2}$,
+\[
+0=4.0+(-3.0)(t-5.0).
+\]
+So,
+\[
+t-5.0=\frac{4.0}{3.0},
+\qquad
+t=\frac{19}{3}\,\mathrm{s}.
+\]
+
+From $t=5.0\,\mathrm{s}$ to $t=19/3\,\mathrm{s}$, the graph is above the axis, giving a triangle of area
+\[
+A_+=\frac{1}{2}\left(\frac{4}{3}\right)(4.0)=\frac{8}{3}\,\mathrm{m}.
+\]
+
+From $t=19/3\,\mathrm{s}$ to $t=7.0\,\mathrm{s}$, the graph is below the axis, giving a triangle with signed area $-\frac{2}{3}\,\mathrm{m}$ and magnitude
+\[
+A_-=\frac{1}{2}\left(\frac{2}{3}\right)(2.0)=\frac{2}{3}\,\mathrm{m}.
+\]
+
+Thus the total distance traveled is
+\[
+d=4.0+12.0+\frac{8}{3}+\frac{2}{3}=16.0+\frac{10}{3}=\frac{58}{3}\,\mathrm{m}\approx 19.3\,\mathrm{m}.
+\]
+
+The average velocity over the full interval is displacement divided by the total elapsed time of $7.0\,\mathrm{s}$:
+\[
+v_{x,\mathrm{avg}}=\frac{18.0\,\mathrm{m}}{7.0\,\mathrm{s}}\approx 2.57\,\mathrm{m/s}.
+\]
+
+The particle moves in the negative $x$-direction when $v_x<0$, which occurs after the graph crosses the axis. Therefore it moves in the negative $x$-direction for
+\[
+\frac{19}{3}\,\mathrm{s}<t\le 7.0\,\mathrm{s}.
+\]
+
+So the interval accelerations are $2.0\,\mathrm{m/s^2}$, $0$, and $-3.0\,\mathrm{m/s^2}$; the displacement is $18.0\,\mathrm{m}$; the total distance is $58/3\,\mathrm{m}$; and the average velocity is about $2.57\,\mathrm{m/s}$. 

concepts/mechanics/u1/m1-5-constant-acceleration.tex

@@ -0,0 +1,147 @@
+\subsection{Constant-Acceleration Motion and Free Fall}
+
+This subsection treats motion over a time interval during which an acceleration component is constant. The standard kinematic formulas are not separate facts; they are the integrated consequences of the local derivative relation between acceleration and velocity.
+
+\dfn{Constant acceleration and the free-fall approximation}{Let $t$ denote time measured from a chosen initial instant $t=0$. Let $x(t)$ and $y(t)$ denote position components, let $v_x(t)=dx/dt$ and $v_y(t)=dy/dt$ denote velocity components, and let $a_x(t)=dv_x/dt$ and $a_y(t)=dv_y/dt$ denote acceleration components.
+
+Motion has \emph{constant acceleration in the $x$-direction} if there is a constant scalar $a_x$ such that $a_x(t)=a_x$ throughout the time interval of interest. Likewise, motion has \emph{constant acceleration in the $y$-direction} if there is a constant scalar $a_y$ such that $a_y(t)=a_y$ throughout the interval.
+
+Near Earth's surface, and neglecting air resistance, \emph{free fall} is motion with constant vertical acceleration of magnitude $g\approx 9.8\,\mathrm{m/s^2}$. If the positive $y$-axis is chosen upward, then $a_y=-g$. If the positive $y$-axis is chosen downward, then $a_y=+g$.}
+
+\thm{Integrated kinematics in component form}{Let $t$ denote elapsed time from the initial instant $t=0$. Let $x_0=x(0)$ and $v_{x0}=v_x(0)$, and let $x=x(t)$ denote the later $x$-coordinate at time $t$. If $a_x$ is constant, then
+\[
+v_x=v_{x0}+a_x t,
+\]
+\[
+x=x_0+v_{x0}t+\tfrac12 a_x t^2,
+\]
+and
+\[
+v_x^2=v_{x0}^2+2a_x(x-x_0).
+\]
+
+Likewise, let $y_0=y(0)$ and $v_{y0}=v_y(0)$, and let $y=y(t)$ denote the later $y$-coordinate at time $t$. If $a_y$ is constant, then
+\[
+v_y=v_{y0}+a_y t,
+\]
+\[
+y=y_0+v_{y0}t+\tfrac12 a_y t^2,
+\]
+and
+\[
+v_y^2=v_{y0}^2+2a_y(y-y_0).
+\]
+If both $a_x$ and $a_y$ are constant, then each component evolves independently according to these formulas.}
+
+\pf{Derivation from a constant acceleration component}{Assume first that $a_x(t)=a_x$ is constant. By the local definition of acceleration,
+\[
+\frac{dv_x}{dt}=a_x.
+\]
+Integrate from the initial time $0$ to a later time $t$:
+\[
+\int_0^t \frac{dv_x}{dt'}\,dt'=\int_0^t a_x\,dt'.
+\]
+The left side is $v_x-v_{x0}$, so
+\[
+v_x-v_{x0}=a_x t,
+\]
+which gives
+\[
+v_x=v_{x0}+a_x t.
+\]
+
+Now use $dx/dt=v_x$ and substitute the result just found:
+\[
+\frac{dx}{dt}=v_{x0}+a_x t.
+\]
+Integrating from $0$ to $t$ gives
+\[
+\int_0^t \frac{dx}{dt'}\,dt'=\int_0^t \bigl(v_{x0}+a_x t'\bigr)\,dt'.
+\]
+The left side is $x-x_0$, so
+\[
+x-x_0=v_{x0}t+\tfrac12 a_x t^2,
+\]
+which gives
+\[
+x=x_0+v_{x0}t+\tfrac12 a_x t^2.
+\]
+
+To eliminate time, apply the chain rule:
+\[
+a_x=\frac{dv_x}{dt}=\frac{dv_x}{dx}\frac{dx}{dt}=v_x\frac{dv_x}{dx}.
+\]
+Integrate from $x_0$ to $x$ and from $v_{x0}$ to $v_x$:
+\[
+\int_{v_{x0}}^{v_x} v\,dv=\int_{x_0}^{x} a_x\,dx'.
+\]
+This gives
+\[
+\tfrac12\bigl(v_x^2-v_{x0}^2\bigr)=a_x(x-x_0),
+\]
+so
+\[
+v_x^2=v_{x0}^2+2a_x(x-x_0).
+\]
+The $y$-component formulas follow in exactly the same way after replacing $x$ by $y$, $v_x$ by $v_y$, and $a_x$ by $a_y$.}
+
+\cor{Vertical free-fall formulas and a sign caution}{Choose the positive $y$-axis upward. Let $y_0=y(0)$, let $v_{y0}=v_y(0)$, and let $y=y(t)$ be the height at time $t$. For free fall near Earth with negligible air resistance, $a_y=-g$, where $g>0$. Therefore,
+\[
+v_y=v_{y0}-gt,
+\]
+\[
+y=y_0+v_{y0}t-\tfrac12 gt^2,
+\]
+and
+\[
+v_y^2=v_{y0}^2-2g(y-y_0).
+\]
+Negative acceleration does not automatically mean an object is slowing down. An object slows down only when its velocity and acceleration point in opposite directions. Thus, in free fall with $a_y=-g$, an object moving upward has decreasing speed, but an object moving downward has increasing speed.}
+
+\qs{Worked example}{Choose the positive $y$-axis upward, and let $y$ denote height above the ground. A ball is thrown straight upward from a balcony. Let the initial time be $t=0$, let the initial height be $y_0=24.0\,\mathrm{m}$, let the initial vertical velocity be $v_{y0}=+12.0\,\mathrm{m/s}$, and let the constant vertical acceleration be $a_y=-9.8\,\mathrm{m/s^2}$. Neglect air resistance.
+
+Find the time when the ball hits the ground and the vertical velocity just before impact.}
+
+\sol The ball hits the ground when its height is $y=0$. Using
+\[
+y=y_0+v_{y0}t+\tfrac12 a_y t^2,
+\]
+we substitute the stated values:
+\[
+0=24.0\,\mathrm{m}+(12.0\,\mathrm{m/s})t+\tfrac12(-9.8\,\mathrm{m/s^2})t^2.
+\]
+So,
+\[
+0=24.0+12.0t-4.9t^2.
+\]
+Rewriting into standard quadratic form gives
+\[
+4.9t^2-12.0t-24.0=0.
+\]
+Using the quadratic formula,
+\[
+t=\frac{12.0\pm\sqrt{(-12.0)^2-4(4.9)(-24.0)}}{2(4.9)}
+=\frac{12.0\pm\sqrt{614.4}}{9.8}.
+\]
+This gives two mathematical roots,
+\[
+t\approx -1.30\,\mathrm{s}\qquad\text{or}\qquad t\approx 3.75\,\mathrm{s}.
+\]
+The negative time does not fit the physical situation after the throw, so the ball hits the ground at
+\[
+t\approx 3.75\,\mathrm{s}.
+\]
+
+Now find the vertical velocity at that time from
+\[
+v_y=v_{y0}+a_y t.
+\]
+Substituting the known values and the physical root gives
+\[
+v_y=(12.0\,\mathrm{m/s})+(-9.8\,\mathrm{m/s^2})(3.75\,\mathrm{s})\approx -24.8\,\mathrm{m/s}.
+\]
+Therefore, just before impact, the ball's vertical velocity is
+\[
+v_y\approx -24.8\,\mathrm{m/s},
+\]
+which means the ball is moving downward with speed $24.8\,\mathrm{m/s}$. The sign is negative because the positive axis was chosen upward. During the descent, both $v_y$ and $a_y$ are negative, so the ball speeds up even though the acceleration is negative.

concepts/mechanics/u1/m1-6-relative-projectile.tex

@@ -0,0 +1,135 @@
+\subsection{Relative Motion and Projectile Motion}
+
+This subsection combines two central AP kinematics ideas in an inertial frame: relative velocity between different observers, and two-dimensional projectile motion with negligible air resistance. In both settings, vectors are handled component-by-component, and careful notation keeps track of what is being measured.
+
+\dfn{Relative velocity and the projectile-motion setup}{Let $A$ and $B$ denote moving objects, and let $E$ denote an inertial reference frame such as Earth. If $\vec{v}_{A/E}$ denotes the velocity of object $A$ measured in frame $E$, then the \emph{relative velocity of $A$ with respect to $B$} is the velocity vector of $A$ as measured in the frame moving with $B$, written $\vec{v}_{A/B}$.
+
+For projectile motion, choose Cartesian axes before writing equations. Let $t$ denote time after launch, let $x(t)$ denote the horizontal coordinate, and let $y(t)$ denote the vertical coordinate measured upward. Let $x_0=x(0)$ and $y_0=y(0)$ denote the initial coordinates, let $v_x(t)$ and $v_y(t)$ denote the velocity components, let $\vec{v}(t)=v_x(t)\hat{\imath}+v_y(t)\hat{\jmath}$ denote the velocity vector, and let $v_{0x}=v_x(0)$ and $v_{0y}=v_y(0)$ denote the initial velocity components. In the AP model, air resistance is neglected and the only acceleration is gravity, so the acceleration vector is
+\[
+\vec{a}=-g\hat{\jmath},
+\]
+where $g$ denotes the positive magnitude of the gravitational acceleration. Thus the component equations are
+\[
+\frac{d^2x}{dt^2}=0,
+\qquad
+\frac{d^2y}{dt^2}=-g.
+\]}
+
+\thm{Relative-velocity addition and projectile component formulas}{Let $A$, $B$, and $E$ denote objects or frames in classical mechanics. Then the relative-velocity addition law is
+\[
+\vec{v}_{A/E}=\vec{v}_{A/B}+\vec{v}_{B/E}.
+\]
+Equivalently,
+\[
+\vec{v}_{A/B}=\vec{v}_{A/E}-\vec{v}_{B/E}.
+\]
+These equations are vector equations, so they may be applied component-by-component in any chosen axes.
+
+For projectile motion with the setup in the definition,
+\[
+\frac{d^2x}{dt^2}=0,
+\qquad
+\frac{d^2y}{dt^2}=-g.
+\]
+Integrating once gives the velocity components
+\[
+v_x(t)=v_{0x},
+\qquad
+v_y(t)=v_{0y}-gt.
+\]
+Integrating again gives the position components
+\[
+x(t)=x_0+v_{0x}t,
+\qquad
+y(t)=y_0+v_{0y}t-\frac{1}{2}gt^2.
+\]
+If $v_0$ denotes the initial speed and $\theta$ denotes the launch angle measured above the positive $x$-axis, then
+\[
+v_{0x}=v_0\cos\theta,
+\qquad
+v_{0y}=v_0\sin\theta.
+\]
+The horizontal and vertical motions are independent in the sense that each component has its own equation, but they are linked by the same time variable $t$.}
+
+\ex{Illustrative relative-motion example}{Let $A$ denote a student walking on a moving walkway, let $B$ denote the walkway, and let $E$ denote the ground frame. Suppose
+\[
+\vec{v}_{A/B}=3.0\hat{\jmath}\,\mathrm{m/s}
+\qquad\text{and}\qquad
+\vec{v}_{B/E}=4.0\hat{\imath}\,\mathrm{m/s}.
+\]
+Then
+\[
+\vec{v}_{A/E}=\vec{v}_{A/B}+\vec{v}_{B/E}=4.0\hat{\imath}+3.0\hat{\jmath}\,\mathrm{m/s}.
+\]
+So the student moves relative to the ground with speed
+\[
+|\vec{v}_{A/E}|=\sqrt{4.0^2+3.0^2}\,\mathrm{m/s}=5.0\,\mathrm{m/s}.
+\]}
+
+\nt{In $\vec{v}_{A/B}$, the first label tells \emph{whose} velocity is being described, and the second label tells \emph{which frame} measures it. Reversing the labels changes the meaning. In projectile motion, choose axes first so that the signs of $v_{0x}$, $v_{0y}$, and $-g$ are clear. The horizontal and vertical motions must be evaluated at the same time $t$; they are not separate motions with separate clocks. Common mistakes include adding speeds instead of velocity vectors in relative-motion problems, forgetting that $v_x$ stays constant only when air resistance is neglected, and setting $v_y=0$ for the entire flight instead of only at the top of the path.}
+
+\qs{Worked example}{A ball is launched from the top of a platform. Let $t$ denote time in seconds after launch. Choose the $x$-axis horizontal and the $y$-axis vertical upward. Let $x(t)$ and $y(t)$ denote the coordinates of the ball in meters. At $t=0$, let $x_0=0$, let $y_0=30.0\,\mathrm{m}$, let $v_{0x}=12.0\,\mathrm{m/s}$, and let $v_{0y}=5.0\,\mathrm{m/s}$. Let $g=10.0\,\mathrm{m/s^2}$.
+
+Find the time when the ball hits the ground, where the ground is given by $y=0$. Then find the horizontal distance traveled and the velocity vector just before impact.}
+
+\sol From the projectile component formulas,
+\[
+x(t)=x_0+v_{0x}t=12.0t,
+\]
+and
+\[
+y(t)=y_0+v_{0y}t-\frac{1}{2}gt^2=30.0+5.0t-5.0t^2.
+\]
+
+The ball hits the ground when $y=0$, so solve
+\[
+30.0+5.0t-5.0t^2=0.
+\]
+Divide by $5.0$:
+\[
+6+t-t^2=0.
+\]
+Rearrange:
+\[
+t^2-t-6=0.
+\]
+Factor:
+\[
+(t-3)(t+2)=0.
+\]
+Thus the two algebraic solutions are $t=3.0\,\mathrm{s}$ and $t=-2.0\,\mathrm{s}$. The negative time is not physically relevant after launch, so the impact time is
+\[
+t=3.0\,\mathrm{s}.
+\]
+
+The horizontal distance traveled is the $x$-coordinate at this time:
+\[
+x(3.0)=12.0(3.0)\,\mathrm{m}=36.0\,\mathrm{m}.
+\]
+So the ball lands $36.0\,\mathrm{m}$ horizontally from the launch point.
+
+Next find the velocity components. The horizontal component is constant:
+\[
+v_x(t)=v_{0x}=12.0\,\mathrm{m/s}.
+\]
+The vertical component is
+\[
+v_y(t)=v_{0y}-gt=5.0-10.0t.
+\]
+At impact,
+\[
+v_y(3.0)=5.0-10.0(3.0)\,\mathrm{m/s}=-25.0\,\mathrm{m/s}.
+\]
+Therefore the velocity vector just before impact is
+\[
+\vec{v}(3.0\,\mathrm{s})=12.0\hat{\imath}-25.0\hat{\jmath}\,\mathrm{m/s}.
+\]
+Its negative $y$-component shows that the ball is moving downward at impact. If the impact speed is also desired, then
+\[
+|\vec{v}(3.0\,\mathrm{s})|=\sqrt{12.0^2+(-25.0)^2}\,\mathrm{m/s}=\sqrt{769}\,\mathrm{m/s}\approx 27.7\,\mathrm{m/s}.
+\]
+
+Thus the ball hits the ground after $3.0\,\mathrm{s}$, lands $36.0\,\mathrm{m}$ from the launch point, and has impact velocity
+\[
+\vec{v}=12.0\hat{\imath}-25.0\hat{\jmath}\,\mathrm{m/s}.
+\]