\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.