Falling objects

$\newenvironment {prompt}{}{} \newcommand {\todo }[0]{} \newcommand {\oiint }[0]{{\large \bigcirc }\kern -1.56em\iint } \newcommand {\mooculus }[0]{\textsf {\textbf {MOOC}\textnormal {\textsf {ULUS}}}} \newcommand {\npnoround }[0]{\nprounddigits {-1}} \newcommand {\npnoroundexp }[0]{\nproundexpdigits {-1}} \newcommand {\npunitcommand }[1]{\ensuremath {\mathrm {#1}}} \newcommand {\RR }[0]{\mathbb R} \newcommand {\R }[0]{\mathbb R} \newcommand {\N }[0]{\mathbb N} \newcommand {\Z }[0]{\mathbb Z} \newcommand {\sagemath }[0]{\textsf {SageMath}} \newcommand {\d }[0]{\mathop {}\!d} \newcommand {\l }[0]{\ell } \newcommand {\ddx }[0]{\frac {d}{\d x}} \newcommand {\zeroOverZero }[0]{\ensuremath {\boldsymbol {\tfrac {0}{0}}}} \newcommand {\inftyOverInfty }[0]{\ensuremath {\boldsymbol {\tfrac {\infty }{\infty }}}} \newcommand {\zeroOverInfty }[0]{\ensuremath {\boldsymbol {\tfrac {0}{\infty }}}} \newcommand {\zeroTimesInfty }[0]{\ensuremath {\small \boldsymbol {0\cdot \infty }}} \newcommand {\inftyMinusInfty }[0]{\ensuremath {\small \boldsymbol {\infty -\infty }}} \newcommand {\oneToInfty }[0]{\ensuremath {\boldsymbol {1^\infty }}} \newcommand {\zeroToZero }[0]{\ensuremath {\boldsymbol {0^0}}} \newcommand {\inftyToZero }[0]{\ensuremath {\boldsymbol {\infty ^0}}} \newcommand {\numOverZero }[0]{\ensuremath {\boldsymbol {\tfrac {\#}{0}}}} \newcommand {\dfn }[0]{\textbf } \newcommand {\unit }[0]{\mathop {}\!\mathrm } \newcommand {\eval }[1]{\bigg [ #1 \bigg ]} \newcommand {\seq }[1]{\left ( #1 \right )} \newcommand {\epsilon }[0]{\varepsilon } \newcommand {\phi }[0]{\varphi } \newcommand {\iff }[0]{\Leftrightarrow } \DeclareMathOperator {\arccot }{arccot} \DeclareMathOperator {\arcsec }{arcsec} \DeclareMathOperator {\arccsc }{arccsc} \DeclareMathOperator {\si }{Si} \DeclareMathOperator {\scal }{scal} \DeclareMathOperator {\sign }{sign} \newcommand {\arrowvec }[1]{{\overset {\rightharpoonup }{#1}}} \newcommand {\vec }[1]{{\overset {\boldsymbol {\rightharpoonup }}{\mathbf {#1}}}} \DeclareMathOperator {\proj }{\vec {proj}} \newcommand {\veci }[0]{{\boldsymbol {\hat {\imath }}}} \newcommand {\vecj }[0]{{\boldsymbol {\hat {\jmath }}}} \newcommand {\veck }[0]{{\boldsymbol {\hat {k}}}} \newcommand {\vecl }[0]{\vec {\boldsymbol {\l }}} \newcommand {\uvec }[1]{\mathbf {\hat {#1}}} \newcommand {\utan }[0]{\mathbf {\hat {t}}} \newcommand {\unormal }[0]{\mathbf {\hat {n}}} \newcommand {\ubinormal }[0]{\mathbf {\hat {b}}} \newcommand {\dotp }[0]{\bullet } \newcommand {\cross }[0]{\boldsymbol \times } \newcommand {\grad }[0]{\boldsymbol \nabla } \newcommand {\divergence }[0]{\grad \dotp } \newcommand {\curl }[0]{\grad \cross } \newcommand {\lto }[0]{\mathop {\longrightarrow \,}\limits } \newcommand {\bar }[0]{\overline } \newcommand {\surfaceColor }[0]{violet} \newcommand {\surfaceColorTwo }[0]{redyellow} \newcommand {\sliceColor }[0]{greenyellow} \newcommand {\vector }[1]{\left \langle #1\right \rangle } \newcommand {\sectionOutcomes }[0]{} \newcommand {\HyperFirstAtBeginDocument }[0]{\AtBeginDocument } \newcommand {\epstopdfsetup }[0]{\setkeys {ETE}} changed the theorem header of amsthm failed\MessageBreak }}} \newcommand {\GetTitleStringSetup }[0]{\setkeys {gettitlestring}} \newcommand {\Sectionformat }[2]{#1}$

We study a special type of differential equation.

A differential equation is simply an equation with a derivative in it like this: $f'(x) = k f(x).$ When a mathematician solves a differential equation, they are finding a function that satisfies the equation.

Consider a falling object. Recall that the acceleration due to gravity is about $-9.8$ m/s $^2$ . Since the first derivative of the velocity function is the acceleration and the second derivative of a position function is the acceleration, we have the differential equations

$\begin{align*} v'(t) &= -9.8,\\ s''(t) &= -9.8. \end{align*}$ From these simple equations, we can derive expressions for the velocity and for the position of the object using antiderivatives.

A ball is tossed into the air with an initial velocity of $15$ m/s. What is the velocity of the ball after 1 second? How about after 2 seconds?

Knowing that the acceleration due to gravity is $-9.8$ m/s $^2$ , we write $v'(t) = \answer [given]{-9.8}.$ To solve this differential equation, take the antiderivative of both sides $\begin{align*} \int v'(t) \d t &= \int \answer[given]{-9.8} \d t\\ v(t) &= \answer[given]{-9.8 t} + C. \end{align*}$ Since it is tossed up with an initial velocity of $15$ m/s, $\answer [given]{15} = v(0) = -9.8\cdot 0 + C,$ and we see that $C=\answer [given]{15}$ . Therefore, $C$ represents the initial velocity of the ball. Hence $v(t) = -9.8t + 15$ .

Now when $t=1$ , $v(1) = 5.2$ m/s, and the ball is rising, and at $t=2$ , $v(2) = -4.6$ m/s, and the ball is falling.

Now let’s do a similar problem, but instead of finding the velocity, we will find the position.

A ball is tossed into the air with an initial velocity of $15$ m/s from a height of 2 meters. When does the ball hit the ground?

Knowing that the acceleration due to gravity is $-9.8$ m/s $^2$ , we write $v'(t) = \answer [given]{-9.8}.$ Start by taking the antiderivative of both sides of the equation $\begin{align*} \int v'(t) \d t &= \int \answer[given]{-9.8} \d t\\ v(t) &= \answer[given]{-9.8 t} + C. \end{align*}$ Here $C$ represents the initial velocity of the ball. Since it is tossed up with an initial velocity of $15$ m/s, $C = 15$ and $v(t) = -9.8t + 15.$ Since the velocity is the derivative of the position, we can write, alternatively $s'(t) = -9.8t + 15.$ Now let’s take the antiderivative again. $\begin{align*} \int s'(t) \d t &= \int \answer[given]{-9.8t +15} \d t\\ s(t) &= \answer[given]{\frac{-9.8t^2}{2} + 15t} + D. \end{align*}$ Since we know the initial height was $2$ meters, write $2 = s(0) = \frac {-9.8\cdot 0^2}{2} + 15\cdot 0 + D.$ Hence $s(t) = \frac {-9.8t^2}{2} + 15t + 2$ . We need to know when the ball hits the ground, this is when $s(t)=0$ . Solving the equation $\frac {-9.8t^2}{2} + 15t + 2 = 0$ we find two solutions $t\approx -0.1$ and $t\approx 3.2$ . Discarding the negative solution, we see the ball will hit the ground after approximately $3.2$ seconds.

The power of calculus is that it frees us from rote memorization of formulas and enables us to derive what we need.