Falling objects

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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 $\begin{align*} \int v'(t) dt &= \int \answer[given]{-9.8} dt\\ v(t) &= \answer[given]{-9.8 t} + C. \end{align*}$ 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 $\begin{align*} \int v'(t) dt &= \int \answer[given]{-9.8} dt\\ v(t) &= \answer[given]{-9.8 t} + C. \end{align*}$ 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. $\begin{align*} \int s'(t) dt &= \int \answer[given]{-9.8t +15} dt\\ s(t) &= \answer[given]{\frac{-9.8t^2}{2} + 15t} + D. \end{align*}$ 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.