Derivatives in Physics

Problem 170: The height above ground of an object moving vertically is given by

(1) $\begin{equation*} s = -6t^2 + 96t + 112, \end{equation*}$

with $s$ in feet and $t$ in seconds. Find

a. the object’s velocity when $t = 0$

b. its maximum height and when it occurs;

c. its velocity when $s = 0$ .

Solution:

a) The object’s velocity is the derivative of the position. That is, when $t = 0$ ,

(2) $\begin{equation*} \frac{ds}{dt} = -32t + 96 \quad \Rightarrow \quad \frac{ds}{dt} = -32t + 96 \quad \Rightarrow \quad \frac{ds}{dt} = 96 \ \text{feet}/\text{sec} .\end{equation*}$

b) The maximum height of the object is given when $ds/dt=0$ . That is, solving for $t$ ,

(3) $\begin{equation*} \frac{ds}{dt} = -32t + 96 = 0 \quad \Rightarrow \quad t = 3. \end{equation*}$

The maximum height happens at the maximum time (t=3). Thus

(4) $\begin{equation*} s(3) = -16(3)^3 + 96(3) + 112 = 256 \ \text{feet}.\end{equation*}$

c) When $s = 0$ ,

(5) $\begin{align*} s& = -6t^2 + 96t + 112 \\ 0 & = -6t^2 + 96t + 112 \\ 0 & = t^2 -6t - 7 \\ 0& = (t-7)(t+1) \quad \Rightarrow \quad t = 7, -1. \end{align*}$

Since $t > 0$ , we take $t = 7$ . That is, the velocity when $s=0$ is given by

(6) $\begin{equation*} \frac{ds}{dt} = -32(7)+96 = -128 \ \text{ft}/\text{sec}. \end{equation*}$

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