Chapter 7, Problem 54RE

To determine

Find the position equation.

Answer to Problem 54RE

  $\boxed{s=-16t^2-20t+220}$

Explanation of Solution

Given information:

An object moving vertically is at the given heights at the specified times. Find the position equation

  $s=\frac{1}{2}a{t}^2+v_{0}t+s_0$ for the object.

At

  $\begin{array}{l}t=1\sec ond,s=184feet\\ t=2\sec ond,s=116feet\\ t=3\sec ond,s=16feet\end{array}$

Calculation:

The position equation is,

  $s=\frac{1}{2}a{t}^2+v_{0}t+s_0$

Substitute $s\left(1\right)=184,s\left(2\right)=116,s\left(3\right)=16$ to generate system of equation in above model.

  $\begin{array}{l}s\left(1\right)=184\to \frac{1}{2}a+v_0+s_0=\mathrm{184....}\left(1\right)\\ s\left(2\right)=116\to 2a+2v_0+s_0=\mathrm{116.....}\left(2\right)\\ s\left(3\right)=16\to \frac{9}{2}a+3v_0+s_0=\mathrm{16......}\left(3\right)\end{array}$

Multiply equation $\left(1\right)$ by $-4$ and add with equation $\left(2\right)$

  $\begin{array}{l}\underset{\_}{\begin{array}{l}-2a-4v_0-4s_0=-736\\ 2a+2v_0+s_0=116\end{array}}\\ -2v_0-3s_0=-620\\ 2v_0+3s_0=\mathrm{620.......}\left(4\right)\end{array}$

Multiply equation $\left(1\right)$ by $-9$ and add with equation $\left(3\right)$

  $\begin{array}{l}\underset{\_}{\begin{array}{l}-\frac{9}{2}a-9v_0-9s_0=-1656\\ \frac{9}{2}a+3v_0+s_0=16\end{array}}\\ -6v_0-8s_0=-1640\\ 3v_0+4s_0=\mathrm{820......}\left(5\right)\end{array}$

Multiply equation $\left(4\right)$ by $-3$ and equation $\left(5\right)$ by $2$ add with equations.

  $\begin{array}{l}\underset{\_}{\begin{array}{l}-6v_0-9s_0=-1860\\ 6v_0+8s_0=1640\end{array}}\\ -s_0=-220\\ s_0=220\end{array}$

Substitute $s_0=220$ in equation $\left(4\right)$

  $\begin{array}{l}2v_0+3\times 220=620\\ 2v_0=-40\\ v_0=-20\end{array}$

Substitute, $s_0=220$ , $v_0=-20$ in equation $\left(1\right)$

  $\begin{array}{l}\frac{1}{2}a-20+220=184\\ \frac{a}{2}=-16\end{array}$

Hence, the position equation is $\boxed{s=-16t^2-20t+220}$