Chapter 7.3, Problem 80E

To determine

To find the position equation $s=\frac{1}{2}a{t}^2+v_{0}t+s_0$ .

Answer to Problem 80E

The position equation of the given object is, $s=-16t^2+16t+132$ .

Explanation of Solution

Given information :

General position equation is, $s=\frac{1}{2}a{t}^2+v_{0}t+s_0$ .The position of the object,

  $\begin{array}{l}t=1,s=132\\ t=2,s=100\\ t=3,s=36\end{array}$

Calculation:

Consider, the general form $s=\frac{1}{2}a{t}^2+v_{0}t+s_0$ and the position of the object

  $\begin{array}{l}t=1,s=132\\ t=2,s=100\\ t=3,s=36\end{array}$ .

Substitute the point $t=1,s=132$ in $s=\frac{1}{2}a{t}^2+v_{0}t+s_0$

  $\begin{array}{l}\Rightarrow 132=\frac{1}{2}a{\left(1\right)}^2+v_0\left(1\right)+s_0\\ \Rightarrow 132=\frac{1}{2}a+v_0+s_0---(1)\end{array}$

Substitute the point $t=2,s=100$ in $s=\frac{1}{2}a{t}^2+v_{0}t+s_0$

  $\begin{array}{l}\Rightarrow 100=\frac{1}{2}a{\left(2\right)}^2+v_0\left(2\right)+s_0\\ \Rightarrow 100=2a+2v_0+s_0---(2)\end{array}$

Substitute the point $t=3,s=36$ in $s=\frac{1}{2}a{t}^2+v_{0}t+s_0$

  $\begin{array}{l}\Rightarrow 36=\frac{1}{2}a{\left(3\right)}^2+v_0\left(3\right)+s_0\\ \Rightarrow 36=\frac{9}{2}a+3v_0+s_0---(3)\end{array}$

Hence,

  $\begin{array}{l}\frac{1}{2}a+v_0+s_0=132---(1)\\ 2a+2v_0+s_0=100---(2)\\ \frac{9}{2}a+3v_0+s_0=36---(3)\end{array}$ are the set of equation

Subtract equations $2\times \left(1\right)-\left(2\right)$ ,

  $\begin{array}{l}\begin{array}{c}a+2v_0+2s_0=264\\ 2a+2v_0+s_0=100\end{array}\\ \Rightarrow -a+s_0=164---(4)\end{array}$

By subtracting the equation $3\times \left(1\right)-\left(3\right)$ ,

  $\begin{array}{l}\begin{array}{c}\frac{3}{2}a+3v_0+3s_0=396\\ \frac{9}{2}a+3v_0+s_0=36\end{array}\\ \Rightarrow -\frac{6}{2}a+2s_0=360\\ \Rightarrow -6a+4s_0=720---(5)\end{array}$

Using $4\times \left(4\right)-\left(5\right)$ in equation (4) and (5),

  $\begin{array}{l}\begin{array}{c}-4a+4s_0=656\\ -6a+4s_0=720\end{array}\\ \Rightarrow 2a=-64\\ \Rightarrow a=-32\end{array}$

Substitute $a=-32$ in (5)then,

  $\begin{array}{l}\Rightarrow -6\left(-32\right)+4s_0=720\\ \Rightarrow 192+4s_0=720\\ \Rightarrow 4s_0=720-192\\ \Rightarrow 4s_0=528\\ \Rightarrow s_0=132\end{array}$

Now substitute $a=-32$ and $s_0=132$ in $2a+2v_0+s_0=100$ ,

  $\begin{array}{l}\Rightarrow 2\left(-32\right)+2v_0+\left(132\right)=100\\ \Rightarrow 2v_0=100-132+64\\ \Rightarrow v_0=16\end{array}$

Therefore, $a=-32,v_0=16,s_0=132$ . Substitute the values in the general form.

  $\begin{array}{l}s=\frac{1}{2}\left(-32\right)t^2+\left(16\right)t+\left(132\right)\\ \Rightarrow s=-16t^2+16t+132\end{array}$

The position equation of the given object is, $s=-16t^2+16t+132$