Chapter 8.5, Problem 62E

To determine

To calculate:Value of $t_1$ , $t_2$ and $a$ .

Answer to Problem 62E

  $t_1=\frac{1152}{7}$ pounds

  $t_2=\frac{576}{7}$ pounds

  $a=-\frac{64}{7}$ feet per second squared.

Explanation of Solution

Given information:The given equations are

  $\begin{array}{l}{t}_1-2t_2=0\\ t_1-3a=192\\ t_2+2a=64\end{array}$

Calculation:

The equations are

  $\begin{array}{l}{t}_1-2t_2+0a=0\\ t_1+0t_2-3a=192\\ 0t_1+t_2+2a=64\end{array}$

Solving by Cramer’s rule we get,

  $t_1=\frac{D_1}{D}$ , $t_2=\frac{D_2}{D}$ , $a=\frac{D_3}{D}$ ……………………..(1)

where

  $D=\left|\begin{array}{ccc}1& -2& 0\\ 1& 0& -3\\ 0& 1& 2\end{array}\right|=1(0+3)+2(2-0)=3+4=7\ne 0$

  $D_1=\left|\begin{array}{ccc}0& -2& 0\\ 192& 0& -3\\ 64& 1& 2\end{array}\right|=2(192\cdot 2+3\cdot 64)=2(384+192)=2\cdot 576=1152$

  $D_2=\left|\begin{array}{ccc}1& 0& 0\\ 1& 192& -3\\ 0& 64& 2\end{array}\right|=1(192\cdot 2+3\cdot 64)=384+192=576$

  $D_3=\left|\begin{array}{ccc}1& -2& 0\\ 1& 0& 192\\ 0& 1& 64\end{array}\right|=1(0-192)+2(64-0)=-192+128=-64$

Hence, from (1) we get,

  $\begin{array}{l}{t}_1=\frac{D_1}{D}\\ =\frac{1152}{7}\end{array}$

  $\begin{array}{l}{t}_2=\frac{D_2}{D}\\ =\frac{576}{7}\end{array}$

  $\begin{array}{l}a=\frac{D_3}{D}\\ =\frac{-64}{7}\\ =-\frac{64}{7}\end{array}$

Thus, the required solution is $t_1=\frac{1152}{7}$ pounds, $t_2=\frac{576}{7}$ pounds, $a=-\frac{64}{7}$ feet per second squared.