Chapter 7, Problem 53RE

To determine

The position equation of $s=\frac{1}{2}a{t}^2+v_{0}t+s_0$

Answer to Problem 53RE

The model is $s=-16t^2+150$

Explanation of Solution

Given:

The position equation of $s=\frac{1}{2}a{t}^2+v_{0}t+s_0$

Formula used:

The substitution method is used

Calculation:

The position equation is

  $s=\frac{1}{2}a{t}^2+v_{0}t+s_0$

Substitute $s\left(1\right)=134,s\left(2\right)=86,s\left(3\right)=6$ in the above model

  $\begin{array}{l}s\left(1\right)=134\to \frac{1}{2}a+v_0+s_0=\mathrm{134.............}\left(1\right)\\ s\left(2\right)=86\to 2a+2v_0+s_0=\mathrm{86.............}\left(2\right)\\ s\left(3\right)=6\to \frac{9}{2}a+3v_0+s_0=\mathrm{6.............}\left(3\right)\end{array}$

Now solve the above three equations

Multiply the 1^st equation by -4 and add it to 2^nd to estimate $a$

  $\begin{array}{l}-2a-4v_0-4s_0=-536\\ 2a+2v_0+s_0=86\\ -----------\\ -2v_0-3s_0=-450\\ 2v_0+3s_0=\mathrm{450.............}\left(4\right)\end{array}$

Multiply the 1^st equation by -9 and add it to 3^rd to eliminate a

  $\begin{array}{l}\frac{-9}{2}a-9v_0-9s_0=-1206\\ \frac{-9}{2}a+3v_0+s_0=6\\ ------------\\ -6v_{0}8s=-1200\end{array}$

  $3v_0+4s_0=\mathrm{600.................}\left(5\right)$

Nowsolve the equations (4) and (5)

Multiply the 4^th equation by -3 and 5^thequation by 2 and add to eliminate $V_0$

  $\begin{array}{l}-6v_0-9s_0=-1350\\ 6v_0+8s_0=1200\\ -----------\\ -s_0=-150\to s_0=150\end{array}$

Substitute $s_0=150$ in (4) then

  $\begin{array}{l}2v_0+3\left(150\right)=150\\ 2v_0=0\\ v_0=0\end{array}$

Substitute $v_0=0,s_0=150$ in (1)

  $\begin{array}{l}\frac{1}{2}a+0+150=134\\ \frac{a}{2}=-16\end{array}$

Hence the model is $s=-16t^2+150$

Conclusion:

The model is $s=-16t^2+150$