Chapter 7.3, Problem 17E

To determine

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

Answer to Problem 17E

The position equation is $s=-32t^2+16t+132$

Explanation of Solution

Given information:

It is given that an object moving vertically is at the given heights at the specified times.

At $t=1$ second, $s=132$ feet

At $t=2$ seconds, $s=100$ feet

At $t=3$ seconds, $s=36$ feet

Calculation:

Substitute each point to produce 3 equations.

For the first equation substitute $t=1$ second, $s=132$ feet, we get,

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

For the second equation substitute $t=2$ seconds, $s=100$ feet, we get,

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

For the third equation substitute $t=3$ seconds, $s=36$ feet, we get,

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

Now subtract equation $\left(i\right)$ and $\left(ii\right)$ , we get,

  $\begin{array}{l}\frac{1}{2}a+v_0+s_0-2a-2v_0-s_0=132-100\\ \frac{-3}{2}a-v_0=32\cdots \left(iv\right)\end{array}$

Now subtract equation $\left(ii\right)$ and $\left(iii\right)$ , we get,

  $\begin{array}{l}2a+2v_0+s_0-\frac{9}{2}a-3v_0-s_0=100-36\\ \frac{-5}{2}a-v_0=64\cdots \left(v\right)\end{array}$

Subtract equations $\left(iv\right)$ and $\left(v\right)$ , we get,

  $\begin{array}{l}\frac{-3}{2}a-v_0+\frac{5}{2}a+v_0=32-64\\ \frac{2}{2}a=-32\\ a=-32\end{array}$

Now by substituting it in equation $\left(iv\right)$ or $\left(v\right)$ , we get,

  $\begin{array}{l}\frac{-3}{2}\left(-32\right)-v_0=32\\ 48-v_0=32\\ v_0=16\end{array}$

Solve for $s_0$ by substituting in $\left(i\right)$

  $\begin{array}{l}\frac{1}{2}a+v_0+s_0=132\\ \frac{1}{2}\left(-32\right)+16+s_0=132\\ -16+16+s_0=132\\ s_0=132\end{array}$

The position equation is:

  $s=-32t^2+16t+132$

Conclusion:

The position equation is $s=-32t^2+16t+132$