Chapter 12, Problem 54SGA

a.

To determine

To find: the distance the object will cover during 12^th second.

Answer to Problem 54SGA

368 feet

Explanation of Solution

Given information:

Let $a_n$ be the distance and $n$ be the time in seconds after the object has fallen. So the distance the object will cover during 12^th seconds is $a_{12}$ .

  $\begin{array}{l}{a}_1=16\text{ feet}\\ a_2=48\text{ feet}\\ a_3=80\text{ feet}\\ \text{.}\\ \text{.}\\ \text{.}\\ a_n\end{array}$

  $n=12$

Calculation:

Evaluate the common difference $\left(d\right)$ as follows:

  $\begin{array}{c}d=a_2-a_1\\ =48-16\\ =32\\ d=a_3-a_2\\ =80-48\\ =32\end{array}$

Thus the sequence is arithmetic series. Now substitute $n=12$ in the formula given below and evaluate $a_{12}$ as follows:

  $\begin{array}{c}{a}_n=a_1+\left(n-1\right)d\\ =16+\left(12-1\right)\left(32\right)\\ =368\end{array}$

Thus the distance the object will cover during 12^th second is 368 feet.

b.

To determine

To find: the distance the object will cover after 12 seconds.

Answer to Problem 54SGA

2304 feet

Explanation of Solution

Given information:

Let $a_n$ be the distance and $n$ be the time in seconds after the object has fallen. So the distance the object will cover during 12^th seconds is $a_{12}$ .

  $\begin{array}{l}{a}_1=16\text{ feet}\\ a_2=48\text{ feet}\\ a_3=80\text{ feet}\\ \text{.}\\ \text{.}\\ \text{.}\\ a_n=368\text{ feet}\end{array}$

  $n=12$

Calculation:

The distance covered by object after 12 seconds will be evaluated by finding the sum of the arithmetic series as follows:

  $\begin{array}{c}{S}_n=\frac{n\left(a_1+a_n\right)}{2}\\ S_{12}=\frac{12\left(a_1+a_{12}\right)}{2}\left[\text{Put }n=12\right]\\ =\frac{12\left(16+368\right)}{2}\left[\text{Put the values of }{a}_1\text{ and }{a}_{12}\right]\\ =6\left(384\right)\\ =2304\text{ feet}\end{array}$

Thus the distance the object will cover after 12 seconds is 2304 feet.