Chapter 12.2, Problem 66E

(a)

To determine

To find: The total distance a ball falls in 6 s.

Answer to Problem 66E

The total distance a ball falls in 6 s is 576 ft.

Explanation of Solution

The distance is increasing with every second.

Here the arithmetic regression is increasing.

Given:

Distance covered by freely falling ball in the first second is 16 ft.

Distance covered by freely falling ball in the next second is 48 ft

Distance covered by freely falling ball in the next second is 80 ft

Total time is 6 s.

Formulas used:

The nth partial sum of an arithmetic sequence is,

$S_n=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$ (1)

Calculation:

Due to the gravitational pull the ball falls with different distance in every second.

In the first second ball falls 16 ft.

So, first term of an arithmetic progression is 16.

In the next second ball falls 48 ft.

So, second term of an arithmetic progression is 48.

And in the next second ball falls 80 ft.

So, third term of an arithmetic progression is 80.

Hence, an arithmetic sequence is formed,

$16,48,80,\cdots$

The common difference is calculated as,

$\begin{array}{c}d=48-16\\ =32\end{array}$

So the common difference is 32.

And total time is 6 s, so, value of n is 6.

Substitute 16 for a, 32 for d and 6 for n in equation (1) to find the total distance a ball falls in 6 s.

$\begin{array}{c}{S}_n=\frac{6}{2}\left[2\times 16+\left(6-1\right)32\right]\\ =3\left(32+160\right)\\ =3\times 192\\ =576\end{array}$

Thus, the total distance a ball falls in 6 s is 576 ft.

(b)

To determine

To find: A formula for the total distance a ball falls in n seconds.

Answer to Problem 66E

The formula for the total distance a ball falls in n seconds is $S_n=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$.

Explanation of Solution

Given:

Distance covered by freely falling ball in the first second is 16 ft.

Distance covered by freely falling ball in the next second is 48 ft

Distance covered by freely falling ball in the next second is 80 ft

Calculation:

The first term of an arithmetic progression is denoted by a and the common difference is denoted by d.

The sum for an arithmetic progression is denoted by $S_n$.

Now the total distance a ball falls in n seconds is calculated by adding all the distance covered by the ball in each second.

$\begin{array}{c}{S}_n=a+\left(a+d\right)+\left(a+2d\right)+\cdots +\left(a+\left(n-1\right)d\right)\\ =\left(a+a+a+\cdots +a\right)n\mathrm{times}+\left[\left(1+2+\cdots +\left(n-1\right)\right)d\right]\\ =n\times a+\frac{\left(n-1\right)n}{2}d\left(\because \mathrm{sum}ofn\mathrm{terms}=\frac{n\left(n+1\right)}{2}\right)\\ =\frac{n}{2}\left[2a+\left(n-1\right)d\right]\end{array}$

Thus, the formula for the total distance a ball falls in n seconds is $S_n=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$.