# Help Me With These Simple Formulae Questions?

## Hi all, first of all I'd like to thank you for your time. Please answer all three of these questions below in as much detail as you can also including detailed steps. Thank you! Question 7 & 8 Question 8 (Review Set)

Question "7" - see below:

#### Explanation:

For the question "7":

$S = \frac{1}{2} a {t}^{2}$

solve for $t$:

$2 S = a {t}^{2}$

${t}^{2} = \frac{2 S}{a}$

$t = \sqrt{\frac{2 S}{a}}$

~~~~~

When $a = 8 , S = 40$

$t = \sqrt{\frac{2 \left(40\right)}{8}} = \sqrt{\frac{80}{8}} = \sqrt{10} \sec$

When $a = 8 , S = 10$

$t = \sqrt{\frac{2 \left(10\right)}{8}} = \sqrt{\frac{20}{8}} = \sqrt{\frac{5}{2}} \sec$

Question "8" - see below:

#### Explanation:

Question "8"

$m = {m}_{o} / \sqrt{1 - {\left(\frac{v}{c}\right)}^{2}}$

Solve for $v$:

$m \sqrt{1 - {\left(\frac{v}{c}\right)}^{2}} = {m}_{o}$

$\sqrt{1 - {\left(\frac{v}{c}\right)}^{2}} = {m}_{o} / m$

$1 - {\left(\frac{v}{c}\right)}^{2} = {\left({m}_{o} / m\right)}^{2}$

${\left(\frac{v}{c}\right)}^{2} = 1 - {\left({m}_{o} / m\right)}^{2}$

$\frac{v}{c} = \sqrt{1 - {\left({m}_{o} / m\right)}^{2}}$

$v = c \sqrt{1 - {\left({m}_{o} / m\right)}^{2}}$

~~~~~

Find $v$ as a fraction of $c$ such that $m = 3 {m}_{o}$

$\frac{v}{c} = \sqrt{1 - {\left({m}_{o} / \left(3 {m}_{o}\right)\right)}^{2}}$

$\frac{v}{c} = \sqrt{1 - {\left(\frac{1}{3}\right)}^{2}}$

$\frac{v}{c} = \sqrt{1 - \left(\frac{1}{9}\right)}$

$\frac{v}{c} = \sqrt{\frac{8}{9}}$

$v = c \frac{\sqrt{8}}{3} \approx .9428 c$

~~~~~

Find $v$, with $m = 30 {m}_{o}$ and $c \approx 3 \times {10}^{8}$

$v = \left(3 \times {10}^{8}\right) \sqrt{1 - {\left({m}_{o} / \left(30 {m}_{o}\right)\right)}^{2}}$

$v = \left(3 \times {10}^{8}\right) \sqrt{1 - {\left(\frac{1}{30}\right)}^{2}}$

$v = \left(3 \times {10}^{8}\right) \sqrt{1 - \left(\frac{1}{900}\right)}$

$v = \left(3 \times {10}^{8}\right) \sqrt{\frac{899}{900}} \approx 299 , 833 , 287 m {s}^{-} 1$

Review Set 8 - see below:

#### Explanation:

Review Set 8:

With the set of consecutive odd integers, we are given:

$1 , 3 , 5 , 7 , \ldots$

The next 3 terms are:

$9 , 11 , 13$

The $n t h$ term can be found:

$2 n - 1$ where $n$ is a natural number ($n > 0$)

For the pattern of ${T}_{1} , {T}_{2} , {T}_{3} , \ldots$

where ${T}_{1} = \frac{1}{1 \times 3} , {T}_{2} = \frac{1}{3 \times 5} , {T}_{3} = \frac{1}{5 \times 7}$

following the pattern:

${T}_{4} = \frac{1}{7 \times 9} , {T}_{5} = \frac{1}{9 \times 11} , {T}_{6} = \frac{1}{11 \times 13}$

notice that, in the denominator, the smaller of the two numbers is the $n t h$ term. For instance, for ${T}_{6} = \frac{1}{11 \times 13}$, the subscript 6 is $n$, and so the $n t h$ term is $2 n - 1 = 2 \left(6\right) - 1 = 11$.

Hence, for the 20th term, we'll have the lesser term in the denominator as:

$2 n - 1 = 2 \left(20\right) - 1 = 39$

and therefore:

${T}_{20} = \frac{1}{39 \times 41}$

and

${T}_{n} = \frac{1}{\left(2 n - 1\right) \left(2 n + 1\right)}$