Tag Archives: Solution Manual

Chemistry in Context| Essential Ideas| Chemistry| Openstax| Problem 2

Identify each of the following statements as being most similar to a hypothesis, a law, or a theory. Explain your reasoning.
(a) Falling barometric pressure precedes the onset of bad weather.
(b) All life on earth has evolved from a common, primitive organism through the process of natural selection.
(c) My truck’s gas mileage has dropped significantly, probably because it’s due for a tune-up.

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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.2, Problem 1

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PROBLEM:

Evaluate \displaystyle \lim _{x\to 2}\left(x^2-4x+3\right).

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SOLUTION:

\begin{align*}

\lim_{x\to 2}\left(x^2-4x+3\right)& = \lim_{x\to 2}\left(x^2\right)-\lim_{x\to 2}\left(4x\right)+\lim_{x\to 2}\left(3\right)\\

& =\left[\lim_{x\to 2}\left(x\right)\right]^2-4\lim_{x\to 2}\left(x\right)+3\\

& =\left(2\right)^2-4\left(2\right)+3\\

& =-1 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\

\end{align*}
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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.1, Problem 10

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PROBLEM:

If  \displaystyle f\left(x\right)=\frac{4}{x+3} and \displaystyle \:g\left(x\right)=x^2-3 , find \displaystyle f\left[g\left(x\right)\right] and \displaystyle g\left[f\left(x\right)\right].

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SOLUTION:

Part A

\begin{align*}

f\left[g\left(x\right)\right] & =\frac{4}{\left(x^2-3\right)+3}\\

& =\frac{4}{x^2} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\

\end{align*}

Part B

\begin{align*}

g\left[f\left(x\right)\right] & =\left(\frac{4}{x+3}\right)^2-3\\

& =\frac{16}{\left(x+3\right)^2}-3\\

& =\frac{16-3\left(x+3\right)^2}{\left(x+3\right)^2}\\

& =\frac{16-3\left(x^2+6x+9\right)}{\left(x+3\right)^2}\\

& =\frac{16-3x^2-18x-27}{\left(x+3\right)^2}\\

& =\frac{-3x^2-18x-11}{\left(x+3\right)^2} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\

\end{align*}
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College Physics by Openstax Chapter 2 Problem 23

a) A light-rail commuter train accelerates at a rate of 1.35 m/s2. How long does it take to reach its top speed of 80.0 km/h, starting from rest?

b) The same train ordinarily decelerates at a rate of 1.65 m/s2. How long does it take to come to a stop from its top speed?

c) In emergencies, the train can decelerate more rapidly, coming to rest from 80.0 km/h in 8.30 s. What is its emergency deceleration in m/s2?

Solution:

Part A

We are given the following: a=1.35 \ \text{m/s}^2; v_f=80.0 \ \text{km/h}; and v_0=0 \ \text{m/s}.

From the formula v_f=v_0+at, we can solve for t as

t=\frac{v_f-v_0}{a}

We need to convert 80.0 km/h to m/s first so that we have a unit uniformity for all the given values.

\begin{align*}
 80\:\text{km/hr} & =\left(80\:\text{km/hr}\right)\left(\frac{1000\:\text{m}}{1\:\text{km}}\right)\left(\frac{1\:\text{hr}}{3600\:\text{s}}\right) \\
& =22.2222\:\text{m/s}
\end{align*}

Substituting the given values into the formula, we have

\begin{align*}
t & =\frac{v_f-v_0}{a}
\\
t & =\frac{22.2222\:\text{m/s}-0\:\text{m/s}}{1.35\:\text{m/s}^2} \\
t & =16.5\:\text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part B

For this problem, we still use the formula used in Part (a). This time, the values of the final and initial velocities interchange and the value of the given deceleration is negative of acceleration. The given values are a=-1.65 \ \text{m/s}^2; v_0=22.2222 \ \text{m/s}[/katex]; and v_f=0 \ \text{m/s}

\begin{align*}
t & =\frac{v_f-v_0}{a} \\
t & =\frac{0\:\text{m/s}-22.2222\:\text{m/s}}{-1.65\:\text{m/s}^2} \\
t & =13.5\:\text{s}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part C

For this problem, we will use the formula

\begin{align*}
a & =\frac{v_f-v_0}{\Delta t}
\end{align*}

Substituting all the given values into the formula, we have

\begin{align*}
a & =\frac{0\:\text{m/s}-22.2222\:\text{m/s}}{8.30\:\text{s}} \\
a & =2.68\:\text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
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