Author Archives: Engineering Math

College Physics by Openstax Chapter 2 Problem 37

Dragsters can actually reach a top speed of 145 m/s in only 4.45 s—considerably less time than given in Example 2.10 and Example 2.11.

(a) Calculate the average acceleration for such a dragster.

(b) Find the final velocity of this dragster starting from rest and accelerating at the rate found in (a) for 402 m (a quarter mile) without using any information on time.

(c) Why is the final velocity greater than that used to find the average acceleration? Hint: Consider whether the assumption of constant acceleration is valid for a dragster. If not, discuss whether the acceleration would be greater at the beginning or end of the run and what effect that would have on the final velocity.

Solution:

We are given the following: v_0=0\ \text{m/s} ; v_f=145 \ \text{m/s}; and \Delta t=4.45 \ \text{sec} .

Part A

To compute for the average acceleration a, we are going to use the formula

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

Substituting the given values, we have

\begin{align*}
a & =\frac{v_f-v_0}{\Delta t} \\
a & =\frac{145\:\text{m/s}-0\:\text{m/s}}{4.45\:\text{s}} \\
a & =32.6\:\text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part B

We are given the following: a=32.6 \ \text{m/s}^2 ; v_0=0 \ \text{m/s}; and \Delta x=402 \ \text{m} .

Since we do not have any information on time, we are going to use the formula

\left(v_f\right)^2=\left(v_0\right)^2+2a\Delta x

To compute for the final velocity, we have

v_f=\sqrt{\left(v_0\right)^2+2a\Delta \:x}

Substituting the given values:

\begin{align*}
v_f & =\sqrt{\left(v_0\right)^2+2a\Delta \:x} \\
v_f & =\sqrt{\left(0\:\text{m/s}\right)^2+2\left(32.6\:\text{m/s}^2\right)\left(402\:\text{m}\right)} \\
v_f & =162\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

\end{align*}

Part C

The final velocity is greater than that used to find the average acceleration because the assumption of constant acceleration is not valid for a dragster. A dragster changes gears and would have a greater acceleration in first gear than second gear than third gear, etc. The acceleration would be greatest at the beginning, so it would not be accelerating at 32.6 m/s2 during the last few meters, but substantially less, and the final velocity would be less than  162\:\text{m/s}.

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Hydrology and Floodplain Analysis by Bedient et.al. Chapter 1 Problem 1

What is the hydrologic cycle? What are the pathways that precipitation falling onto the land surface of the Earth is dispersed to the hydrologic cycle?

Solution:

The hydrologic cycle is a continuous process in which water is evaporated from water surfaces and the oceans, moves inland as moist are masses, and produces precipitation if the correct vertical lifting conditions exist.

A portion of precipitation (rainfall) is retained in the soil near where it falls and returns to the atmosphere via evaporation (the conversion of water vapor from a water surface) and transpiration (the loss of water vapor through plant tissue and leaves). Combined loss is called evapotranspiration and is a maximum value if the water supply in the soil moisture conditions and soil may reenter channels layer as interflow or may percolate to recharge the shallow groundwater. The remaining portion of the precipitation becomes overland flow or direct runoff which flows generally in a downgradient direction to accumulate in local streams that then flow into rivers.

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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.3, Problem 1

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PROBLEM:

Evaluate \displaystyle \lim\limits_{x\to 4}\left(\frac{x^3-64}{x^2-16}\right)

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SOLUTION:

A straight substitution of  x=4 leads to the indeterminate form  \frac{0}{0} which is meaningless.

Therefore, to evaluate the limit of the given function, we proceed as follows

\begin{align*}

\lim\limits_{x\to 4}\left(\frac{x^3-64}{x^2-16}\right)& =\lim\limits_{x\to 4}\left(\frac{\left(x-4\right)\left(x^2+4x+16\right)}{\left(x+4\right)\left(x-4\right)}\right)\\
\\
& =\lim\limits_{x\to 4}\left(\frac{x^2+4x+16}{x+4}\right)\\
\\
& =\frac{\left(4\right)^2+4\left(4\right)+16}{4+4}\\
\\
& =\frac{48}{8}\\
\\
& =6 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\
\\
\end{align*}
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College Physics by Openstax Chapter 2 Problem 36

An express train passes through a station. It enters with an initial velocity of 22.0 m/s and decelerates at a rate of  0.150 m/s2 as it goes through.  The station is 210 m long.

(a) How long is the nose of the train in the station?

(b) How fast is it going when the nose leaves the station?

(c) If the train is 130 m long, when does the end of the train leave the station?

(d) What is the velocity of the end of the train as it leaves?

Solution:

Part A

We are given the following: v_0=22.0\:\text{m/s}; a=-0.150\:\text{m/s}^2; and \Delta x=210\:\text{m}

We are required to solve for time, t. We are going to use the formula

\Delta x=v_0t+\frac{1}{2}at^2

Substituting the given values, we have

\begin{align*}
\Delta x & =v_0t+\frac{1}{2}at^2 \\
210\:\text{m} & =\left(22.0\:\text{m/s}\right)t+\frac{1}{2}\left(-0.150\:\text{m/s}^2\right)t^2
\end{align*}

If we simplify and rearrange the terms into a general quadratic equation, we have

0.075t^2-0.22t+210=0

Solve for t using the quadratic formula. We are given a=0.075;\:b=-22t;\:c=210.

\begin{align*}
t & =\frac{-b\pm \sqrt{b^2-4ac}}{2a} \\
t& =\frac{-\left(-22\right)\pm \sqrt{\left(-22\right)^2-4\left(0.075\right)\left(210\right)}}{2\left(0.075\right)}\\
\end{align*}

There are two values of t that can satisfy the quadratic equation.

t=9.88\ \text{s} \qquad \text{and} \qquad t=283.46 \ \text{s}

Discard t=283.46 \ \text{s} as it can be seen from the problem that this is not a feasible solution. So, we have

t=9.88\:\text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

Part B

We have the same given values from part a. We are going to solve v_f in the formula

\left(v_f\right)^2=\left(v_0\right)^2+2a\Delta x

So we have

\begin{align*}
\left(v_f\right)^2 & =\left(v_0\right)^2+2a\Delta x \\
v_f & =\sqrt{\left(v_0\right)^2+2a\Delta x} \\
v_f & =\sqrt{\left(22.0\:\text{m/s}\right)^2+2\left(-0.150\:\text{m/s}^2\right)\left(210\:\text{m}\right)} \\
v_f & =20.6\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part C

We are given the same values as in the previous parts, except that for the value of \Delta x since we should incorporate the length of the train. For the distance, \Delta x, we have

\begin{align*}
\Delta x & =210\:\text{m}+130\:\text{m} \\
\Delta x & = 340 \ \text{m}
\end{align*}

To solve for time t, we are going to use the formula

\Delta x=v_ot+\frac{1}{2}at^2

If we rearrange the formula into a general quadratic equation and solve for t using the quadratic formula, we come up with

t=\frac{-v_0\pm \sqrt{v_0^2+2ax}}{a}

Substituting the given values:

\begin{align*}
t & =\frac{-v_0\pm \sqrt{v_0^2+2ax}}{a} \\
t & =\frac{-\left(22.0\:\text{m/s}\right)\pm \sqrt{\left(22.0\:\text{m/s}\right)^2+2\left(-0.150\:\text{m/s}^2\right)\left(340\:\text{m}\right)}}{-0.150\:\text{m/s}^2} \\
t &=16.4\:\text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part D

We have the same given values. We are going to solve for v_f in the equation

\left(v_f\right)^2=\left(v_0\right)^2+2a\Delta x

Substituting the given values:

\begin{align*}
\left(v_f\right)^2 & =\left(v_0\right)^2+2a\Delta x \\
v_f & =\sqrt{\left(v_0\right)^2+2a\Delta x} \\
v_f & =\sqrt{\left(22.0\:\text{m/s}\right)^2+2\left(-0.150\:\text{m/s}^2\right)\left(340\:\text{m}\right)} \\
v_f & =19.5\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.2, Problem 8

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PROBLEM:

Evaluate \displaystyle \lim\limits_{x\to 0}\left(\frac{3x+2}{x^2-2x+4}\right).

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SOLUTION:

Plug in the value x=0. 

\begin{align*}

\lim\limits_{x\to 0}\left(\frac{3x+2}{x^2-2x+4}\right)& =\frac{3\left(0\right)+2}{\left(0\right)^2-2\left(0\right)+4}\\
\\
& =\frac{0+2}{0-0+4}\\
\\
& =\frac{2}{4}\\
\\
& =\frac{1}{2} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\
\\

\end{align*}
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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.2, Problem 7

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PROBLEM:

Evaluate \displaystyle \lim\limits_{x\to 3}\left(\frac{\sqrt{3x}}{x\sqrt{x+1}}\right).

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SOLUTION:

Plug the value x=3.

\begin{align*}

\lim\limits_{x\to 3}\left(\frac{\sqrt{3x}}{x\sqrt{x+1}}\right)&=\frac{\sqrt{3\left(3\right)}}{3\sqrt{3+1}} \\
\\
&=\frac{\sqrt{9}}{3\sqrt{4}} \\
\\
& =\frac{3}{3\cdot 2}\\
\\
& =\frac{3}{6}\\
\\
&=\frac{1}{2} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\
\\
\end{align*}
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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.2, Problem 6

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PROBLEM:

Evaluate \displaystyle \lim_{x\to 2}\left(4x-3\right)\left(x^2+5\right).

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SOLUTION:

Plug the value x=2.

\begin{align*}

\lim\limits_{x\to 2}\left(4x-3\right)\left(x^2+5\right) & =\left[\left(4\cdot 2\right)-3\right]\left[\left(2\right)^2+5\right]\\

& =\left[8-3\right]\left[4+5\right]\\

& =\left(5\right)\left(9\right)\\

& =45 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\

\end{align*}
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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.2, Problem 5

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PROBLEM:

Evaluate \displaystyle \lim\limits_{x\to 8}\left(2x+\sqrt[3]{x}-4\right).

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SOLUTION:

Plug in the value x=8.

\begin{align*}

\lim\limits_{x\to 8}\left(2x+\sqrt[3]{x}-4\right) & = \left[2\left(8\right)+\sqrt[3]{8}-4\right]\\
& =\left[16+2-4\right]\\
& =14 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\

\end{align*}
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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.2, Problem 4

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PROBLEM:

Evaluate \displaystyle \lim\limits _{x\to \frac{\pi }{3}}\left(\frac{\sin\:2x}{\sin\:x}\right).

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SOLUTION:

Plug in the value \displaystyle x=\frac{\pi }{3}.

\begin{align*}

\lim\limits_{x\to \frac{\pi }{3}}\left(\frac{\sin\:2x}{\sin\:x}\right) & =\frac{\sin\left(2\cdot \frac{\pi }{3}\right)}{\sin\:\left(\frac{\pi }{3}\right)} \\

& =\frac{\frac{\sqrt{3}}{2}}{\frac{\sqrt{3}}{2}}\\

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

\end{align*}
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