Author Archives: Engineering Math

College Physics by Openstax Chapter 2 Problem 43


A basketball referee tosses the ball straight up for the starting tip-off. At what velocity must a basketball player leave the ground to rise 1.25 m above the floor in an attempt to get the ball?


Solution:

It is our assumption that the player attempts to get the ball at the top where the velocity is zero.

The given are the following: v_{fy}=0 \ \text{m/s}; \Delta y=1.25 \ \text{m}; and a=-9.80 \ \text{m/s}^2.

We are required to solve for the initial velocity v_{0y} of the player. We are going to use the formula

\left(v_{fy}\right)^2=\left(v_{oy}\right)^2+2a\Delta y

Solving for v_{oy} in terms of the other variables:

v_{oy}=\sqrt{\left(v_{fy}\right)^2-2a\Delta y}

Substituting the given values:

\begin{align*}
v_{oy} & =\sqrt{\left(v_{fy}\right)^2-2a\Delta y} \\
v_{oy} & = \sqrt{\left(0\:\text{m/s}\right)^2-2\left(-9.80\:\text{m/s}^2\right)\left(1.25\:\text{m}\right)} \\
v_{oy} & =4.95 \ \text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

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College Physics by Openstax Chapter 2 Problem 42


Calculate the displacement and velocity at times of (a) 0.500, (b) 1.00, (c) 1.50, (d) 2.00, and (e) 2.50 s for a rock thrown straight down with an initial velocity of 14.0 m/s from the Verrazano Narrows Bridge in New York City. The roadway of this bridge is 70.0 m above the water.


Solution:

The given known quantities are: a=-9.8\:\text{m/s}^2; y_0=0 \ \text{m}; and v_0=-14 \ \text{m/s}.

To compute for the displacement, we use the formula

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

and to compute for the final velocity, we use the formula

v_f=v_0+at

Part A

The displacement at t=0.500 \ \text{s} is

\begin{align*}
\Delta y & =v_0t+\frac{1}{2}at^2 \\
\Delta y &=\left(-14.0\:\text{m/s}\right)\left(0.500\:\text{s}\right)+\frac{1}{2}\left(-9.8\:\text{m/s}^2\right)\left(0.500\:\text{s}\right)^2 \\
\Delta y & =-8.23\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

The velocity at t=0.500 \ \text{s} is

\begin{align*}
v_f & =v_0+at \\
&= \left(-14.0\:\text{m/s}\right)+\left(-9.8\:\text{m/s}^2\right)\left(0.500\:\text{s}\right) \\
& =-18.9\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part B

The displacement at t=1.00\ \text{s} is

\begin{align*}
\Delta y & =v_0t+\frac{1}{2}at^2 \\
\Delta y &=\left(-14.0\:\text{m/s}\right)\left(1.00\:\text{s}\right)+\frac{1}{2}\left(-9.8\:\text{m/s}^2\right)\left(1.00\:\text{s}\right)^2 \\
\Delta y & =-18.9\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

The velocity at t=1.00\ \text{s} is

\begin{align*}
v_f & =v_0+at \\
&= \left(-14.0\:\text{m/s}\right)+\left(-9.8\:\text{m/s}^2\right)\left(1.00\:\text{s}\right) \\
& =-23.8\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part C

The displacement at t=1.50\ \text{s} is

\begin{align*}
\Delta y & =v_0t+\frac{1}{2}at^2 \\
\Delta y &=\left(-14.0\:\text{m/s}\right)\left(1.50\:\text{s}\right)+\frac{1}{2}\left(-9.8\:\text{m/s}^2\right)\left(1.50\:\text{s}\right)^2 \\
\Delta y & =-32.0\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

The velocity at t=1.50\ \text{s} is

\begin{align*}
v_f & =v_0+at \\
&= \left(-14.0\:\text{m/s}\right)+\left(-9.8\:\text{m/s}^2\right)\left(1.50\:\text{s}\right) \\
& =-28.7\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part D

The displacement at t=2.00\ \text{s} is

\begin{align*}
\Delta y & =v_0t+\frac{1}{2}at^2 \\
\Delta y &=\left(-14.0\:\text{m/s}\right)\left(2.00\:\text{s}\right)+\frac{1}{2}\left(-9.8\:\text{m/s}^2\right)\left(2.00\:\text{s}\right)^2 \\
\Delta y & =-47.6\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

The velocity at t= 2.00 \ \text{s} is

\begin{align*}
v_f & =v_0+at \\
&= \left(-14.0\:\text{m/s}\right)+\left(-9.8\:\text{m/s}^2\right)\left(2.00\:\text{s}\right) \\
& =-33.6\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part E

The displacement at t=2.50\ \text{s} is

\begin{align*}
\Delta y & =v_0t+\frac{1}{2}at^2 \\
\Delta y &=\left(-14.0\:\text{m/s}\right)\left(2.50\:\text{s}\right)+\frac{1}{2}\left(-9.8\:\text{m/s}^2\right)\left(2.50\:\text{s}\right)^2 \\
\Delta y & =-65.6\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

The velocity at t= 2.50 \ \text{s} is

\begin{align*}
v_f & =v_0+at \\
&= \left(-14.0\:\text{m/s}\right)+\left(-9.8\:\text{m/s}^2\right)\left(2.50\:\text{s}\right) \\
& =-38.5\:\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 1.3, Problem 8

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PROBLEM:

Evaluate \displaystyle \lim\limits_{x\to 8}\:\frac{\sqrt[3]{x}-2}{x-8}.


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SOLUTION:

A straight substitution of x=8 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 8}\:\frac{\sqrt[3]{x}-2}{x-8}& =\lim\limits_{x\to \:8}\:\frac{\sqrt[3]{x}-2}{x-8}\cdot \frac{\sqrt[3]{x^2}+2\sqrt[3]{x}+4}{\sqrt[3]{x^2}+2\sqrt[3]{x}+4}\\
\\
& =\lim\limits_{x\to 8}\:\frac{x-8}{\left(x-8\right)\left(\sqrt[3]{x^2}+2\sqrt[3]{x}+4\right)}\\
\\

& =\lim\limits_{x\to 8}\:\frac{1}{\left(\sqrt[3]{x^2}+2\sqrt[3]{x}+4\right)}\\
\\

& =\frac{1}{\left(\sqrt[3]{8^2}+2\sqrt[3]{8}+4\right)}\\
\\

& =\frac{1}{4+4+4}\\
\\

& =\frac{1}{12} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\
\\
\end{align*}

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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.3, Problem 7

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PROBLEM:

Evaluate \displaystyle \lim\limits_{x\to 1}\:\frac{x-1}{\sqrt{x+3}-2}.


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 SOLUTION:

A straight substitution of  x=1 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 1}\:\frac{x-1}{\sqrt{x+3}-2}& =\lim\limits_{x\to \:1}\:\frac{x-1}{\sqrt{x+3}-2}\cdot \frac{\sqrt{x+3}+2}{\sqrt{x+3}+2}\\
\\
& =\lim\limits_{x\to 1}\frac{\left(x-1\right)\left(\sqrt{x+3}+2\right)}{\left(x+3\right)-2^2}\\
\\
& =\lim\limits_{x\to 1}\frac{\left(x-1\right)\left(\sqrt{x+3}+2\right)}{x-1}\\
\\
& =\lim\limits_{x\to 1}\sqrt{x+3}+2&\\
\\
& =\sqrt{1+3}+2\\
\\
&=\sqrt{4}+2\\
\\
& =2+2\\
\\
& =4 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\
\\
\end{align*}

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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.3, Problem 6

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PROBLEM:

Evaluate \displaystyle\lim\limits_{x\to 0}\:\frac{\sqrt{x+16}-4}{x}.


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 SOLUTION:

A straight substitution of x=0 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 0}\:\frac{\sqrt{x+16}-4}{x} & =\lim\limits_{x\to 0}\:\frac{\sqrt{x+16}-4}{x}\cdot \frac{\sqrt{x+16}+4}{\sqrt{x+16}+4}\\
\\
 &  =\lim\limits_{x\to 0}\:\frac{\left(x+16\right)-4^2}{x\left(\sqrt{x+16}+4\right)}\\
\\
 & =\lim\limits_{x\to 0}\:\frac{x+16-16}{x\left(\sqrt{x+16}+4\right)}\\
\\
 &  =\lim\limits_{x\to 0}\:\frac{x}{x\left(\sqrt{x+16}+4\right)}\\
\\
 &  =\lim\limits_{x\to 0}\:\frac{1}{\sqrt{x+16}+4}\\
\\
 &  =\:\frac{1}{\sqrt{0+16}+4}\\
\\
 &  =\:\frac{1}{4+4}\\
\\
 &  =\:\frac{1}{8} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\
\\
\end{align*}

 

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College Physics by Openstax Chapter 2 Problem 41


Calculate the displacement and velocity at times of (a) 0.500, (b) 1.00, (c) 1.50, and (d) 2.00 s for a ball thrown straight up with an initial velocity of 15.0 m/s. Take the point of release to be y0=0.


Solution:

The given known quantities are:a=-9.8\:\text{m/s}^2; y_o=0\:\text{m}; and v_{oy}=+15\:\text{m/s}.

To compute for the displacement, we use the formula

\Delta y=v_{oy}t+\frac{1}{2}at^2

and to compute for the final velocity, we use the formula

v_{fy}=v_{oy}+at

Part A

The displacement at t=0.500 \ \text{s} is

\begin{align*}
\Delta y & =v_ot+\frac{1}{2}at^2 \\
\Delta y  & =0\:\text{m}+\left(15.0\:\text{m/s}\right)\left(0.500\:\text{s}\right)+\frac{1}{2}\left(-9.8\:\text{m/s}^2\right)\left(0.500\:\text{s}\right)^2 \\
\Delta y & =6.28\:\text{m}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

The velocity at t=0.500 \ \text{s} is

\begin{align*}
v_{fy} & = v_{oy}+at \\
v_{fy} & =\left(15.0\:\text{m/s}\right)+\left(-9.8\:\text{m/s}^2\right)\left(0.500\:\text{s}\right) \\
v_{fy} & =10.1\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

\end{align*}

Part B

The displacement at t=1.000 \ \text{s} is

\begin{align*}
\Delta y & =v_ot+\frac{1}{2}at^2 \\
\Delta y  & =0\:\text{m}+\left(15.0\:\text{m/s}\right)\left(1.000\:\text{s}\right)+\frac{1}{2}\left(-9.8\:\text{m/s}^2\right)\left(1.000\:\text{s}\right)^2 \\
\Delta y & =10.1\:\text{m}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

The velocity at t=1.000\ \text{s} is

\begin{align*}
v_{fy} & = v_{oy}+at \\
v_{fy} & =\left(15.0\:\text{m/s}\right)+\left(-9.8\:\text{m/s}^2\right)\left(1.000\:\text{s}\right) \\
v_{fy} & =5.20\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

\end{align*}

Part C

The displacement at t=1.500\ \text{s} is

\begin{align*}
\Delta y & =v_ot+\frac{1}{2}at^2 \\
\Delta y  & =0\:\text{m}+\left(15.0\:\text{m/s}\right)\left(1.500\:\text{s}\right)+\frac{1}{2}\left(-9.8\:\text{m/s}^2\right)\left(1.500\:\text{s}\right)^2 \\
\Delta y & =11.5\:\text{m}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

The velocity at t=1.500\ \text{s} is

\begin{align*}
v_{fy} & = v_{oy}+at \\
v_{fy} & =\left(15.0\:\text{m/s}\right)+\left(-9.8\:\text{m/s}^2\right)\left(1.500\:\text{s}\right) \\
v_{fy} & =0.300\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

\end{align*}

Part D

The displacement at t=2.000\ \text{s} is

\begin{align*}
\Delta y & =v_ot+\frac{1}{2}at^2 \\
\Delta y  & =0\:\text{m}+\left(15.0\:\text{m/s}\right)\left(2.000\:\text{s}\right)+\frac{1}{2}\left(-9.8\:\text{m/s}^2\right)\left(2.000\:\text{s}\right)^2 \\
\Delta y & =10.4\:\text{m}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

The velocity at t=2.000\ \text{s} is

\begin{align*}
v_{fy} & = v_{oy}+at \\
v_{fy} & =\left(15.0\:\text{m/s}\right)+\left(-9.8\:\text{m/s}^2\right)\left(2.000\:\text{s}\right) \\
v_{fy} & =-4.600\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

\end{align*}

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College Physics by Openstax Chapter 2 Problem 40


(a) A world record was set for the men’s 100-m dash in the 2008 Olympic Games in Beijing by Usain Bolt of Jamaica. Bolt “coasted” across the finish line with a time of 9.69 s. If we assume that Bolt accelerated for 3.00 s to reach his maximum speed, and maintained that speed for the rest of the race, calculate his maximum speed and his acceleration.

(b) During the same Olympics, Bolt also set the world record in the 200-m dash with a time of 19.30 s. Using the same assumptions as for the 100-m dash, what was his maximum speed for this race?


Solution:

Part A

There are two parts to the race and must be treated separately since acceleration is not uniform over the race. We will divide the race into \Delta x_1 (while accelerating) and \Delta x_2 (with constant speed), where \Delta x_1 + \Delta x_2 = 100 \ \text{m} .

For \Delta x_1:

During the accelerating period, we are going to use the formula \Delta x=v_0t+\frac{1}{2}at^2, since we know that \displaystyle a=\frac{\Delta v}{t}=\frac{v_{max}-v_0}{t}=\frac{v_{max}}{t}; and t=3.00 \ \text{s}.

\begin{align*}
\Delta x & =v_0t+\frac{1}{2}at^2 \\
\Delta x_1 & =0+\frac{1}{2}at^2 \\
\Delta x_1 & =\frac{1}{2}at^2 \\
\Delta x_1 & =\frac{1}{2}\left(\frac{v_{max}}{t}\right)t^2 \\
\Delta x_1 & =\frac{1}{2}\left(v_{max}\right)t \\
\Delta x _1&=\frac{1}{2}\left(v_{max}\right)\left(3.00\:\text{s}\right) \\
\Delta x _1&=1.5v_{max} 
\end{align*}

When the speed is constant, t=6.69 \ \text{s}, so

\begin{align*}
\Delta x_2 & = v_{max}t \\
\Delta x_2 & = v_{max}\left(6.69\:\text{s}\right) \\
\Delta x_2 & =6.69v_{max}
\end{align*}

Plugging-in the two equations in the equation \Delta x_1 + \Delta x_2 = 100 \ \text{m} .

\begin{align*}
\Delta x_1 + \Delta x_2  & = 100 \ \text{m} \\
1.5v_{max}  + 6.69v_{max} & =100 \ \text{m} \\
8.19\:v_{max} & =100 \\
v_{max} & =\frac{100}{8.19} \\
v_{max} & =12.2\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Therefore, his acceleration can be computed using the formula

a=\frac{v_{max}}{t}

Plugging in the given values

\begin{align*}
a & =\frac{v_{max}}{t} \\
a & = \frac{12.2\:\text{m/s}}{3.00\:\text{s}} \\
a & = 4.07\:\text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part B

Similar to part (a), we can plug in the different values for time and total distance:

\begin{align*}
\Delta x_1+ \Delta x_2 &  =200 \\
1.5\:v_{max}+\left(19.30-3.00\right)v_{max} & =200 \\
1.5\:v_{max}+16.30v_{max} & =200 \\
17.80v_{max} & =200 \\
v_{max} & =\frac{200}{17.80} \\
v_{max} & = 11.2\:\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 1.3, Problem 5

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PROBLEM:

Evaluate \displaystyle \lim\limits_{x\to 0}\:\frac{\left(x+3\right)^2-9}{2x}.


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 SOLUTION:

A straight substitution of x=0 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 0}\:\frac{\left(x+3\right)^2-9}{2x} & =\:\lim\limits_{x\to 0}\:\frac{\left(x+3\right)^2-\left(3\right)^2}{2x}\\
\\
& =\:\lim\limits_{x\to 0}\:\frac{\left(x+3-3\right)\left(x+3+3\right)}{2x}\\
\\
& =\lim\limits_{x\to 0}\:\frac{x\left(x+6\right)}{2x}\\
\\
& =\lim\limits_{x\to 0}\:\frac{x+6}{2}\\
\\
& =\frac{0+6}{2}\\
\\
& =\frac{6}{2}\\
\\
& =3 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\
\end{align*}

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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.3, Problem 4

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PROBLEM:

Evaluate  \displaystyle \lim\limits_{x\to 2}\left(\frac{x^3-x^2-x-2}{2x^3-5x^2+5x-6}\right).


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 SOLUTION:

A straight substitution of x=2 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 2}\left(\frac{x^3-x^2-x-2}{2x^3-5x^2+5x-6}\right)&=\lim\limits_{x\to 2}\left(\frac{\left(x-2\right)\left(x^2+x+1\right)}{\left(x-2\right)\left(2x^2-x+3\right)}\right)\\
\\
& =\lim\limits_{x\to 2}\left(\frac{x^2+x+1}{2x^2-x+3}\right)\\
\\
& =\frac{2^2+2+1}{2\left(2\right)^2-2+3}\\
\\
& =\frac{4+2+1}{8-2+3}\\
\\
& =\frac{7}{9} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\
\\
\end{align*}

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