Author Archives: Engineering Math

College Physics by Openstax Chapter 2 Problem 45


A dolphin in an aquatic show jumps straight up out of the water at a velocity of 13.0 m/s.(a) List the knowns in this problem. (b) How high does his body rise above the water? To solve this part, first note that the final velocity is now a known and identify its value. Then identify the unknown, and discuss how you chose the appropriate equation to solve for it. After choosing the equation, show your steps in solving for the unknown, checking units, and discuss whether the answer is reasonable.(c) How long is the dolphin in the air? Neglect any effects due to his size or orientation.


Solution:

We will treat the downward direction as negative, and the upward direction as positive.

Part A

The known values are:a=-9.80\:\text{m/s}^2; v_0=13\:\text{m/s}; and y_0=0\:\text{m}.

Part B

At the highest point of the jump, the velocity is equal to 0. For this part, we will treat the initial position at the moment it jumps out of the water, and the final position at the highest point. Therefore, v_f=0 \text{m/s}.

The unknown is the final position, y_f. We are going to use the formula

\left(v_f\right)^2=\left(v_0\right)^2+2a\Delta y \\
\text{or} \\
\left(v_f\right)^2=\left(v_0\right)^2+2a\left(y_f-y_0\right)

Solving for y_f in terms of the other variables:

y_f=\frac{\left(v_f\right)^2-\left(v_0\right)^2}{2a}+y_0

Substituting the given values:

\begin{align*}
y_f & =\frac{\left(v_f\right)^2-\left(v_0\right)^2}{2a}+y_0 \\
y_f & =\frac{\left(0\:\text{m/s}\right)^2-\left(13.0\:\text{m/s}\right)^2}{2\left(-9.80\:\text{m/s}^2\right)}+0\:\text{m} \\
y_f & =8.62\:\text{m}+0\:\text{m} \\
y_f & =8.62\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

This value is reasonable since dolphins can jump several meters high out of the water. Usually, a dolphin measures about 2 meters and they can jump several times their length.

Part C

The unknown is time, \Delta t. We are going to use the formula

v_f=v_0+at

Solving for time, \Delta t in terms of the other variables:

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

Substituting the given values:

\begin{align*}
t & =\frac{v_f-v_0}{a} \\
t & =\frac{0\:\text{m/s}-13.0\:\text{m/s}}{-9.80\:\text{m/s}^2} \\
t &=1.3625\:\text{s}
\end{align*}

This value is the time it takes the dolphin to reach the highest point. Since the time it takes to reach this point is equal to the time it takes to go back to the water, the time it is in the air is:

\begin{align*}
t_{air} & =2\times t \\
t_{air}&=2\times 1.3625\:\text{s} \\
t_{air}&=2.65\:\text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

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College Physics by Openstax Chapter 2 Problem 44


A rescue helicopter is hovering over a person whose boat has sunk. One of the rescuers throws a life preserver straight down to the victim with an initial velocity of 1.40 m/s and observes that it takes 1.8 s to reach the water. (a) List the knowns in this problem. (b) How high above the water was the preserver released? Note that the downdraft of the helicopter reduces the effects of air resistance on the falling life preserver, so that an acceleration equal to that of gravity is reasonable.


Solution:

We will treat the upward direction as positive, and the downward direction as negative.

Part A

The known values are: a=-9.80 \text{m/s}^2; v_0=-1.40\:\text{m/s}; \Delta t=1.8\:\text{s}; and y_f=0\:\text{m}

Part B

We are looking for the initial position, y_0. We are going to use the formula

\Delta y=v_{0y}t+\frac{1}{2}at^2 
\\
\text{or}
\\
y_f-y_0=v_{0y}t+\frac{1}{2}at^2

Solving for y_0 in terms of the other variables:

y_0=y_f-v_{0y}t-\frac{1}{2}at^2

Substituting the given values:

\begin{align*}
y_0 & =y_f-v_{0y}t-\frac{1}{2}at^2 \\
y_0& =0-\left(-1.4\:\text{m/s}\right)\left(1.8\:\text{s}\right)-\frac{1}{2}\left(-9.80\:\text{m/s}^2\right)\left(1.8\:\text{s}\right)^2 \\
y_0&= 0-\left(-1.4\:\text{m/s}\right)\left(1.8\:\text{s}\right)-\frac{1}{2}\left(-9.80\:\text{m/s}^2\right)\left(1.8\:\text{s}\right)^2 \\ 
y_0& = 0+2.52\:\text{m}+15.876\:\text{m} \\
y_0& =18.396\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\
\end{align*}

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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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