Author Archives: Engineering Math

College Physics by Openstax Chapter 2 Problem 30

A fireworks shell is accelerated from rest to a velocity of 65.0 m/s over a distance of 0.250 m.

(a) How long did the acceleration last?

(b) Calculate the acceleration.

Solution:

We are given the following values:v_0=0\:\text{m/s}; v_f=65.0\:\text{m/s}; and \Delta x=0.250\:\text{m}.

We can immediately solve for the acceleration using the given values, so we are going to answer Part B first.

Part B

Solve for the acceleration first using the formula

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

We solve for acceleration in terms of the other variables.

a=\frac{\left(v_f\right)^2-\left(v_0\right)^2}{2\Delta x}

Substitute the given values

\begin{align*}
a & =\frac{\left(v_f\right)^2-\left(v_0\right)^2}{2\Delta x} \\
a & = \frac{\left(65.0\:\text{m/s}\right)^2-\left(0\:\text{m/s}\right)^2}{2\left(0.250\:\text{m}\right)} \\
a & =8450\:\text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part A

To solve for the time of this motion, we shall use the formula

v_f=v_0+at

Solving for time, t, in terms of the other variables we have.

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

We now substitute the values given, and the computed acceleration to find the time.

\begin{align*}
t & =\frac{v_f-v_0}{a} \\
t & =\frac{65.0\:\text{m/s}-0\:\text{m/s}}{8450\:\text{m/s}^2} \\
t & =7.6922\:\times 10^{-3}\:\text{s}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
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Mortgage Project

You can buy the solution to this project for only $10 right here.

A home will likely be the biggest purchase a person ever makes, along with being the most intimidating purchase. But the suitable news is most of the problems homebuyers face have a quick solution, if accomplished before trying to get a mortgage.

Also the purpose of this project is help you to overcome the imitating home buying process, definitely it will gave a chance to get familiar with the process:

  • Check your credit report so you are mindful of what your current credit score is before for a loan. Credit reporting agencies must give you one free report annually.
  • Try to work closely with your banker to figure out how much you can borrow and which loan exactly fits you.
  • Acquire what current mortgage rates are. Bankers are there to help you understand how that translates into monthly mortgage payments and that’s the purpose of this project.

The following formula is used for figuring out a monthly home mortgage payment:

\displaystyle M=\frac{Lr\left[1+\frac{r}{12}\right]^{12t}}{12\left[\left(1+\frac{r}{12}\right)^{12t}-1\right]}

where:           L=the loan amount in dollars
                        r = the annual interest rate expressed as decimal
                        t = the number of years of the loan
                       M = the monthly payment in dollars

You are looking to buy a $325,000 home in Haverhill. If Bank of America will give them a 30-year mortgage at 6% annual interest rate for the cost of the house after they receive a 10% down payment.

A. Determine the loan amount?

B. How much their monthly payment will be?  

C. At the end of the 30-years, how much total money will you have paid to Bank of America for your home? In another word how much did the $325,000 house really cost the couple? 

D. How much interest will they have paid?       

E. How many of her monthly payment go toward the interest?

F. What percent increase over the cost of the home does this interest represent? 

G. Redo and re-answer all questions, but this time for 15 years?

Do analysis comparison between 30 and 15 years mortgage (at least one page not double spacing).

 

College Physics by Openstax Chapter 2 Problem 29

Freight trains can produce only relatively small accelerations and decelerations.

(a) What is the final velocity of a freight train that accelerates at a rate of 0.0500 m/s2 for 8.00 minutes, starting with an initial velocity of 4.00 m/s?

(b) If the train can slow down at a rate of 0.550 m/s2, how long will it take to come to a stop from this velocity?

(c) How far will it travel in each case?

Solution:

Part A

We are given the the following: a=0.0500 \ \text{m/s}^2; t=8.00 \ \text{mins}; and v_0=4.00 \ \text{m/s}.

The final velocity can be solved using the formula v_f=v_0+at. We substitute the given values.

\begin{align*}
v_f&  = v_0+at \\
v_f & = 4.00\:\text{m/s}+\left(0.0500\:\text{m/s}^2\right)\left(8.00\:\text{min}\times \frac{60\:\sec }{1\:\min }\right) \\
v_f & = 28.0 \ \text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part B

Rearrange the equation we used in part (a) by solving in terms of t, we have

\begin{align*}
t & =\frac{{v_f}-v_0}{a} \\
t & = \frac{0\:\text{m/s}-28\:\text{m/s}}{-0.550\:\text{m/s}^2} \\
t & = 50.91\:\sec\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part C

The change in position for part (a), \Delta x, or distance traveled is computed using the formula  \Delta x=v_0 t+\frac{1}{2} at^2.

\begin{align*}
 \Delta x & =v_0 t+\frac{1}{2} at^2 \\
\Delta x & =\left(4.0\:\text{m/s}\right)\left(480\:\text{s}\right)+\frac{1}{2}\left(0.0500\:\text{m/s}^2\right)\left(480\:\text{s}\right)^2 \\
 \Delta x & = 7680\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

For the situation in part (b), the distance traveled is computed using the formula \Delta x=\frac{v_f^2-v_0^2}{2 a}.

\begin{align*}
\Delta x & =\frac{\left(0\:\text{m/s}\right)^2-\left(28.0\:\text{m/s}\right)^2}{2\left(-0.550\:\text{m/s}^2\right)} \\
\Delta x  & =712.73\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
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College Physics by Openstax Chapter 2 Problem 28

A powerful motorcycle can accelerate from rest to 26.8 m/s (100 km/h) in only 3.90 s.

(a) What is its average acceleration?

(b) How far does it travel in that time?

Solution:

We are given the following: v_0=0 \ \text{m/s}; v_f=26.8 \ \text{m/s}; and t=3.90\ \text{s}.

Part A

The average acceleration of the motorcycle can be solved using the equation \overline{a}=\frac{\Delta v}{\Delta t}. Substitute the given into the equation. That is,

\begin{align*}
\overline{a} & =\frac{\Delta v}{\Delta t} \\
\overline{a} & =\frac{26.8\:\text{m/s}-0\:\text{m/s}}{3.90\:\text{s}} \\
\overline{a} & =6.872\:\text{m/s}^2\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part B

The distance traveled is equal to the average velocity multiplied by the time of travel. That is,

\begin{align*}
\Delta x & =v_{ave}t\\
\Delta x & =\left(\frac{0\:\text{m/s}+26.8\:\text{m/s}}{2}\right)\left(3.90\:\text{s}\right) \\
\Delta x & =52.26\:\text{m}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\
\end{align*}
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College Physics by Openstax Chapter 2 Problem 27

In a slap shot, a hockey player accelerates the puck from a velocity of 8.00 m/s to 40.0 m/s in the same direction. If this shot takes 3.33×10-2 s, calculate the distance over which the puck accelerates.

Solution:

The best equation that can be used to solve this problem is \Delta x=v_{ave} t. That is,

\begin{align*}
\Delta x & = v_{ave} t \\
\Delta x & = \left(\frac{8\:\text{m/s}+40\:\text{m/s}}{2}\right)\left(3.33\times 10^{-2}\:\text{s}\right) \\
\Delta x & = 0.7992\:\text{m}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Therefore, the distance over which the puck accelerates is 0.7992 meters.

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College Physics by Openstax Chapter 2 Problem 26

Blood is accelerated from rest to 30.0 cm/s in a distance of 1.80 cm by the left ventricle of the heart.

(a) Make a sketch of the solution.

(b) List the knowns in this problem.

(c) How long does the acceleration take? To solve this part, identify the unknown, and then discuss how you chose the appropriate equation to solve for it. After choosing the equation, show your steps in solving for the unknown, checking your units.

(d) Is the answer reasonable when compared with the time for a heartbeat?

Solution:

Part A

The sketch should contain the starting point and the final point. This will be done by connecting a straight line from the starting point to the final point. The sketch is shown below.

Part B

The list of known variables are:

Initial velocity: v_0=0\:\text{m/s}
Final Velocity: v_f=30.0\:\text{cm/s}
Distance Traveled: x-x_0=1.80\:\text{cm}

Part C

The best equation to solve for this is \Delta \text{x}=\text{v}_{\text{ave}}\text{t} where v_{ave} is the average velocity, and t is time. That is

\begin{align*}
\Delta x & =v_{ave} t \\
t &=\frac{\Delta x}{v_{ave}} \\
t & =\frac{1.80\:\text{cm}}{\frac{\left(0\:\text{cm/s}+30\:\text{cm/s}\right)}{2}}\\
t & =0.12\:\text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\
\end{align*}

Part D

Since the computed value of the time for the acceleration of blood out of the ventricle is only 0.12 seconds (only a fraction of a second), the answer seems reasonable. This is due to the fact that an entire heartbeat cycle takes about one second. So, the answer is yes, the answer is reasonable.

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College Physics by Openstax Chapter 2 Problem 25

At the end of a race, a runner decelerates from a velocity of 9.00 m/s at a rate of 2.00 m/s2.

(a) How far does she travel in the next 5.00 s?

(b) What is her final velocity?

(c) Evaluate the result. Does it make sense?

Solution:

We are given the following: v_0=9.00 \ \text{m/s}; and a=2.00 \ \text{m/s}^2.

Part A

For this part, we are given t=5.00 \ \text{s} and we shall use the formula  x=x_0+v_0 t+\frac{1}{2}at^2.

\begin{align*}
x & =x_0+v_0 t+\frac{1}{2}at^2 \\
x & =0\:\text{m}+\left(9.00\:\text{m/s}\right)\left(5.00\:\text{s}\right)+\frac{1}{2}\left(-2.00\:\text{m/s}^2\right)\left(5.00\:\text{s}\right)^2 \\
x & =20\:\text{meters} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\
\end{align*}

Part B

The final velocity can be determined using the formula v_f=v_0+at.

\begin{align*}
v_f & =v_0+at \\
v_f & =9.00\:\text{m/s}+\left(-2.00\:\text{m/s}^2\right)\left(5.00\:\text{s}\right) \\
v_f & =-1\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\
\end{align*}

Part C

The result says that the runner starts at the rate of 9 m/s and decelerates at 2 m/s2. After some time, the velocity is already negative. This does not make sense because if the velocity is negative, that means that the runner is already running backwards.

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