Category Archives: Engineering Mathematics Blog

College Physics by Openstax Chapter 2 Problem 21

A well-thrown ball is caught in a well-padded mitt. If the deceleration of the ball is 2.10×104 m/s2, and 1.85 ms (1 ms = 10-3 s) elapses from the time the ball first touches the mitt until it stops, what was the initial velocity of the ball?

Solution:

We are given the following values: a=-2.10 \times 10^4 \ \text{m/s}^2; t=1.85 \times 10^{-3} \ \text{s}; v_f=0 \ \text{m/s}.

The formula in solving for the initial velocity is

v_0=v_f-at

Substitute the given values

\begin{align*}
v_0 & =0\:\text{m/s}-\left(-2.10\times 10^4\text{ m/s}^2\right)\left(1.85\times 10^{-3}\:\text{s}\right) \\
v_0 & =38.85\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
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College Physics by Openstax Chapter 2 Problem 20

An Olympic-class sprinter starts a race with an acceleration of 4.50 m/s2.

(a) What is her speed 2.40 s later?

(b) Sketch a graph of her position vs. time for this period.

Solution:

We are given \overline{a}=4.50\:\text{m/s}^2, \ \Delta t=2.40\:\sec ,\:\text{and}\: v_0=0\:\text{m/s}

Part A

The unknown is v_f. The formula in solving for v_f is

v_f=v_0+at

Substituting the given values,

\begin{align*}
v_f & =0\:\text{m/s}+\left(4.50\:\text{m/s}^2\right)\left(2.40\:\text{s}\right) \\
v_f & = 108\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part B

The relationship between position and time can be calculated using the formula

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

Then, with the given, we can express position in terms of time

\begin{align*}
x & =0+\frac{1}{2}\left(4.50\:\text{m/s}^2\right)\left(\text{t}^2\right) \\
x & =2.52\text{t}^2 \\
\end{align*}

The values of the position given the time are tabulated below

[wpdatatable id=2]

The values are plotted in the coordinate axes 

Time vs Position: College Physics 2.20 - Acceleration of an Olympic-class Sprinter
Time vs Position
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College Physics by Openstax Chapter 2 Problem 19

Assume that an intercontinental ballistic missile goes from rest to a suborbital speed of 6.50 km/s in 60.0 s (the actual speed and time are classified). What is its average acceleration in m/s2 and in multiples of g (9.80 m/s2) ?

Solution:

The formula for acceleration is 

\overline{a}=\frac{\Delta v}{\Delta t}

Substituting the given values

\begin{align*}
\overline{a} & = \frac{v_f-v_0}{\Delta t} \\
\overline{a} & =\frac{6.5\times 10^3\:\text{m/s}-0\:\text{m/s}}{60.0\:\text{sec}}\\
\overline{a} & =108.33\:\text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

This can be expressed in multiples of g

\begin{align*}
\overline{a} & = \frac{108.33\:\text{m/s}^2}{9.80\:\text{m/s}^2}\\
\overline{a} &  =11.05\text{g} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Therefore, the average acceleration is 108.33 m/s2 and can be expressed as 11.05g.

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College Physics by Openstax Chapter 2 Problem 18

A commuter backs her car out of her garage with an acceleration of 1.40 m/s2 .

(a) How long does it take her to reach a speed of 2.00 m/s?

(b) If she then brakes to a stop in 0.800 s, what is her deceleration?

Solution:

Part A

The formula for acceleration is

\overline{a}=\frac{\Delta v}{\Delta t}

If we rearrange the formula by solving for \Delta t, in terms of velocity and acceleration, we come up with

\Delta t=\frac{\Delta v}{\overline{a}}

Substituting the given values, we have

\begin{align*}
\Delta t & =\frac{\Delta v}{\overline{a}} \\
\Delta t & = \frac{v_f-v_0}{\overline{a}} \\
\Delta t & =\frac{2.00 \ \text{m/s}-0 \ \text{m/s}}{1.40 \ \text{m/s}^2} \\
\Delta t & =1.43 \ \text{seconds} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part B

The formula for acceleration (deceleration) is

\overline{a}=\frac{\Delta v}{\Delta t}

Then substituting all the given values, we have

\begin{align*}
\overline{a} & = \frac{v_f-v_0}{\Delta t} \\
\overline{a} & = \frac{0 \ \text{m/s}-2\ \text{m/s}}{0.8 \ \text{m/s}^2} \\
\overline{a} & = -2.50 \ \text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
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College Physics by Openstax Chapter 2 Problem 17

Dr. John Paul Stapp was U.S. Air Force officer who studied the effects of extreme deceleration on the human body. On December 10, 1954, Stapp rode a rocket sled, accelerating from rest to a top speed of 282 m/s (1015 km/h) in 5.00 s, and was brought jarringly back to rest in only 1.40 s! Calculate his

(a) acceleration and

(b) deceleration.

Express each in multiples of g (9.80 m/s2) by taking its ratio to the acceleration of gravity.

Solution:

Part A

The formula for acceleration is

\begin{align*}
\overline{a} & =\frac{\Delta v}{\Delta t} \\
\overline{a} & = \frac{v_f-v_0}{t_f-t_0} \\
\end{align*}

Substituting the given values

\begin{align*}
\overline{a} & =\frac{282\:\text{m/s}-0\:\text{m/s}}{5.00\:\sec } \\
\overline{a} & =56.4\:\text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part B

The deceleration is 

\begin{align*}
\overline{a} & =\frac{0\:\text{m/s}-282\:\text{m/s}}{1.40\:\text{s}} \\
\overline{a} & =-201.43\:\text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

In expressing the computed values in terms of g, we just divide them by 9.80.

The acceleration is

\overline{a}=\frac{56.4}{9.80}=5.76\text{g} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

The deceleration is

\overline{a}=\frac{201.43}{9.80}=20.55\text{g} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
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College Physics by Openstax Chapter 2 Problem 16

A cheetah can accelerate from rest to a speed of 30.0 m/s in 7.00 s. What is its acceleration?

Solution:

The formula for acceleration is 

\begin{align*}
\overline{a} & =\frac{\text{change in velocity}}{\text{change in time}}\\
\overline{a}  & =\frac{\Delta \text{v}}{\Delta t} \\
\overline{a}  & =\frac{v_f-v_0}{t_f-t_0} \\
\end{align*}

Substituting the given values

\begin{align*}
\overline{a} & =\frac{30.0\:\text{m/s}-0\:\text{m/s}}{7.00\:\text{s}} \\
& =4.29\:\text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
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College Physics by Openstax Chapter 2 Problem 15

The planetary model of the atom pictures electrons orbiting the atomic nucleus much as planets orbit the Sun. In this model, you can view hydrogen, the simplest atom, as having a single electron in a circular orbit 1.06×10-10 m in diameter.

(a) If the average speed of the electron in this orbit is known to be 2.20×106 m/s, calculate the number of revolutions per second it makes about the nucleus.

(b) What is the electron’s average velocity?

Solution:

Part A

The formula to be used is

\begin{align*}
\text{average speed} & =\frac{\text{distance}}{\text{time}} \\
r & = \frac{d}{t}
\end{align*}

Rearranging the formula–solving for the distance

\begin{align*}
d=r\times t
\end{align*}

Substituting the given values for 1 second period

\begin{align*}
d & = \left(2.20\times 10^6\:\text{m/s}\right)\left(1\:\text{s}\right) \\
& =2.20\times 10^6\:\text{meters}
\end{align*}

This is the total distance traveled in 1 sec.

With the given radius, the total distance traveled in 1 revolution is

\begin{align*}
1\:\text{revolution} & =2\pi \text{r} \\
& =\pi \text{d} \\
&=\pi \left(1.06\times 10^{-10}\text{m}\right)
\end{align*}

Therefore, the total number of revolutions traveled in 1 second is

\begin{align*}
\text{no. of revolutions} & = \frac{\text{total distance}}{\text{distance in 1 revolution}} \\
& = \frac{2.20\times 10^6}{\pi \left(1.06\times 10^{-10}\right)} \\
& =6.61\times 10^{15} \ \text{revolutions} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part B

In one complete revolution, the electron will go back to its original position. Thus, there is no net displacement. Therefore,

\begin{align*}
\overline{v} & =\frac{\Delta x}{\Delta t} \\
\overline{v} & =0 \ \text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
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College Physics by Openstax Chapter 2 Problem 14

A football quarterback runs 15.0 m straight down the playing field in 2.50 s. He is then hit and pushed 3.00 m straight backward in 1.75 s. He breaks the tackle and runs straight forward another 21.0 m in 5.20 s. Calculate his average velocity

(a) for each of the three intervals and

(b) for the entire motion.

Solution:

Part A

The average velocity for each interval is computed using the formula

\overline{v}=\frac{\Delta x}{\Delta t}

For the first interval

 \overline{v_1}=\frac{15.0\:\text{meters}}{2.50\:\sec }=6.00\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

For the second interval

\overline{v_2}=\frac{-3.00\:\text{meters}}{1.75\:\sec }=-1.71\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

For the third interval

\overline{v_3}=\frac{21.0\:\text{m}}{5.20\:\text{sec}}=4.04\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

Part B

For the entire motion, we need displacement from the beginning to the end.

\begin{align*}
\overline{v}& =\frac{\Delta x}{\Delta t} \\
& =\frac{15\:\text{m}-3\:\text{m}+21\:\text{m}}{2.50\:\text{s}+1.75\:\text{s}+5.20\:\text{s}} \\
& =\frac{33\:\text{m}}{9.45\:\text{s}} \\
& =3.49\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
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College Physics by Openstax Chapter 2 Problem 13

Conversations with astronauts on the lunar surface were characterized by a kind of echo in which the earthbound person’s voice was so loud in the astronaut’s space helmet that it was picked up by the astronaut’s microphone and transmitted back to Earth. It is reasonable to assume that the echo time equals the time necessary for the radio wave to travel from the Earth to the Moon and back (that is, neglecting any time delays in the electronic equipment). Calculate the distance from Earth to the Moon given that the echo time was 2.56 s and that radio waves travel at the speed of light 3.00×108 m/s.

Solution:

The total distance traveled is computed using the formula

\text{distance}=\text{speed}\times \text{time}

Therefore, the total distance traveled is

\begin{align*}
\text{distance} & =\left(3.00\times 10^8\:\text{m/s}\right)\left(2.56\:\sec \right) \\
& =768\:000\:000\:\text{meters}
\end{align*}

Hence, the distance between the Earth and Moon is equal to the total distance traveled divided by 2.

\begin{align*}
\text{Earth-Moon Distance} & =\frac{768\:000\:000\:\text{m}}{2} \\
& =384\:000\:000\:\text{meters} \\
& = 384 \ 000 \  \text{km} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

The distance between the Earth and Moon is approximately 384 thousand kilometers.

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College Physics by Openstax Chapter 2 Problem 12

The speed of propagation of the action potential (an electrical signal) in a nerve cell depends (inversely) on the diameter of the axon (nerve fiber). If the nerve cell connecting the spinal cord to your feet is 1.1 m long, and the nerve impulse speed is 18 m/s, how long does it take for the nerve signal to travel this distance?

Solution:

The time of travel is computed based on the formula

\text{time}=\frac{\text{distance}}{\text{speed}}

Therefore, the time of travel is

\begin{align*}
\text{time} & =\frac{1.1\:\text{m}}{18\:\text{m/s}} \\
& =0.0611\:\text{seconds} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
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