Tag Archives: College Physics

College Physics by Openstax Chapter 2 Problem 23

a) A light-rail commuter train accelerates at a rate of 1.35 m/s2. How long does it take to reach its top speed of 80.0 km/h, starting from rest?

b) The same train ordinarily decelerates at a rate of 1.65 m/s2. How long does it take to come to a stop from its top speed?

c) In emergencies, the train can decelerate more rapidly, coming to rest from 80.0 km/h in 8.30 s. What is its emergency deceleration in m/s2?

Solution:

Part A

We are given the following: a=1.35 \ \text{m/s}^2; v_f=80.0 \ \text{km/h}; and v_0=0 \ \text{m/s}.

From the formula v_f=v_0+at, we can solve for t as

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

We need to convert 80.0 km/h to m/s first so that we have a unit uniformity for all the given values.

\begin{align*}
 80\:\text{km/hr} & =\left(80\:\text{km/hr}\right)\left(\frac{1000\:\text{m}}{1\:\text{km}}\right)\left(\frac{1\:\text{hr}}{3600\:\text{s}}\right) \\
& =22.2222\:\text{m/s}
\end{align*}

Substituting the given values into the formula, we have

\begin{align*}
t & =\frac{v_f-v_0}{a}
\\
t & =\frac{22.2222\:\text{m/s}-0\:\text{m/s}}{1.35\:\text{m/s}^2} \\
t & =16.5\:\text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part B

For this problem, we still use the formula used in Part (a). This time, the values of the final and initial velocities interchange and the value of the given deceleration is negative of acceleration. The given values are a=-1.65 \ \text{m/s}^2; v_0=22.2222 \ \text{m/s}[/katex]; and v_f=0 \ \text{m/s}

\begin{align*}
t & =\frac{v_f-v_0}{a} \\
t & =\frac{0\:\text{m/s}-22.2222\:\text{m/s}}{-1.65\:\text{m/s}^2} \\
t & =13.5\:\text{s}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part C

For this problem, we will use the formula

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

Substituting all the given values into the formula, we have

\begin{align*}
a & =\frac{0\:\text{m/s}-22.2222\:\text{m/s}}{8.30\:\text{s}} \\
a & =2.68\:\text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
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College Physics by Openstax Chapter 2 Problem 22

A bullet in a gun is accelerated from the firing chamber to the end of the barrel at an average rate of 6.20×105 m/s2 for 8.10×10-4 s . What is its muzzle velocity (that is, its final velocity)?

Solution:

We are given the following: a=6.20 \times 10^{5} \ \text{m/s}^2; \Delta t=8.10 \times 10^{-4} \ \text{s}; and v_0=0 \text{m/s}.

The muzzle velocity of the bullet is computed as follows:

\begin{align*}
v_f & =v_0+at \\
v_f & = 0\:\text{m/s}+\left(6.20\times 10^5\text{ m/s}^2\right)\left(8.10\times 10^{-4}\:\text{s}\right) \\
v_f & =502\:\text{m/s}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Therefore, the muzzle velocity, or final velocity, is 502 m/s. 

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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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