Tag Archives: College Physics by Openstax Solution Manual

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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College Physics by Openstax Chapter 2 Problem 24

While entering a freeway, a car accelerates from rest at a rate of 2.40 m/s² for 12.0 s.

(a) Draw a sketch of the situation.

(b) List the knowns in this problem.

(c) How far does the car travel in those 12.0 s? To solve this part, first 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, check your units, and discuss whether the answer is reasonable.

(d) What is the car’s final velocity? Solve for this unknown in the same manner as in part (c), showing all steps explicitly.

Solution:

Part A

The sketch of the situation is shown below. Also, the knowns and unknowns are in the illustration.

College Physics Problem 2.24 Illustration

From the illustration above, we can see that the initial velocity is 0 m/s, the initial time and initial distance are also zero. The final velocity and the final distance are unknowns. The time at the final location is 12 seconds and the acceleration is constant all throughout the trip at 2.40 meters per second square.

Part B

The knowns are: $a=2.40\:\text{m/s}^2$ ; $t=12.0\:\sec$ ; $v_0=0\:\text{m/s}$ ; and $x_0=0\:\text{m}$

Part C

For this part, the unknown is the value of x. If we examine the equations for constant acceleration and the given values in this problem, we can readily use the formula $x=x_0+v_0 t+\frac{1}{2}at ^2$ . That is

\begin{align*} x & =x_0+v_0 t+\frac{1}{2}at^2 \\ x & =0\:\text{m}+\left(0\:\text{m/s}\right)\left(12.0\:\text{s}\right)+\frac{1}{2}\left(2.40\:\text{m/s}^2\right)\left(12.0\:\text{s}\right)^2 \\ x & =172.8\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\ \end{align*}

Part D

For this part, the unknown is the value of v. The equation that can be used based from the known variables is v=v_0+at. That is

\begin{align*} v & =v_0+at \\ v & =0\:\text{m/s}+\left(2.40\:\text{m/s}^2\right)\left(12.0\:\text{s}\right) \\ v & =28.8\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

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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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Tag Archives: College Physics by Openstax Solution Manual

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.