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College Physics by Openstax Chapter 2 Problem 55

Suppose you drop a rock into a dark well and, using precision equipment, you measure the time for the sound of a splash to return.
(a) Neglecting the time required for sound to travel up the well, calculate the distance to the water if the sound returns in 2.0000 s.
(b) Now calculate the distance taking into account the time for sound to travel up the well. The speed of sound is 332.00 m/s in this well.

Solution:

Part A

Consider Figure A.

We shall consider two points for our solution. First, position 1 is the top of the well. In this position, we know that y₁=0, t₁=0 and v_y1=0.

Position 2 is located at the top of the water table where the rock will meet the water. Since we neglect the time for the sound to travel from position 2 to position 1, we can say that t₂=2.0000 s, the time of the rock to reach this position.

Solving for the value of y₂ will determine the distance between the two positions.

\begin{align*} \Delta y & = v_{y_1}t+\frac{1}{2}at^2 \\ y_2-y_1 & = v_{y_1}t+\frac{1}{2}at^2 \\ y_2 & = y_1 +v_{y_1}t+\frac{1}{2}at^2 \\ y_2 & = 0+0+\frac{1}{2}\left( -9.81\ \text{m/s}^2 \right)\left( 2.0000\ \text{s} \right)^2 \\ y_2 & = -19.6\ \text{m} \qquad {\color{DarkOrange} \left( \text{Answer }\right)} \end{align*}

Position 2 is 19.6 meters measured downward from position 1.

$\therefore$ The distance to the water is about 19.6 meters.

Part B

For this case, the 2.0000 seconds that is given includes the time that the rock travels from position 1 to position 2, t_r, and the time that the sound travels from position 2 to position 1, t_s.

\begin{align*} t_r+t_s & =2.0000\ \text{s} \\ t_s & = 2.0000\ \text{s}-t_r \end{align*}

Considering the motion of the rock from position 1 to position 2.

\begin{align*} \Delta y & = v_{y_1}t_r+\frac{1}{2}a\left( t_r \right)^2 \\ y_2-y_1 & = v_{y_1}t_r+\frac{1}{2}a\left( t_r \right)^2 \\ y_2 & = y_1 +v_{y_1}t_r+\frac{1}{2}a\left( t_r \right)^2 \\ y_2 & = 0+0+\frac{1}{2}\left( -9.81\ \text{m/s}^2 \right)\left( t_r \right)^2 \\ y_2 & = -4.905\left( t_r \right)^2 \qquad {\color{Blue} \text{Equation 1}}\\ \end{align*}

Now, let us consider the motion of the sound from position 2 to position 1. Sound is assumed to have a constant velocity of 322.00 m/s.

\begin{align*} \Delta y & = v_s \times t_s \\ y_1-y_2 & =\left( 322.00\ \text{m/s} \right)\left( t_s \right) \\ 0-y_2 & =\left( 322.00 \right)\left( 2.0000-t_r \right) \\ y_2 & = -322.00\left( 2.0000-t_r \right) \qquad {\color{Blue} \text{Equation 2}} \end{align*}

So, we have two equations from the two motions. We can solve the equations simultaneously.

\begin{align*} -4.905 \left( t_r \right)^2 & = -322.00\left( 2.0000-t_r \right) \\ 4.905 \left( t_r \right)^2 & =322.00\left( 2.0000-t_r \right) \\ 4.905 \left( t_r \right)^2 & = 644.00-322.00t_r \\ 4.905\left( t_r \right)^2 + 322.00t_r-644.00 & = 0 \\ \end{align*}

We can solve the quadratic formula using the quadratic equation.

\begin{align*} t_r & = \frac{-b \pm\sqrt{b^2-4ac}}{2a} \\ t_r & = \frac{-322.00\pm\sqrt{\left( 322.00 \right)^2-4\left( 4.905 \right)\left( -644.00 \right)}}{2\left( 4.905 \right)}\\ t_r & =1.9425 \ \text{s} \end{align*}

Now that we have solved for the value of t_r, we can use this to solve for y₂ using either Equation 1 or Equation 2. We will use equation 1.

\begin{align*} y_2 & = -4.905\left( t_r \right)^2 \\ y_2 & = -4.905 \left( 1.9425 \right)^2 \\ y_2 & =-18.5 \ \text{m} \qquad {\color{DarkOrange} \left( \text{Answer} \right)} \end{align*}

Position 2 is about 18.5 meters below position 1.

$\therefore$ In this case, the distance between the two positions is 18.5 meters.

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College Physics by Openstax Chapter 2 Problem 54

A ball is thrown straight up. It passes a 2.00-m-high window 7.50 m off the ground on its path up and takes 0.312 s to go past the window. What was the ball’s initial velocity? Hint: First consider only the distance along the window, and solve for the ball’s velocity at the bottom of the window. Next, consider only the distance from the ground to the bottom of the window, and solve for the initial velocity using the velocity at the bottom of the window as the final velocity.

Solution:

First, we have the position 1 where the motion starts. Here, we know that y<sub>1</sub>=0, t<sub>1</sub>=0, and v<sub>y1</sub>=0.

Position 2 is at the bottom of the window. We know that it is 7.50 meters from where the motion started. So we have y<sub>2</sub>=7.50 meters. We do not know the time and velocity at this point.

Then we have position 3 at the top of the window where the overall height is 9.50 meters, y<sub>3</sub>=9.50. We also do not know the velocity and time elapsed in this position. — *Figure A*

Consider Figure A. We shall be considering the three positions shown.

First, we have position 1 where the motion starts. Here, we know that y₁=0 and t₁=0, but we do not know v_y1.

Position 2 is at the bottom of the window. We know that it is 7.50 meters from where the motion started. So we have y₂=7.50 meters. We do not know the time and velocity at this point.

Then we have position 3 at the top of the window where the overall height is 9.50 meters, y₃=9.50. We also do not know the velocity and time elapsed in this position.

Consider positions 2 and 3. The initial position in this case is at position 2 and the final position is at position 3. We know that the difference of time between this two positions is 0.312 seconds. We can say that

t_3 =t_2+0.312 \ \text{s} \\ t_3-t_2 = 0.312\ \text{s}

Using the same 2 positions still, we have

\begin{align*} y_3 & = y_2 + v_{y_2} \Delta t+\frac{1}{2}a\left( \Delta t \right)^2 \\ 9.50\ \text{m} & = 7.50\ \text{m} + v_{y_2} \left( t_3-t_2 \right)+\frac{1}{2}a\left( t_3-t_2 \right)^2 \\ 9.50\ \text{m}-7.50\ \text{m} & = v_{y_2}\left( 0.312\ \text{s} \right)+\frac{1}{2}\left( -9.81\ \text{m/s}^2 \right)\left( 0.312\ \text{s} \right)^2\\ 2.00\ \text{m} & = 0.312\ \text{s} \left( v_{y_2} \right)-0.4775\ \text{m} \\ 0.312\ \text{s} \left( v_{y_2} \right) & = 2.00\ \text{m}+0.4775\ \text{m} \\ 0.312\ \text{s} \left( v_{y_2} \right) & = 2.4775\ \text{m} \\ v_{y_2}& =\frac{2.4775\ \text{m}}{0.312\ \text{s}} \\ v_{y_2}& = 7.94\ \text{m/s} \end{align*}

We have computed the velocity of the ball at the bottom of the window.

Next, we shall consider positions 1 and 2. In this consideration, position 1 will be considered the initial position while position2 is the final position.

\begin{align*} \left( v_{y_2} \right)^2 & = \left( v_{y_1} \right)^2 +2a \Delta y \\ \left( 7.94\ \text{m/s} \right)^2 & = \left( v_{y_1} \right)^2 + 2\left( -9.81 \ \text{m/s}^2 \right)\left( 7.50\ \text{m}-0 \right)\\ \left( v_{y_1} \right)^2 & = \left( 7.94\ \text{m/s} \right)^2- 2\left( -9.81 \ \text{m/s}^2 \right)\left( 7.50\ \text{m}-0 \right)\\ v_{y_1} & = + \sqrt{ \left( 7.94\ \text{m/s} \right)^2- 2\left( -9.81 \ \text{m/s}^2 \right)\left( 7.50\ \text{m}-0 \right)} \\ v_{y_1} & = + 14.5\ \text{m/s}\qquad {\color{DarkOrange} \left( \text{Answer} \right) } \end{align*}

$\therefore$ The ball’s initial velocity is about 14.5 m/s upward.

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College Physics by Openstax Chapter 2 Problem 53

There is a 250-m-high cliff at Half Dome in Yosemite National Park in California. Suppose a boulder breaks loose from the top of this cliff.
(a) How fast will it be going when it strikes the ground?
(b) Assuming a reaction time of 0.300 s, how long will a tourist at the bottom have to get out of the way after hearing the sound of the rock breaking loose (neglecting the height of the tourist, which would become negligible anyway if hit)? The speed of sound is 335 m/s on this day.

Solution:

Part A

Consider the Figure A. We shall consider two position points — position 1 at the top of the cliff, and position 2 on the ground.

Position 1 is 250 meters above the ground (y=250 m), and since this is the initial position, t=0, and the initial velocity is v_y=0.

Position 2 is on the ground (y=0), and we do not know the time and velocity at this point.

For this part, we will solve for the value of v_y at position 2.

\begin{align*} \left( v_{y_{2}} \right)^2 & = \left( v_{y_{1}} \right)^2+2a \Delta y \\ \left( v_{y_{2}} \right)^2 & = \left( 0\ \text{m/s} \right)^2+2\left( -9.81\ \text{m/s}^2 \right)\left( 0\ \text{m}-250\ \text{m} \right) \\ \left( v_{y_{2}} \right)^2 & = 4905\ \text{m}^2/\text{s}^2 \\ v_{y_{2}} & = \pm \sqrt{4905\ \text{m}^2/\text{s}^2} \\ v_{y_{2}} & = \pm \ 70.0\ \text{m/s}\\ v_{y_{2}} & = - 70.0\ \text{m/s} \qquad {\color{gold}\left( \text{Answer} \right) }\\ \end{align*}

Since the boulder is moving downward at this point, we shall consider the negative sign of the velocity as an indication of the downward direction of the velocity.

To answer the question, the boulder is going at about 70.0 m/s downward when it strikes the ground.

Part B

Let:
$t_s$ be the time for the sound to travel from the top of the cliff to the tourist;
$t$ be the total time elapsed before the tourist can react;
$t_b$ be the time for the rock to travel from the top of the cliff to the ground
$t_t$ be the amount of time the tourist has to get out of the way after hearing the sound of the rock breaking loose

Since we are given the speed of sound, we have

\begin{align*} t_s & = \frac{\text{height of the cliff}}{\text{speed of sound}} \\ t_s & = \frac{250\ \text{m}}{335\ \text{m/s}}\\ t_s & =0.746 \ \text{s} \end{align*}

So, the tourist can react after

\begin{align*} t & =t_s + \text{reaction time} \\ t & = 0.746\ \text{s}+0.300\ \text{s} \\ t &= 1.046\ \text{s} \end{align*}

Then, we can compute for the total time it takes for the rock to travel from top to the ground.

\begin{align*} t_b & =\frac{v_{y_2}-v_{y_1}}{a} \\ t_b & =\frac{-70\ \text{m/s}-0\ \text{m/s}}{-9.81\ \text{m/s}^2} \\ t_b & = 7.14\ \text{s} \end{align*}

Therefore, the tourist still has the time to get out of the way. That is,

\begin{align*} t_t & =7.14\ \text{s}-1.046\ \text{s} \\ t_t & = 6.09\ \text{s} \qquad {\color{gold}\left( \text{Answer} \right) } \\ \end{align*}

$\therefore$ the tourist has about 6.09 seconds to get out of the way.

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College Physics by Openstax Chapter 2 Problem 52

An object is dropped from a height of 75.0 m above ground level. (a) Determine the distance traveled during the first second. (b) Determine the final velocity at which the object hits the ground. (c) Determine the distance traveled during the last second of motion before hitting the ground.

Solution:

Consider Figure 1. The object was dropped from a height of 75.0 m. At the start of motion, the velocity is zero, $v_{oy}=0$ .

The object traveled for a period of time t for the whole 75.0 m distance to the ground.

Part A

We are solving for the distance traveled by the object for the first 1 second. So, we have

\begin{align*} \Delta y & = v_{oy} t + \frac{1}{2}at^2\\ \Delta y & = \left( 0 \ \text{m/s} \right)\left( 1 \ \text{s} \right)+\frac{1}{2}\left( -9.81 \ \text{m/s}^2 \right)\left( 1 \ \text{s} \right)^2 \\ \Delta y & = 0 -4.905 \ \text{m} \\ \Delta y & = -4.91 \ \text{m} \\ |\Delta y| & =4.91 \ \text{m} \end{align*}

The negative sign of Δy indicates that the direction of the displacement is downward. Since we are looking for the scalar value of the distance, the answer is 4.91 m.

Part B

So we now consider the two positions of the object as shown in the figure to the right. The initial height of the object is 75.0 m above the ground, and the initial velocity is 0.

At the ground, we know that the position of the object is 0 m above the ground, but we do not know the time and velocity. Therefore, to determine the velocity of the object at this point, we proceed as follows:

\begin{align*} \left( v_2 \right)^2 & =\left( v_1 \right)^2+2a \Delta y \\ \left( v_2 \right)^2 & =\left( v_1 \right)^2+2a \left( y_2-y_1 \right) \\ v_2 & = \pm \sqrt{\left( v_1 \right)^2+2a \left( y_2-y_1 \right)}\\ v_2 & = \pm \sqrt{\left( 0\ \text{m/s} \right)^2+2\left( -9.81\ \text{m/s}^2 \right)\left( 0 \ \text{m}-75\ \text{m} \right)} \\ v_2 & = \pm \ 38.4\ \text{m/s}\\ v_2 & =- 38.4\ \text{m/s}\\ \end{align*}

Since the object is directing downwards when it hit the ground, the velocity is negative.

Part C

First, we calculate the total time of the object’s motion from the beginning to the ground.

\begin{align*} \Delta y & =\bcancel{v_{oy}t}+ \frac{1}{2}at^2 \\ \Delta y & = \frac{1}{2}at^2 \\ 0 \text{m}-75\ \text{m} & = \frac{1}{2}\left( -9.81\ \text{m/s}^2 \right)t^2 \\ t^2& =\frac{-75\ \text{m}}{-4.905 \text{m/s}^2}\\ t&=\sqrt{\frac{75\ \text{m}}{4.905 \text{m/s}^2}}\\ t&=3.91 \ \text{s} \end{align*}

Second, determine the total distance traveled from 0 s to 2.91 s, leaving out the last 1 s of the motion.

\begin{align*} \Delta y & = v_{oy} t + \frac{1}{2}at^2\\ \Delta y & = \left( 0 \ \text{m/s} \right)\left( 1 \ \text{s} \right)+\frac{1}{2}\left( -9.81 \ \text{m/s}^2 \right)\left( 2.91 \ \text{s} \right)^2 \\ \Delta y & = 0 -41.5 \ \text{m} \\ \Delta y & = -41.5 \ \text{m} \\ |\Delta y| & =41.5 \ \text{m} \end{align*}

Finally, subtract this distance from the total distance traveled to get the distance traveled in the last 1 second.

\begin{align*} y_{_{\text{last 1 sec}}} & = 75.0 \ \text{m}-41.5 \ \text{m} \\ y_{_{\text{last 1 sec}}} & = 33.5\ \text{m} \end{align*}

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Acceleration of a Revolving Ball – Uniform Circular Motion Example

A 150-g ball at the end of a string is revolving uniformly in a horizontal circle of radius 0.600 m, as in the Figure 1 below. The ball makes 2.00 revolutions in a second. What is its centripetal acceleration?

Figure 1: A small object moving in a circle, showing how the velocity changes. At each point, the instantaneous velocity is in a direction tangent to the circular path.

Solution:

The linear velocity of the ball can be computed by dividing the total arc length traveled by the total time of travel. That is, the ball traveled 2 revolutions (twice the circumference of the circle) for 1 second. Thus,

\begin{align*} \text{v} &= \frac{2\cdot2 \pi \text{r}}{\text{t}} \\ \\ & = \frac{4\pi \text{r}}{\text{t}} \\ \\ & = \frac{4\pi\left( 0.600\ \text{m} \right)}{1 \ \text{s}} \\ \\ & = 7.54 \ \text{m/s} \end{align*}

Since the linear velocity has already been computed, we can now compute for the centripetal acceleration, a_c.

\begin{align*} \text{a}_\text{c} & = \frac{\text{v}^{2}}{\text{r}} \\ \\ & = \frac{\left( 7.54\ \text{m/s} \right)^{2}}{0.600\ \text{m}}\\ \\ & =94.8 \ \text{m/s}^{2} \end{align*}

College Physics 2.51 – Time of the hiker to move out from a falling rock

Standing at the base of one of the cliffs of Mt. Arapiles in Victoria, Australia, a hiker hears a rock break loose from a height of 105 m. He can’t see the rock right away but then does, 1.50 s later. (a) How far above the hiker is the rock when he can see it? (b) How much time does he have to move before the rock hits his head?

Solution:

Part A

We know that the initial height, $y_0$ of the rock is 105 meters, and the initial velocity, $v_0$ is zero. We shall solve for the distance traveled by the rock for 1.5 seconds from the initial position first to find the height at detection.

The change in height is

\displaystyle \begin{aligned} \Delta \text{y}&=\text{v}_0\text{t}+\frac{1}{2}\text{at}^2 \\ &=\left( 0 \right)\left( 1.50 \ \text{s} \right)+\frac{1}{2}\left( 9.81\ \text{m/s}^{2} \right)\left( 1.50\ \text{s} \right)^{2}\\ &=0+11.036\ \text{m} \\ &=11.04 \ \text{m} \end{aligned}

So, the rock falls about 11.04 m from the initial height for 1.50 seconds. Therefore, the height of the rock above his head at this point is

\displaystyle \begin{aligned} \text{y}&=\text{y}_{0}-\Delta \text{y} \\ &=105\ \text{m}-11.04\ \text{m} \\ &=93.96 \ \text{m} \end{aligned}

Part B

We shall solve for the total time of travel, that is, from the initial position to his head. Then we shall subtract 1.50 s from that to solve for the unknown time of moving out. The total time of travel is

\begin{aligned} \text{y} & =\frac{1}{2}\text{at}^{2} \\ &\text {Solving for t, we have}\\ \text{t}&=\sqrt{\frac{\text{2y}}{\text{a}}} \\ &=\sqrt{\frac{2\left( 105\ \text{m} \right)}{9.81 \ \text{m/s}^{2}}} \\ &=4.63 \ \text{s} \end{aligned}

Therefore, to move out the hiker has about

\begin{aligned} \text{t}&=4.63 \ \text{s}-1.50\ \text{s}\\ &=3.13\ \text{s} \end{aligned}

College Physics 2.50 – Motion of a Jumping Kangaroo

A kangaroo can jump over an object 2.50 m high. (a) Calculate its vertical speed when it leaves the ground. (b) How long is it in the air?

Part A

The motion of the kangaroo is under free-fall. We are looking for the initial velocity, and we know that the velocity in the highest position is zero.

From

\begin{aligned} \text{v}^2 &=\left (\text{v}_0 \right )^2+2\text{ay},\\ \end{aligned}

we have

\begin{aligned} \text{v}^2 &=\left (\text{v}_0 \right )^2+2\text{ay}\\ \text{v}^2-2\text{ay} &= \left ( \text{v}_0\right)^2\\ \text{v}_0&=\sqrt{\text{v}^2-2\text{ay}} \end{aligned}

Substituting the known values,

\begin{aligned} \text{v}_0&=\sqrt{\text{v}^2-2\text{ay}} \\ \text{v}_0&=\sqrt{0^2-2\left(-9.81 \text{m/s}^2\right)\left(2.50 \text{m}\right)}\\ \text{v}_0&= {\color{green}7.00 \ \text{m/s}} \end{aligned}

Therefore, the vertical speed of the kangaroo when it leaves the ground is 7.00 m/s.

Part B

Since the motion of the kangaroo has uniform acceleration, we can use the formula

\text{y}=\text{v}_o\text{t}+\frac{1}{2}\text{a}\text{t}^2

The initial and final position of the kangaroo will be the same, so $y$ is equal to zero. The initial velocity is 7.00 m/s, and the acceleration is -9.81 m/s².

\begin{aligned} \text{y} & =\text{v}_0\text{t}+\frac{1}{2}\text{a}\text{t}^2\\ 0 & = \left( 7.00\ \text{m/s} \right)\text{t}+\frac{1}{2}\left( -9.81\ \text{m/s}^{2} \right)\text{t}^2\\ 0 & =7\text{t}-4.905\text{t}^{2}\\ 7\text{t}-4.905\text{t}^{2}&=0 \\ \text{t}\left( 7-4.905\text{t} \right) & =0 \\ \text{t}=0 \qquad &\text{or} \qquad 7-4.905\text{t}=0 \\ \end{aligned}

Discard the time 0 since this refers to the beginning of motion. Therefore, we have

\begin{aligned} 7-4.905\text{t} &=0 \\ 4.905\text{t} & = 7 \\ \text{t} & =\frac{7}{4.905} \\ \text{t}&={\color{green}1.43 \ \text{s}} \end{aligned}

The kangaroo is about 1.43 seconds long in the air.

$Solution Guides to College Physics by Openstax Chapter 13 Banner$

Chapter 13: Temperature, Kinetic Theory, and the Gas Laws

Temperature

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Thermal Expansion of Solids and Liquids

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Kinetic Theory: Atomic and Molecular Explanation of Pressure and Temperature

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Humidity, Evaporation, and Boiling

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$Solution Guides to College Physics by Openstax Chapter 12 Banner$

Chapter 12: Fluid Dynamics and Its Biological and Medical Applications

Flow Rate and Its Relation to Velocity

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The Most General Applications of Bernoulli’s Equation

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The Onset of Turbulence

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