Tag Archives: College Physics by Openstax Solution Manual

College Physics by Openstax Chapter 2 Problem 43

A basketball referee tosses the ball straight up for the starting tip-off. At what velocity must a basketball player leave the ground to rise 1.25 m above the floor in an attempt to get the ball?

Solution:

It is our assumption that the player attempts to get the ball at the top where the velocity is zero.

The given are the following: $v_{fy}=0 \ \text{m/s}$ ; $\Delta y=1.25 \ \text{m}$ ; and $a=-9.80 \ \text{m/s}^2$ .

We are required to solve for the initial velocity $v_{0y}$ of the player. We are going to use the formula

\left(v_{fy}\right)^2=\left(v_{oy}\right)^2+2a\Delta y

Solving for $v_{oy}$ in terms of the other variables:

v_{oy}=\sqrt{\left(v_{fy}\right)^2-2a\Delta y}

Substituting the given values:

\begin{align*} v_{oy} & =\sqrt{\left(v_{fy}\right)^2-2a\Delta y} \\ v_{oy} & = \sqrt{\left(0\:\text{m/s}\right)^2-2\left(-9.80\:\text{m/s}^2\right)\left(1.25\:\text{m}\right)} \\ v_{oy} & =4.95 \ \text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Advertisements

College Physics by Openstax Chapter 2 Problem 42

Calculate the displacement and velocity at times of (a) 0.500, (b) 1.00, (c) 1.50, (d) 2.00, and (e) 2.50 s for a rock thrown straight down with an initial velocity of 14.0 m/s from the Verrazano Narrows Bridge in New York City. The roadway of this bridge is 70.0 m above the water.

Solution:

The given known quantities are: $a=-9.8\:\text{m/s}^2$ ; $y_0=0 \ \text{m}$ ; and $v_0=-14 \ \text{m/s}$ .

To compute for the displacement, we use the formula

\Delta y=v_0t+\frac{1}{2}at^2

and to compute for the final velocity, we use the formula

v_f=v_0+at

Part A

The displacement at $t=0.500 \ \text{s}$ is

\begin{align*} \Delta y & =v_0t+\frac{1}{2}at^2 \\ \Delta y &=\left(-14.0\:\text{m/s}\right)\left(0.500\:\text{s}\right)+\frac{1}{2}\left(-9.8\:\text{m/s}^2\right)\left(0.500\:\text{s}\right)^2 \\ \Delta y & =-8.23\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

The velocity at $t=0.500 \ \text{s}$ is

\begin{align*} v_f & =v_0+at \\ &= \left(-14.0\:\text{m/s}\right)+\left(-9.8\:\text{m/s}^2\right)\left(0.500\:\text{s}\right) \\ & =-18.9\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B

The displacement at $t=1.00\ \text{s}$ is

\begin{align*} \Delta y & =v_0t+\frac{1}{2}at^2 \\ \Delta y &=\left(-14.0\:\text{m/s}\right)\left(1.00\:\text{s}\right)+\frac{1}{2}\left(-9.8\:\text{m/s}^2\right)\left(1.00\:\text{s}\right)^2 \\ \Delta y & =-18.9\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

The velocity at $t=1.00\ \text{s}$ is

\begin{align*} v_f & =v_0+at \\ &= \left(-14.0\:\text{m/s}\right)+\left(-9.8\:\text{m/s}^2\right)\left(1.00\:\text{s}\right) \\ & =-23.8\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part C

The displacement at $t=1.50\ \text{s}$ is

\begin{align*} \Delta y & =v_0t+\frac{1}{2}at^2 \\ \Delta y &=\left(-14.0\:\text{m/s}\right)\left(1.50\:\text{s}\right)+\frac{1}{2}\left(-9.8\:\text{m/s}^2\right)\left(1.50\:\text{s}\right)^2 \\ \Delta y & =-32.0\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

The velocity at $t=1.50\ \text{s}$ is

\begin{align*} v_f & =v_0+at \\ &= \left(-14.0\:\text{m/s}\right)+\left(-9.8\:\text{m/s}^2\right)\left(1.50\:\text{s}\right) \\ & =-28.7\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part D

The displacement at $t=2.00\ \text{s}$ is

\begin{align*} \Delta y & =v_0t+\frac{1}{2}at^2 \\ \Delta y &=\left(-14.0\:\text{m/s}\right)\left(2.00\:\text{s}\right)+\frac{1}{2}\left(-9.8\:\text{m/s}^2\right)\left(2.00\:\text{s}\right)^2 \\ \Delta y & =-47.6\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

The velocity at $t= 2.00 \ \text{s}$ is

\begin{align*} v_f & =v_0+at \\ &= \left(-14.0\:\text{m/s}\right)+\left(-9.8\:\text{m/s}^2\right)\left(2.00\:\text{s}\right) \\ & =-33.6\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part E

The displacement at $t=2.50\ \text{s}$ is

\begin{align*} \Delta y & =v_0t+\frac{1}{2}at^2 \\ \Delta y &=\left(-14.0\:\text{m/s}\right)\left(2.50\:\text{s}\right)+\frac{1}{2}\left(-9.8\:\text{m/s}^2\right)\left(2.50\:\text{s}\right)^2 \\ \Delta y & =-65.6\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

The velocity at $t= 2.50 \ \text{s}$ is

\begin{align*} v_f & =v_0+at \\ &= \left(-14.0\:\text{m/s}\right)+\left(-9.8\:\text{m/s}^2\right)\left(2.50\:\text{s}\right) \\ & =-38.5\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Advertisements

College Physics by Openstax Chapter 2 Problem 41

Calculate the displacement and velocity at times of (a) 0.500, (b) 1.00, (c) 1.50, and (d) 2.00 s for a ball thrown straight up with an initial velocity of 15.0 m/s. Take the point of release to be y₀=0.

Solution:

The given known quantities are: $a=-9.8\:\text{m/s}^2$ ; $y_o=0\:\text{m}$ ; and $v_{oy}=+15\:\text{m/s}$ .

To compute for the displacement, we use the formula

\Delta y=v_{oy}t+\frac{1}{2}at^2

and to compute for the final velocity, we use the formula

v_{fy}=v_{oy}+at

Part A

The displacement at $t=0.500 \ \text{s}$ is

\begin{align*} \Delta y & =v_ot+\frac{1}{2}at^2 \\ \Delta y & =0\:\text{m}+\left(15.0\:\text{m/s}\right)\left(0.500\:\text{s}\right)+\frac{1}{2}\left(-9.8\:\text{m/s}^2\right)\left(0.500\:\text{s}\right)^2 \\ \Delta y & =6.28\:\text{m}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

The velocity at $t=0.500 \ \text{s}$ is

\begin{align*} v_{fy} & = v_{oy}+at \\ v_{fy} & =\left(15.0\:\text{m/s}\right)+\left(-9.8\:\text{m/s}^2\right)\left(0.500\:\text{s}\right) \\ v_{fy} & =10.1\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B

The displacement at $t=1.000 \ \text{s}$ is

\begin{align*} \Delta y & =v_ot+\frac{1}{2}at^2 \\ \Delta y & =0\:\text{m}+\left(15.0\:\text{m/s}\right)\left(1.000\:\text{s}\right)+\frac{1}{2}\left(-9.8\:\text{m/s}^2\right)\left(1.000\:\text{s}\right)^2 \\ \Delta y & =10.1\:\text{m}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

The velocity at $t=1.000\ \text{s}$ is

\begin{align*} v_{fy} & = v_{oy}+at \\ v_{fy} & =\left(15.0\:\text{m/s}\right)+\left(-9.8\:\text{m/s}^2\right)\left(1.000\:\text{s}\right) \\ v_{fy} & =5.20\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part C

The displacement at $t=1.500\ \text{s}$ is

\begin{align*} \Delta y & =v_ot+\frac{1}{2}at^2 \\ \Delta y & =0\:\text{m}+\left(15.0\:\text{m/s}\right)\left(1.500\:\text{s}\right)+\frac{1}{2}\left(-9.8\:\text{m/s}^2\right)\left(1.500\:\text{s}\right)^2 \\ \Delta y & =11.5\:\text{m}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

The velocity at $t=1.500\ \text{s}$ is

\begin{align*} v_{fy} & = v_{oy}+at \\ v_{fy} & =\left(15.0\:\text{m/s}\right)+\left(-9.8\:\text{m/s}^2\right)\left(1.500\:\text{s}\right) \\ v_{fy} & =0.300\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part D

The displacement at $t=2.000\ \text{s}$ is

\begin{align*} \Delta y & =v_ot+\frac{1}{2}at^2 \\ \Delta y & =0\:\text{m}+\left(15.0\:\text{m/s}\right)\left(2.000\:\text{s}\right)+\frac{1}{2}\left(-9.8\:\text{m/s}^2\right)\left(2.000\:\text{s}\right)^2 \\ \Delta y & =10.4\:\text{m}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

The velocity at $t=2.000\ \text{s}$ is

\begin{align*} v_{fy} & = v_{oy}+at \\ v_{fy} & =\left(15.0\:\text{m/s}\right)+\left(-9.8\:\text{m/s}^2\right)\left(2.000\:\text{s}\right) \\ v_{fy} & =-4.600\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Advertisements

College Physics by Openstax Chapter 2 Problem 40

(a) A world record was set for the men’s 100-m dash in the 2008 Olympic Games in Beijing by Usain Bolt of Jamaica. Bolt “coasted” across the finish line with a time of 9.69 s. If we assume that Bolt accelerated for 3.00 s to reach his maximum speed, and maintained that speed for the rest of the race, calculate his maximum speed and his acceleration.

(b) During the same Olympics, Bolt also set the world record in the 200-m dash with a time of 19.30 s. Using the same assumptions as for the 100-m dash, what was his maximum speed for this race?

Solution:

Part A

There are two parts to the race and must be treated separately since acceleration is not uniform over the race. We will divide the race into $\Delta x_1$ (while accelerating) and $\Delta x_2$ (with constant speed), where $\Delta x_1 + \Delta x_2 = 100 \ \text{m}$ .

For $\Delta x_1$ :

During the accelerating period, we are going to use the formula $\Delta x=v_0t+\frac{1}{2}at^2$ , since we know that $\displaystyle a=\frac{\Delta v}{t}=\frac{v_{max}-v_0}{t}=\frac{v_{max}}{t}$ ; and $t=3.00 \ \text{s}$ .

\begin{align*} \Delta x & =v_0t+\frac{1}{2}at^2 \\ \Delta x_1 & =0+\frac{1}{2}at^2 \\ \Delta x_1 & =\frac{1}{2}at^2 \\ \Delta x_1 & =\frac{1}{2}\left(\frac{v_{max}}{t}\right)t^2 \\ \Delta x_1 & =\frac{1}{2}\left(v_{max}\right)t \\ \Delta x _1&=\frac{1}{2}\left(v_{max}\right)\left(3.00\:\text{s}\right) \\ \Delta x _1&=1.5v_{max} \end{align*}

When the speed is constant, $t=6.69 \ \text{s}$ , so

\begin{align*} \Delta x_2 & = v_{max}t \\ \Delta x_2 & = v_{max}\left(6.69\:\text{s}\right) \\ \Delta x_2 & =6.69v_{max} \end{align*}

Plugging-in the two equations in the equation $\Delta x_1 + \Delta x_2 = 100 \ \text{m}$ .

\begin{align*} \Delta x_1 + \Delta x_2 & = 100 \ \text{m} \\ 1.5v_{max} + 6.69v_{max} & =100 \ \text{m} \\ 8.19\:v_{max} & =100 \\ v_{max} & =\frac{100}{8.19} \\ v_{max} & =12.2\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Therefore, his acceleration can be computed using the formula

a=\frac{v_{max}}{t}

Plugging in the given values

\begin{align*} a & =\frac{v_{max}}{t} \\ a & = \frac{12.2\:\text{m/s}}{3.00\:\text{s}} \\ a & = 4.07\:\text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B

Similar to part (a), we can plug in the different values for time and total distance:

\begin{align*} \Delta x_1+ \Delta x_2 & =200 \\ 1.5\:v_{max}+\left(19.30-3.00\right)v_{max} & =200 \\ 1.5\:v_{max}+16.30v_{max} & =200 \\ 17.80v_{max} & =200 \\ v_{max} & =\frac{200}{17.80} \\ v_{max} & = 11.2\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Advertisements

College Physics by Openstax Chapter 2 Problem 39

In 1967, New Zealander Burt Munro set the world record for an Indian motorcycle, on the Bonneville Salt Flats in Utah, of 183.58 mi/h. The one-way course was 5.00 mi long. Acceleration rates are often described by the time it takes to reach 60.0 mi/h from rest. If this time was 4.00 s, and Burt accelerated at this rate until he reached his maximum speed, how long did it take Burt to complete the course?

Solution:

There are two parts to the race: an acceleration part and a constant speed part.

For the acceleration part:

We are given the following values: $v_0=0 \ \text{mph}$ ; $v_f=60 \ \text{mph}$ ; and $\Delta t=4.00 \ \text{s}$ .

First, we need to determine how long (both in distance and time) it takes the motorcycle to finish accelerating. During acceleration, the value of the acceleration is given by

a=\frac{60\:\text{mph}}{4\:\text{s}}

To compute for the time it takes to reach its maximum velocity, we are going to use the formula

v_f=v_0+at

Solving for time $t$ in terms of the other variables

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

Substituting the given values to solve for $t_1$ , the time it takes to accelerate from rest to maximum velocity:

\begin{align*} t_1 & =\frac{v_f-v_0}{a} \\ t_1 & =\frac{183\:\text{mph}-0\:\text{mph}}{\left(\frac{60\:\text{mph}}{4\:\text{s}}\right)} \\ t_1 & =12.2\:\text{s} \end{align*}

Since we have a constant acceleration, the distance traveled $\Delta x_1$ during this period is computed using the formula

\begin{align*} \Delta x_1 & =v_{ave}t \\ \Delta x_1 & =\left(\frac{v_f+v_0}{2}\right)t \\ \end{align*}

Substituting the given values:

\begin{align*} \Delta x_1 & =\left(\frac{v_f+v_0}{2}\right)t \\ \Delta x_1 & =\left(\frac{183\:\text{mph}+0\:\text{mph}}{2}\right)\left(12.2\:\text{s}\right) \\ \Delta x_1 & =\left(91.5\:\text{mph}\right)\left(\frac{1\:\text{hr}}{3600\:\text{s}}\right)\left(12.2\:\text{s}\right) \\ \Delta x_1 & =0.31\:\text{mi} \end{align*}

For the constant speed part:

For the next part of the motion, the speed is constant.

We are given the following values: $\Delta x_2=5.0\:\text{mi}-0.3\:\text{mi}=4.7\:\text{mi}$ .

We are going to solve $t_2$ , the time spent on the course at max speed using the formula

\Delta x_2=v_{max}t_2

Solving for $t_2$ in terms of the other variables:

t_2=\frac{\Delta x_2}{v_{max}}

Substituting the given values:

\begin{align*} t_2 & =\frac{\Delta x_2}{v_{max}} \\ t_2 & =\frac{4.7\:\text{mi}}{183\:\text{mph}} \\ t_2 & =\left(0.026\:\text{h}\right)\left(\frac{3600\:\text{s}}{1\:\text{h}}\right) \\ t_2 &=92\:\text{s} \end{align*}

For the whole course:

So, the total time is

\begin{align*} t_{total}&=t_1+t_2 \\ t_{total}& =12.2\:\text{s}+\:92\:\text{s} \\ t_{total}& =104\:\text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Advertisements

College Physics by Openstax Chapter 2 Problem 38

A bicycle racer sprints at the end of a race to clinch a victory. The racer has an initial velocity of 11.5 m/s and accelerates at the rate of 0.500 m/s² for 7.00 s.

(a) What is his final velocity?

(b) The racer continues at this velocity to the finish line. If he was 300 m from the finish line when he started to accelerate, how much time did he save?

(c) One other racer was 5.00 m ahead when the winner started to accelerate, but he was unable to accelerate, and traveled at 11.8 m/s until the finish line. How far ahead of him (in meters and in seconds) did the winner finish?

Solution:

We are given the following: $v_0=11.5 \ \text{m/s}$ ; $a=0.500 \ \text{m/s}^2$ ; and $\Delta t=7.00 \ \text{s}$ .

Part A

To solve for the final velocity, we are going to use the formula

v_f=v_0+at

Substituting the given values:

\begin{align*} v_f &=v_0+at\\ v_f&=11.5\ \text{m/s}+\left( 0.500\ \text{m/s}^2 \right)\left( 7.00\ \text{s} \right)\\ v_f&=15.0\ \text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B

Let $t_{const}$ be the time it takes to reach the finish line without accelerating:

\begin{align*} t_{const}&=\frac{x}{v_0}\\ t_{const}&=\frac{300\ \text{m}}{11.5\ \text{m/s}}\\ t_{const}&=26.1\ \text{m/s} \end{align*}

Now let $d$ be the distance traveled during the 7 seconds of acceleration. We know $t=7.00 \ \text{s}$ so

\begin{align*} d&=v_0t+\frac{1}{2}at^2\\ d&=\left( 11.5\ \text{m/s} \right)\left( 7.00\ \text{s} \right)+\frac{1}{2}\left( 0.500\ \text{m/s} ^2\right)\left( 7.00\ \text{s} \right)^2\\ d&=92.8\ \text{m} \end{align*}

Let $t'$ be the time it will take the rider at the constant final velocity to complete the race:

\begin{align*} t'&=\frac{x-d}{v}\\ t'&=\frac{300\ \text{m}-92.8\ \text{m}}{15.0\ \text{m/s}}\\ t'&=13.8\ \text{s} \end{align*}

So the total time $T$ it will take the accelerating rider to reach the finish line is

\begin{align*} T&=t+t'\\ T&=7.00\ \text{s}+13.8\ \text{s}\\ T&=20.8\ \text{s} \end{align*}

Finally, let $T^{*}$ be the time saved. So

\begin{align*} T^{*}&=26.1\ \text{s}-20.8\ \text{s}\\ T^{*}&={\color{DarkGreen} 5.3\ \text{s}} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part C

For rider 2, we are given the following values: $\Delta x'=295 \ \text{m}$ ; $v'=11.8 \ \text{m/s}$

Let $t_2$ be the time it takes for rider 2 to reach the finish line.

We are going to use the formula

t_2=\frac{\Delta x'}{v'}

Substituting the given values:

\begin{align*} t_2 & =\frac{x'}{v'} \\ t_2 & =\frac{295\:\text{m}}{11.8\:\text{m/s}} \\ t_2 & =25.0\:\text{s} \end{align*}

The time difference is

\begin{align*} \text{time difference} & =t_2-T \\ \text{time difference} & =25.0\:\text{s}-20.817\:\text{s} \\ \text{time difference} & =4.2\:\text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Therefore, he finishes 4.2 s after the winner.

When the other racer reaches the finish line, he has been traveling at 11.8 m/s for 4.2 seconds, so the other racer finishes

\begin{align*} \Delta x & =\left(11.8\:\text{m/s}\right)\left(4.2\:\text{s}\right) \\ \Delta x & =49.56\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

behind the other racer.

Advertisements

College Physics by Openstax Chapter 2 Problem 37

Dragsters can actually reach a top speed of 145 m/s in only 4.45 s—considerably less time than given in Example 2.10 and Example 2.11.

(a) Calculate the average acceleration for such a dragster.

(b) Find the final velocity of this dragster starting from rest and accelerating at the rate found in (a) for 402 m (a quarter mile) without using any information on time.

(c) Why is the final velocity greater than that used to find the average acceleration? Hint: Consider whether the assumption of constant acceleration is valid for a dragster. If not, discuss whether the acceleration would be greater at the beginning or end of the run and what effect that would have on the final velocity.

Solution:

We are given the following: $v_0=0\ \text{m/s}$ ; $v_f=145 \ \text{m/s}$ ; and $\Delta t=4.45 \ \text{sec}$ .

Part A

To compute for the average acceleration $a$ , we are going to use the formula

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

Substituting the given values, we have

\begin{align*} a & =\frac{v_f-v_0}{\Delta t} \\ a & =\frac{145\:\text{m/s}-0\:\text{m/s}}{4.45\:\text{s}} \\ a & =32.6\:\text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B

We are given the following: $a=32.6 \ \text{m/s}^2$ ; $v_0=0 \ \text{m/s}$ ; and $\Delta x=402 \ \text{m}$ .

Since we do not have any information on time, we are going to use the formula

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

To compute for the final velocity, we have

v_f=\sqrt{\left(v_0\right)^2+2a\Delta \:x}

Substituting the given values:

\begin{align*} v_f & =\sqrt{\left(v_0\right)^2+2a\Delta \:x} \\ v_f & =\sqrt{\left(0\:\text{m/s}\right)^2+2\left(32.6\:\text{m/s}^2\right)\left(402\:\text{m}\right)} \\ v_f & =162\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part C

The final velocity is greater than that used to find the average acceleration because the assumption of constant acceleration is not valid for a dragster. A dragster changes gears and would have a greater acceleration in first gear than second gear than third gear, etc. The acceleration would be greatest at the beginning, so it would not be accelerating at 32.6 m/s² during the last few meters, but substantially less, and the final velocity would be less than $162\:\text{m/s}$ .

Advertisements

College Physics by Openstax Chapter 2 Problem 36

An express train passes through a station. It enters with an initial velocity of 22.0 m/s and decelerates at a rate of 0.150 m/s² as it goes through. The station is 210 m long.

(a) How long is the nose of the train in the station?

(b) How fast is it going when the nose leaves the station?

(c) If the train is 130 m long, when does the end of the train leave the station?

(d) What is the velocity of the end of the train as it leaves?

Solution:

Part A

We are given the following: $v_0=22.0\:\text{m/s}$ ; $a=-0.150\:\text{m/s}^2$ ; and $\Delta x=210\:\text{m}$

We are required to solve for time, $t$ . We are going to use the formula

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

Substituting the given values, we have

\begin{align*} \Delta x & =v_0t+\frac{1}{2}at^2 \\ 210\:\text{m} & =\left(22.0\:\text{m/s}\right)t+\frac{1}{2}\left(-0.150\:\text{m/s}^2\right)t^2 \end{align*}

If we simplify and rearrange the terms into a general quadratic equation, we have

0.075t^2-0.22t+210=0

Solve for $t$ using the quadratic formula. We are given $a=0.075;\:b=-22t;\:c=210$ .

\begin{align*} t & =\frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ t& =\frac{-\left(-22\right)\pm \sqrt{\left(-22\right)^2-4\left(0.075\right)\left(210\right)}}{2\left(0.075\right)}\\ \end{align*}

There are two values of $t$ that can satisfy the quadratic equation.

t=9.88\ \text{s} \qquad \text{and} \qquad t=283.46 \ \text{s}

Discard $t=283.46 \ \text{s}$ as it can be seen from the problem that this is not a feasible solution. So, we have

t=9.88\:\text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

Part B

We have the same given values from part a. We are going to solve $v_f$ in the formula

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

So we have

\begin{align*} \left(v_f\right)^2 & =\left(v_0\right)^2+2a\Delta x \\ v_f & =\sqrt{\left(v_0\right)^2+2a\Delta x} \\ v_f & =\sqrt{\left(22.0\:\text{m/s}\right)^2+2\left(-0.150\:\text{m/s}^2\right)\left(210\:\text{m}\right)} \\ v_f & =20.6\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part C

We are given the same values as in the previous parts, except that for the value of $\Delta x$ since we should incorporate the length of the train. For the distance, $\Delta x$ , we have

\begin{align*} \Delta x & =210\:\text{m}+130\:\text{m} \\ \Delta x & = 340 \ \text{m} \end{align*}

To solve for time $t$ , we are going to use the formula

\Delta x=v_ot+\frac{1}{2}at^2

If we rearrange the formula into a general quadratic equation and solve for $t$ using the quadratic formula, we come up with

t=\frac{-v_0\pm \sqrt{v_0^2+2ax}}{a}

Substituting the given values:

\begin{align*} t & =\frac{-v_0\pm \sqrt{v_0^2+2ax}}{a} \\ t & =\frac{-\left(22.0\:\text{m/s}\right)\pm \sqrt{\left(22.0\:\text{m/s}\right)^2+2\left(-0.150\:\text{m/s}^2\right)\left(340\:\text{m}\right)}}{-0.150\:\text{m/s}^2} \\ t &=16.4\:\text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part D

We have the same given values. We are going to solve for $v_f$ in the equation

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

Substituting the given values:

\begin{align*} \left(v_f\right)^2 & =\left(v_0\right)^2+2a\Delta x \\ v_f & =\sqrt{\left(v_0\right)^2+2a\Delta x} \\ v_f & =\sqrt{\left(22.0\:\text{m/s}\right)^2+2\left(-0.150\:\text{m/s}^2\right)\left(340\:\text{m}\right)} \\ v_f & =19.5\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Advertisements

College Physics by Openstax Chapter 2 Problem 35

Consider a grey squirrel falling out of a tree to the ground.

(a) If we ignore air resistance in this case (only for the sake of this problem), determine a squirrel’s velocity just before hitting the ground, assuming it fell from a height of 3.0 m.

(b) If the squirrel stops in a distance of 2.0 cm through bending its limbs, compare its deceleration with that of the airman in the previous problem.

Solution:

Part A

We are given the following: $v_0=0 \ \text{m/s}$ ; $a=-9.80 \ \text{m/s}^2$ ; and $\Delta x=-3.0\ \text{m}$ .

Note that the acceleration is due to gravity and its value is constant at $a=-9.80 \ \text{m/s}^2$ . Also, the distance $x$ , is negative because of the direction of motion. For free-fall, downward motion is considered negative. To solve for the velocity just before it hits the ground, we will solve $v_f$ in the formula

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

Substituting the given values:

\begin{align*} \left(v_f\right)^2 & =\left(v_0\right)^2+2a\Delta x \\ v_f & =\sqrt{\left(v_0\right)^2+2a\Delta x} \\ v_f & = \sqrt{\left(0\:\text{m/s}\right)^2+2\left(-9.80\:\text{m/s}^2\right)\left(-3.0\:\text{m}\right)} \\ v_f & = 7.7\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B

We are given the following: $v_f=0 \ \text{m/s}$ ; $v_0=7.7 \ \text{m/s}$ ; and $\Delta x=0.02 \ \text{m}$

To solve for the acceleration we shall use the same formula as that in Part A. Solving for the acceleration $a$ :

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

Substituting the given values:

\begin{align*} a & =\frac{\left(v_f\right)^2-\left(v_0\right)^2}{2\Delta x} \\ a & =\frac{\left(0\:\text{m/s}\right)^2-\left(7.7\:\text{m/s}\right)^2}{2\left(0.02\:\text{m}\right)} \\ a & =-1.5\times 10^3\:\text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

This is approximately 3 times the deceleration of the airman from the previous problem, who was falling from thousands of meters high.

Advertisements

College Physics by Openstax Chapter 2 Problem 34

In World War II, there were several reported cases of airmen who jumped from their flaming airplanes with no parachute to escape certain death. Some fell about 20,000 feet (6000 m), and some of them survived, with few life-threatening injuries. For these lucky pilots, the tree branches and snow drifts on the ground allowed their deceleration to be relatively small. If we assume that a pilot’s speed upon impact was 123 mph (54 m/s), then what was his deceleration? Assume that the trees and snow stopped him over a distance of 3.0 m.

Solution:

We are given the following: $x=3\ \text{m}$ ; $v_0=54\ \text{m/s}$ ; and $v_f=0 \ \text{m/s}$ .

We are required to solve for the acceleration, and we are going to use the formula

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

Solving for acceleration $a$ in terms of the other variables:

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

Substituting the given values:

\begin{align*} a & = \frac{\left(v_f\right)^2-\left(v_0\right)^2}{2\Delta x} \\ a & =\frac{\left(0\:\text{m/s}\right)^2-\left(54\:\text{m/s}\right)^2}{2\left(3.0\:\text{m}\right)} \\ a & =-486\:\text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

The negative acceleration means that the pilot was decelerating at a rate of 486 m/s every second.

Advertisements

Engineering Mathematics and Sciences

Your virtual study buddy in Mathematics, Engineering Sciences, and Civil Engineering

Tag Archives: College Physics by Openstax Solution Manual

College Physics by Openstax Chapter 2 Problem 43

A basketball referee tosses the ball straight up for the starting tip-off. At what velocity must a basketball player leave the ground to rise 1.25 m above the floor in an attempt to get the ball?

College Physics by Openstax Chapter 2 Problem 42

Calculate the displacement and velocity at times of (a) 0.500, (b) 1.00, (c) 1.50, (d) 2.00, and (e) 2.50 s for a rock thrown straight down with an initial velocity of 14.0 m/s from the Verrazano Narrows Bridge in New York City. The roadway of this bridge is 70.0 m above the water.

College Physics by Openstax Chapter 2 Problem 41

Calculate the displacement and velocity at times of (a) 0.500, (b) 1.00, (c) 1.50, and (d) 2.00 s for a ball thrown straight up with an initial velocity of 15.0 m/s. Take the point of release to be y₀=0.

College Physics by Openstax Chapter 2 Problem 40

(b) During the same Olympics, Bolt also set the world record in the 200-m dash with a time of 19.30 s. Using the same assumptions as for the 100-m dash, what was his maximum speed for this race?

College Physics by Openstax Chapter 2 Problem 39

Solution:

College Physics by Openstax Chapter 2 Problem 38

A bicycle racer sprints at the end of a race to clinch a victory. The racer has an initial velocity of 11.5 m/s and accelerates at the rate of 0.500 m/s² for 7.00 s.

(a) What is his final velocity?

(b) The racer continues at this velocity to the finish line. If he was 300 m from the finish line when he started to accelerate, how much time did he save?

(c) One other racer was 5.00 m ahead when the winner started to accelerate, but he was unable to accelerate, and traveled at 11.8 m/s until the finish line. How far ahead of him (in meters and in seconds) did the winner finish?

College Physics by Openstax Chapter 2 Problem 37

Dragsters can actually reach a top speed of 145 m/s in only 4.45 s—considerably less time than given in Example 2.10 and Example 2.11.

(a) Calculate the average acceleration for such a dragster.

(b) Find the final velocity of this dragster starting from rest and accelerating at the rate found in (a) for 402 m (a quarter mile) without using any information on time.

College Physics by Openstax Chapter 2 Problem 36

An express train passes through a station. It enters with an initial velocity of 22.0 m/s and decelerates at a rate of 0.150 m/s² as it goes through. The station is 210 m long.

(a) How long is the nose of the train in the station?

(b) How fast is it going when the nose leaves the station?

(c) If the train is 130 m long, when does the end of the train leave the station?

(d) What is the velocity of the end of the train as it leaves?

College Physics by Openstax Chapter 2 Problem 35

Consider a grey squirrel falling out of a tree to the ground.

(a) If we ignore air resistance in this case (only for the sake of this problem), determine a squirrel’s velocity just before hitting the ground, assuming it fell from a height of 3.0 m.

(b) If the squirrel stops in a distance of 2.0 cm through bending its limbs, compare its deceleration with that of the airman in the previous problem.

College Physics by Openstax Chapter 2 Problem 34

A basketball referee tosses the ball straight up for the starting tip-off. At what velocity must a basketball player leave the ground to rise 1.25 m above the floor in an attempt to get the ball?

Calculate the displacement and velocity at times of (a) 0.500, (b) 1.00, (c) 1.50, (d) 2.00, and (e) 2.50 s for a rock thrown straight down with an initial velocity of 14.0 m/s from the Verrazano Narrows Bridge in New York City. The roadway of this bridge is 70.0 m above the water.

Calculate the displacement and velocity at times of (a) 0.500, (b) 1.00, (c) 1.50, and (d) 2.00 s for a ball thrown straight up with an initial velocity of 15.0 m/s. Take the point of release to be y0=0.

(b) During the same Olympics, Bolt also set the world record in the 200-m dash with a time of 19.30 s. Using the same assumptions as for the 100-m dash, what was his maximum speed for this race?

Solution:

A bicycle racer sprints at the end of a race to clinch a victory. The racer has an initial velocity of 11.5 m/s and accelerates at the rate of 0.500 m/s2 for 7.00 s.

(a) What is his final velocity?

(b) The racer continues at this velocity to the finish line. If he was 300 m from the finish line when he started to accelerate, how much time did he save?

(c) One other racer was 5.00 m ahead when the winner started to accelerate, but he was unable to accelerate, and traveled at 11.8 m/s until the finish line. How far ahead of him (in meters and in seconds) did the winner finish?

Dragsters can actually reach a top speed of 145 m/s in only 4.45 s—considerably less time than given in Example 2.10 and Example 2.11.

(a) Calculate the average acceleration for such a dragster.

(b) Find the final velocity of this dragster starting from rest and accelerating at the rate found in (a) for 402 m (a quarter mile) without using any information on time.

An express train passes through a station. It enters with an initial velocity of 22.0 m/s and decelerates at a rate of 0.150 m/s2 as it goes through. The station is 210 m long.

(a) How long is the nose of the train in the station?

(b) How fast is it going when the nose leaves the station?

(c) If the train is 130 m long, when does the end of the train leave the station?

(d) What is the velocity of the end of the train as it leaves?

Consider a grey squirrel falling out of a tree to the ground.

(a) If we ignore air resistance in this case (only for the sake of this problem), determine a squirrel’s velocity just before hitting the ground, assuming it fell from a height of 3.0 m.

(b) If the squirrel stops in a distance of 2.0 cm through bending its limbs, compare its deceleration with that of the airman in the previous problem.

Calculate the displacement and velocity at times of (a) 0.500, (b) 1.00, (c) 1.50, and (d) 2.00 s for a ball thrown straight up with an initial velocity of 15.0 m/s. Take the point of release to be y₀=0.

A bicycle racer sprints at the end of a race to clinch a victory. The racer has an initial velocity of 11.5 m/s and accelerates at the rate of 0.500 m/s² for 7.00 s.

An express train passes through a station. It enters with an initial velocity of 22.0 m/s and decelerates at a rate of 0.150 m/s² as it goes through. The station is 210 m long.