The speed of a sky diver, whose acceleration is given by the equation a y = g − b v y 2 is given by v y ( t ) = v t tanh ( t T ) ,when expressed as a function of time, where v t is the terminal speed which is given by v t = g b and T is the time scale parameter given by T = v t g .

Question

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Answer 1

Question

Chapter 2, Problem 119P

(a)

To determine

To show:The speed of a sky diver, whose acceleration is given by the equation $ay=g−bvy2$ is given by $vy(t)=vttanh(tT)$ ,when expressed as a function of time, where $vt$ is the terminal speed which is given by $vt=gb$ and $T$ is the time scale parameter given by $T=vtg$ .

(a)

Expert Solution

Explanation of Solution

Given:

The equation for the acceleration $ay$ of the sky diver as a function of its speed $vy$

$ay=g−bvy2$ ........(1)

The expressions for the terminal speed $vt$ and the time scale parameter $T$ :

$vt=gb$ ........(2)

$T=vtg$ ........(3)

The speed of the sky diver at time $t=0$

$(vy)t=0=0$

Calculation:

Rewrite equation (1) using the expression for acceleration $ay=dvydt$ ,as follows:

$dvydt=g−bvy2$

Rearrange the above equation as follows:

$dvyg−bvy2=dt$

Integrate both sides of the equation.

$∫d v yg−b v y 2=∫dt$

Therefore,

$∫d v yg−b v y 2=t+C$ ........(4)

Evaluate the integral $∫d v yg−b v y 2$ as shown below:

$∫ d v y g−b v y 2 =1g∫ d v y 1− b g v y 2 =1g∫ d v y 1− ( b g v y ) 2$ ........(5)

Let $u=bgvy$ . ........(6)

Therefore,

$du=bgdvydvy=gbdu$

Rewrite equation (5) as shown:

$∫ d v y g−b v y 2 =1g∫ d v y 1− ( b g v y ) 2 =(1g)( g b )∫ du 1− u 2 =1 gb∫ du 1− u 2$ ........(7)

The expression $11−u2$ can be written as

$11−u2=12(11+u+11−u)$

Therefore, equation (7) takes the form,

$∫ d v y g−b v y 2 =1 gb∫ du 1− u 2 =1 gb∫ 1 2( 1 1+u + 1 1−u )du=12 gb[ln(1+u)−ln(1−u)]=(1 gb )[12ln1+u1−u]$ ........(8)

Use the identity,

$arctanh u=12ln1+u1−u$

Hence write equation (8) as,

$∫ d v y g−b v y 2 =(1 gb )[12ln1+u1−u]=(1 gb )(arctanhu)$

Therefore, equation (4) can be written as ,

$(1 gb)(arctanhu)=t+C$ .....(9)

Substitute the value of $u$ in equation (9).

$(1 gb)(arctanh( b g vy))=t+C$ ........(10)

Apply the initial conditions of the sky diver $(vy)t=0=0$ in equation (10).

$(1 gb )(arctanh( b g ×0))=0+CC=0$

Therefore, equation (10) can be written as,

$(1 gb)(arctanh( b g vy))=t$ ........(11)

Therefore,

$arctanh(bgvy)=gbt$

$tanh(gbt)=bgvy$ ........(12)

From equations (2) $vt=gb$ and (3) $T=vtg$ ,

$T=vtg=( g b )(1g)=1 gb$

Therefore, equation (12) can be written as,

$tanh( gbt)=bgvytanh(t 1 gb )=vy g b tanh(tT)=vyvt$

Hence,

$vy=vttanh(tT)$

Conclusion:

Thus, it is proved that the speed of a sky diver, whose acceleration is given by the equation $ay=g−bvy2$ is given by $vy(t)=vttanh(tT)$ .

(b)

To determine

To Find:The fraction of the terminal speed is the speed of the sky diver at time $t=T$ if $T$ is the time scale parameter given by $T=vtg$ .

(b)

Expert Solution

Answer to Problem 119P

At time $t=T$ the speed of the sky diver is $0.762$ times the terminal speed.

Explanation of Solution

Given:

The equation for the speed of the sky diver

$vy=vttanh(tT)$

Calculation:

Substitute $t=T$ in the given equation:

$vy=vttanh(tT)=vttanh(1)=vt(0.762)$

Conclusion:

Thus, at time $t=T$ the speed of the sky diver is $0.762$ times the terminal speed.

(c)

To determine

To graph: $vy$ as a function of time assuming the value of terminal velocity as $56 m/s$ .

(c)

Expert Solution

Explanation of Solution

Given:

The equation for the speed of the sky diver

$vy=vttanh(tT)$

Calculation:

In a spread sheet, calculate the values of $vy$ by substituting $56 m/s$ for $vt$ and plot a graph of $vy$ as a function of time as shown below:

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 2, Problem 119P

Conclusion:

Thus, it can be seen from the graph that the sky diver attains the terminal velocity at around 20 s.

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Answer 2

Question

Chapter 2, Problem 119P

(a)

To determine

To show:The speed of a sky diver, whose acceleration is given by the equation $ay=g−bvy2$ is given by $vy(t)=vttanh(tT)$ ,when expressed as a function of time, where $vt$ is the terminal speed which is given by $vt=gb$ and $T$ is the time scale parameter given by $T=vtg$ .

(a)

Expert Solution

Explanation of Solution

Given:

The equation for the acceleration $ay$ of the sky diver as a function of its speed $vy$

$ay=g−bvy2$ ........(1)

The expressions for the terminal speed $vt$ and the time scale parameter $T$ :

$vt=gb$ ........(2)

$T=vtg$ ........(3)

The speed of the sky diver at time $t=0$

$(vy)t=0=0$

Calculation:

Rewrite equation (1) using the expression for acceleration $ay=dvydt$ ,as follows:

$dvydt=g−bvy2$

Rearrange the above equation as follows:

$dvyg−bvy2=dt$

Integrate both sides of the equation.

$∫d v yg−b v y 2=∫dt$

Therefore,

$∫d v yg−b v y 2=t+C$ ........(4)

Evaluate the integral $∫d v yg−b v y 2$ as shown below:

$∫ d v y g−b v y 2 =1g∫ d v y 1− b g v y 2 =1g∫ d v y 1− ( b g v y ) 2$ ........(5)

Let $u=bgvy$ . ........(6)

Therefore,

$du=bgdvydvy=gbdu$

Rewrite equation (5) as shown:

$∫ d v y g−b v y 2 =1g∫ d v y 1− ( b g v y ) 2 =(1g)( g b )∫ du 1− u 2 =1 gb∫ du 1− u 2$ ........(7)

The expression $11−u2$ can be written as

$11−u2=12(11+u+11−u)$

Therefore, equation (7) takes the form,

$∫ d v y g−b v y 2 =1 gb∫ du 1− u 2 =1 gb∫ 1 2( 1 1+u + 1 1−u )du=12 gb[ln(1+u)−ln(1−u)]=(1 gb )[12ln1+u1−u]$ ........(8)

Use the identity,

$arctanh u=12ln1+u1−u$

Hence write equation (8) as,

$∫ d v y g−b v y 2 =(1 gb )[12ln1+u1−u]=(1 gb )(arctanhu)$

Therefore, equation (4) can be written as ,

$(1 gb)(arctanhu)=t+C$ .....(9)

Substitute the value of $u$ in equation (9).

$(1 gb)(arctanh( b g vy))=t+C$ ........(10)

Apply the initial conditions of the sky diver $(vy)t=0=0$ in equation (10).

$(1 gb )(arctanh( b g ×0))=0+CC=0$

Therefore, equation (10) can be written as,

$(1 gb)(arctanh( b g vy))=t$ ........(11)

Therefore,

$arctanh(bgvy)=gbt$

$tanh(gbt)=bgvy$ ........(12)

From equations (2) $vt=gb$ and (3) $T=vtg$ ,

$T=vtg=( g b )(1g)=1 gb$

Therefore, equation (12) can be written as,

$tanh( gbt)=bgvytanh(t 1 gb )=vy g b tanh(tT)=vyvt$

Hence,

$vy=vttanh(tT)$

Conclusion:

Thus, it is proved that the speed of a sky diver, whose acceleration is given by the equation $ay=g−bvy2$ is given by $vy(t)=vttanh(tT)$ .

(b)

To determine

To Find:The fraction of the terminal speed is the speed of the sky diver at time $t=T$ if $T$ is the time scale parameter given by $T=vtg$ .

(b)

Expert Solution

Answer to Problem 119P

At time $t=T$ the speed of the sky diver is $0.762$ times the terminal speed.

Explanation of Solution

Given:

The equation for the speed of the sky diver

$vy=vttanh(tT)$

Calculation:

Substitute $t=T$ in the given equation:

$vy=vttanh(tT)=vttanh(1)=vt(0.762)$

Conclusion:

Thus, at time $t=T$ the speed of the sky diver is $0.762$ times the terminal speed.

(c)

To determine

To graph: $vy$ as a function of time assuming the value of terminal velocity as $56 m/s$ .

(c)

Expert Solution

Explanation of Solution

Given:

The equation for the speed of the sky diver

$vy=vttanh(tT)$

Calculation:

In a spread sheet, calculate the values of $vy$ by substituting $56 m/s$ for $vt$ and plot a graph of $vy$ as a function of time as shown below:

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 2, Problem 119P

Conclusion:

Thus, it can be seen from the graph that the sky diver attains the terminal velocity at around 20 s.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Question

Chapter 2, Problem 119P

(a)

To determine

To show:The speed of a sky diver, whose acceleration is given by the equation $ay=g−bvy2$ is given by $vy(t)=vttanh(tT)$ ,when expressed as a function of time, where $vt$ is the terminal speed which is given by $vt=gb$ and $T$ is the time scale parameter given by $T=vtg$ .

(a)

Expert Solution

Explanation of Solution

Given:

The equation for the acceleration $ay$ of the sky diver as a function of its speed $vy$

$ay=g−bvy2$ ........(1)

The expressions for the terminal speed $vt$ and the time scale parameter $T$ :

$vt=gb$ ........(2)

$T=vtg$ ........(3)

The speed of the sky diver at time $t=0$

$(vy)t=0=0$

Calculation:

Rewrite equation (1) using the expression for acceleration $ay=dvydt$ ,as follows:

$dvydt=g−bvy2$

Rearrange the above equation as follows:

$dvyg−bvy2=dt$

Integrate both sides of the equation.

$∫d v yg−b v y 2=∫dt$

Therefore,

$∫d v yg−b v y 2=t+C$ ........(4)

Evaluate the integral $∫d v yg−b v y 2$ as shown below:

$∫ d v y g−b v y 2 =1g∫ d v y 1− b g v y 2 =1g∫ d v y 1− ( b g v y ) 2$ ........(5)

Let $u=bgvy$ . ........(6)

Therefore,

$du=bgdvydvy=gbdu$

Rewrite equation (5) as shown:

$∫ d v y g−b v y 2 =1g∫ d v y 1− ( b g v y ) 2 =(1g)( g b )∫ du 1− u 2 =1 gb∫ du 1− u 2$ ........(7)

The expression $11−u2$ can be written as

$11−u2=12(11+u+11−u)$

Therefore, equation (7) takes the form,

$∫ d v y g−b v y 2 =1 gb∫ du 1− u 2 =1 gb∫ 1 2( 1 1+u + 1 1−u )du=12 gb[ln(1+u)−ln(1−u)]=(1 gb )[12ln1+u1−u]$ ........(8)

Use the identity,

$arctanh u=12ln1+u1−u$

Hence write equation (8) as,

$∫ d v y g−b v y 2 =(1 gb )[12ln1+u1−u]=(1 gb )(arctanhu)$

Therefore, equation (4) can be written as ,

$(1 gb)(arctanhu)=t+C$ .....(9)

Substitute the value of $u$ in equation (9).

$(1 gb)(arctanh( b g vy))=t+C$ ........(10)

Apply the initial conditions of the sky diver $(vy)t=0=0$ in equation (10).

$(1 gb )(arctanh( b g ×0))=0+CC=0$

Therefore, equation (10) can be written as,

$(1 gb)(arctanh( b g vy))=t$ ........(11)

Therefore,

$arctanh(bgvy)=gbt$

$tanh(gbt)=bgvy$ ........(12)

From equations (2) $vt=gb$ and (3) $T=vtg$ ,

$T=vtg=( g b )(1g)=1 gb$

Therefore, equation (12) can be written as,

$tanh( gbt)=bgvytanh(t 1 gb )=vy g b tanh(tT)=vyvt$

Hence,

$vy=vttanh(tT)$

Conclusion:

Thus, it is proved that the speed of a sky diver, whose acceleration is given by the equation $ay=g−bvy2$ is given by $vy(t)=vttanh(tT)$ .

(b)

To determine

To Find:The fraction of the terminal speed is the speed of the sky diver at time $t=T$ if $T$ is the time scale parameter given by $T=vtg$ .

(b)

Expert Solution

Answer to Problem 119P

At time $t=T$ the speed of the sky diver is $0.762$ times the terminal speed.

Explanation of Solution

Given:

The equation for the speed of the sky diver

$vy=vttanh(tT)$

Calculation:

Substitute $t=T$ in the given equation:

$vy=vttanh(tT)=vttanh(1)=vt(0.762)$

Conclusion:

Thus, at time $t=T$ the speed of the sky diver is $0.762$ times the terminal speed.

(c)

To determine

To graph: $vy$ as a function of time assuming the value of terminal velocity as $56 m/s$ .

(c)

Expert Solution

Explanation of Solution

Given:

The equation for the speed of the sky diver

$vy=vttanh(tT)$

Calculation:

In a spread sheet, calculate the values of $vy$ by substituting $56 m/s$ for $vt$ and plot a graph of $vy$ as a function of time as shown below:

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 2, Problem 119P

Conclusion:

Thus, it can be seen from the graph that the sky diver attains the terminal velocity at around 20 s.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter 2, Problem 119P

(a)

To determine

To show:The speed of a sky diver, whose acceleration is given by the equation $ay=g−bvy2$ is given by $vy(t)=vttanh(tT)$ ,when expressed as a function of time, where $vt$ is the terminal speed which is given by $vt=gb$ and $T$ is the time scale parameter given by $T=vtg$ .

(a)

Expert Solution

Explanation of Solution

Given:

The equation for the acceleration $ay$ of the sky diver as a function of its speed $vy$

$ay=g−bvy2$ ........(1)

The expressions for the terminal speed $vt$ and the time scale parameter $T$ :

$vt=gb$ ........(2)

$T=vtg$ ........(3)

The speed of the sky diver at time $t=0$

$(vy)t=0=0$

Calculation:

Rewrite equation (1) using the expression for acceleration $ay=dvydt$ ,as follows:

$dvydt=g−bvy2$

Rearrange the above equation as follows:

$dvyg−bvy2=dt$

Integrate both sides of the equation.

$∫d v yg−b v y 2=∫dt$

Therefore,

$∫d v yg−b v y 2=t+C$ ........(4)

Evaluate the integral $∫d v yg−b v y 2$ as shown below:

$∫ d v y g−b v y 2 =1g∫ d v y 1− b g v y 2 =1g∫ d v y 1− ( b g v y ) 2$ ........(5)

Let $u=bgvy$ . ........(6)

Therefore,

$du=bgdvydvy=gbdu$

Rewrite equation (5) as shown:

$∫ d v y g−b v y 2 =1g∫ d v y 1− ( b g v y ) 2 =(1g)( g b )∫ du 1− u 2 =1 gb∫ du 1− u 2$ ........(7)

The expression $11−u2$ can be written as

$11−u2=12(11+u+11−u)$

Therefore, equation (7) takes the form,

$∫ d v y g−b v y 2 =1 gb∫ du 1− u 2 =1 gb∫ 1 2( 1 1+u + 1 1−u )du=12 gb[ln(1+u)−ln(1−u)]=(1 gb )[12ln1+u1−u]$ ........(8)

Use the identity,

$arctanh u=12ln1+u1−u$

Hence write equation (8) as,

$∫ d v y g−b v y 2 =(1 gb )[12ln1+u1−u]=(1 gb )(arctanhu)$

Therefore, equation (4) can be written as ,

$(1 gb)(arctanhu)=t+C$ .....(9)

Substitute the value of $u$ in equation (9).

$(1 gb)(arctanh( b g vy))=t+C$ ........(10)

Apply the initial conditions of the sky diver $(vy)t=0=0$ in equation (10).

$(1 gb )(arctanh( b g ×0))=0+CC=0$

Therefore, equation (10) can be written as,

$(1 gb)(arctanh( b g vy))=t$ ........(11)

Therefore,

$arctanh(bgvy)=gbt$

$tanh(gbt)=bgvy$ ........(12)

From equations (2) $vt=gb$ and (3) $T=vtg$ ,

$T=vtg=( g b )(1g)=1 gb$

Therefore, equation (12) can be written as,

$tanh( gbt)=bgvytanh(t 1 gb )=vy g b tanh(tT)=vyvt$

Hence,

$vy=vttanh(tT)$

Conclusion:

Thus, it is proved that the speed of a sky diver, whose acceleration is given by the equation $ay=g−bvy2$ is given by $vy(t)=vttanh(tT)$ .

(b)

To determine

To Find:The fraction of the terminal speed is the speed of the sky diver at time $t=T$ if $T$ is the time scale parameter given by $T=vtg$ .

(b)

Expert Solution

Answer to Problem 119P

At time $t=T$ the speed of the sky diver is $0.762$ times the terminal speed.

Explanation of Solution

Given:

The equation for the speed of the sky diver

$vy=vttanh(tT)$

Calculation:

Substitute $t=T$ in the given equation:

$vy=vttanh(tT)=vttanh(1)=vt(0.762)$

Conclusion:

Thus, at time $t=T$ the speed of the sky diver is $0.762$ times the terminal speed.

(c)

To determine

To graph: $vy$ as a function of time assuming the value of terminal velocity as $56 m/s$ .

(c)

Expert Solution

Explanation of Solution

Given:

The equation for the speed of the sky diver

$vy=vttanh(tT)$

Calculation:

In a spread sheet, calculate the values of $vy$ by substituting $56 m/s$ for $vt$ and plot a graph of $vy$ as a function of time as shown below:

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 2, Problem 119P

Conclusion:

Thus, it can be seen from the graph that the sky diver attains the terminal velocity at around 20 s.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 2, Problem 119P

(a)

To determine

To show:The speed of a sky diver, whose acceleration is given by the equation $ay=g−bvy2$ is given by $vy(t)=vttanh(tT)$ ,when expressed as a function of time, where $vt$ is the terminal speed which is given by $vt=gb$ and $T$ is the time scale parameter given by $T=vtg$ .

(a)

Expert Solution

Explanation of Solution

Given:

The equation for the acceleration $ay$ of the sky diver as a function of its speed $vy$

$ay=g−bvy2$ ........(1)

The expressions for the terminal speed $vt$ and the time scale parameter $T$ :

$vt=gb$ ........(2)

$T=vtg$ ........(3)

The speed of the sky diver at time $t=0$

$(vy)t=0=0$

Calculation:

Rewrite equation (1) using the expression for acceleration $ay=dvydt$ ,as follows:

$dvydt=g−bvy2$

Rearrange the above equation as follows:

$dvyg−bvy2=dt$

Integrate both sides of the equation.

$∫d v yg−b v y 2=∫dt$

Therefore,

$∫d v yg−b v y 2=t+C$ ........(4)

Evaluate the integral $∫d v yg−b v y 2$ as shown below:

$∫ d v y g−b v y 2 =1g∫ d v y 1− b g v y 2 =1g∫ d v y 1− ( b g v y ) 2$ ........(5)

Let $u=bgvy$ . ........(6)

Therefore,

$du=bgdvydvy=gbdu$

Rewrite equation (5) as shown:

$∫ d v y g−b v y 2 =1g∫ d v y 1− ( b g v y ) 2 =(1g)( g b )∫ du 1− u 2 =1 gb∫ du 1− u 2$ ........(7)

The expression $11−u2$ can be written as

$11−u2=12(11+u+11−u)$

Therefore, equation (7) takes the form,

$∫ d v y g−b v y 2 =1 gb∫ du 1− u 2 =1 gb∫ 1 2( 1 1+u + 1 1−u )du=12 gb[ln(1+u)−ln(1−u)]=(1 gb )[12ln1+u1−u]$ ........(8)

Use the identity,

$arctanh u=12ln1+u1−u$

Hence write equation (8) as,

$∫ d v y g−b v y 2 =(1 gb )[12ln1+u1−u]=(1 gb )(arctanhu)$

Therefore, equation (4) can be written as ,

$(1 gb)(arctanhu)=t+C$ .....(9)

Substitute the value of $u$ in equation (9).

$(1 gb)(arctanh( b g vy))=t+C$ ........(10)

Apply the initial conditions of the sky diver $(vy)t=0=0$ in equation (10).

$(1 gb )(arctanh( b g ×0))=0+CC=0$

Therefore, equation (10) can be written as,

$(1 gb)(arctanh( b g vy))=t$ ........(11)

Therefore,

$arctanh(bgvy)=gbt$

$tanh(gbt)=bgvy$ ........(12)

From equations (2) $vt=gb$ and (3) $T=vtg$ ,

$T=vtg=( g b )(1g)=1 gb$

Therefore, equation (12) can be written as,

$tanh( gbt)=bgvytanh(t 1 gb )=vy g b tanh(tT)=vyvt$

Hence,

$vy=vttanh(tT)$

Conclusion:

Thus, it is proved that the speed of a sky diver, whose acceleration is given by the equation $ay=g−bvy2$ is given by $vy(t)=vttanh(tT)$ .

(b)

To determine

To Find:The fraction of the terminal speed is the speed of the sky diver at time $t=T$ if $T$ is the time scale parameter given by $T=vtg$ .

(b)

Expert Solution

Answer to Problem 119P

At time $t=T$ the speed of the sky diver is $0.762$ times the terminal speed.

Explanation of Solution

Given:

The equation for the speed of the sky diver

$vy=vttanh(tT)$

Calculation:

Substitute $t=T$ in the given equation:

$vy=vttanh(tT)=vttanh(1)=vt(0.762)$

Conclusion:

Thus, at time $t=T$ the speed of the sky diver is $0.762$ times the terminal speed.

(c)

To determine

To graph: $vy$ as a function of time assuming the value of terminal velocity as $56 m/s$ .

(c)

Expert Solution

Explanation of Solution

Given:

The equation for the speed of the sky diver

$vy=vttanh(tT)$

Calculation:

In a spread sheet, calculate the values of $vy$ by substituting $56 m/s$ for $vt$ and plot a graph of $vy$ as a function of time as shown below:

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 2, Problem 119P

Conclusion:

Thus, it can be seen from the graph that the sky diver attains the terminal velocity at around 20 s.

Answer 6

Chapter 2, Problem 119P

(a)

To determine

To show:The speed of a sky diver, whose acceleration is given by the equation $ay=g−bvy2$ is given by $vy(t)=vttanh(tT)$ ,when expressed as a function of time, where $vt$ is the terminal speed which is given by $vt=gb$ and $T$ is the time scale parameter given by $T=vtg$ .

(a)

Expert Solution

Explanation of Solution

Given:

The equation for the acceleration $ay$ of the sky diver as a function of its speed $vy$

$ay=g−bvy2$ ........(1)

The expressions for the terminal speed $vt$ and the time scale parameter $T$ :

$vt=gb$ ........(2)

$T=vtg$ ........(3)

The speed of the sky diver at time $t=0$

$(vy)t=0=0$

Calculation:

Rewrite equation (1) using the expression for acceleration $ay=dvydt$ ,as follows:

$dvydt=g−bvy2$

Rearrange the above equation as follows:

$dvyg−bvy2=dt$

Integrate both sides of the equation.

$∫d v yg−b v y 2=∫dt$

Therefore,

$∫d v yg−b v y 2=t+C$ ........(4)

Evaluate the integral $∫d v yg−b v y 2$ as shown below:

$∫ d v y g−b v y 2 =1g∫ d v y 1− b g v y 2 =1g∫ d v y 1− ( b g v y ) 2$ ........(5)

Let $u=bgvy$ . ........(6)

Therefore,

$du=bgdvydvy=gbdu$

Rewrite equation (5) as shown:

$∫ d v y g−b v y 2 =1g∫ d v y 1− ( b g v y ) 2 =(1g)( g b )∫ du 1− u 2 =1 gb∫ du 1− u 2$ ........(7)

The expression $11−u2$ can be written as

$11−u2=12(11+u+11−u)$

Therefore, equation (7) takes the form,

$∫ d v y g−b v y 2 =1 gb∫ du 1− u 2 =1 gb∫ 1 2( 1 1+u + 1 1−u )du=12 gb[ln(1+u)−ln(1−u)]=(1 gb )[12ln1+u1−u]$ ........(8)

Use the identity,

$arctanh u=12ln1+u1−u$

Hence write equation (8) as,

$∫ d v y g−b v y 2 =(1 gb )[12ln1+u1−u]=(1 gb )(arctanhu)$

Therefore, equation (4) can be written as ,

$(1 gb)(arctanhu)=t+C$ .....(9)

Substitute the value of $u$ in equation (9).

$(1 gb)(arctanh( b g vy))=t+C$ ........(10)

Apply the initial conditions of the sky diver $(vy)t=0=0$ in equation (10).

$(1 gb )(arctanh( b g ×0))=0+CC=0$

Therefore, equation (10) can be written as,

$(1 gb)(arctanh( b g vy))=t$ ........(11)

Therefore,

$arctanh(bgvy)=gbt$

$tanh(gbt)=bgvy$ ........(12)

From equations (2) $vt=gb$ and (3) $T=vtg$ ,

$T=vtg=( g b )(1g)=1 gb$

Therefore, equation (12) can be written as,

$tanh( gbt)=bgvytanh(t 1 gb )=vy g b tanh(tT)=vyvt$

Hence,

$vy=vttanh(tT)$

Conclusion:

Thus, it is proved that the speed of a sky diver, whose acceleration is given by the equation $ay=g−bvy2$ is given by $vy(t)=vttanh(tT)$ .

Answer 7

Chapter 2, Problem 119P

(a)

To determine

To show:The speed of a sky diver, whose acceleration is given by the equation $ay=g−bvy2$ is given by $vy(t)=vttanh(tT)$ ,when expressed as a function of time, where $vt$ is the terminal speed which is given by $vt=gb$ and $T$ is the time scale parameter given by $T=vtg$ .

Answer 8

(a)

Expert Solution

Explanation of Solution

Given:

The equation for the acceleration $ay$ of the sky diver as a function of its speed $vy$

$ay=g−bvy2$ ........(1)

The expressions for the terminal speed $vt$ and the time scale parameter $T$ :

$vt=gb$ ........(2)

$T=vtg$ ........(3)

The speed of the sky diver at time $t=0$

$(vy)t=0=0$

Calculation:

Rewrite equation (1) using the expression for acceleration $ay=dvydt$ ,as follows:

$dvydt=g−bvy2$

Rearrange the above equation as follows:

$dvyg−bvy2=dt$

Integrate both sides of the equation.

$∫d v yg−b v y 2=∫dt$

Therefore,

$∫d v yg−b v y 2=t+C$ ........(4)

Evaluate the integral $∫d v yg−b v y 2$ as shown below:

$∫ d v y g−b v y 2 =1g∫ d v y 1− b g v y 2 =1g∫ d v y 1− ( b g v y ) 2$ ........(5)

Let $u=bgvy$ . ........(6)

Therefore,

$du=bgdvydvy=gbdu$

Rewrite equation (5) as shown:

$∫ d v y g−b v y 2 =1g∫ d v y 1− ( b g v y ) 2 =(1g)( g b )∫ du 1− u 2 =1 gb∫ du 1− u 2$ ........(7)

The expression $11−u2$ can be written as

$11−u2=12(11+u+11−u)$

Therefore, equation (7) takes the form,

$∫ d v y g−b v y 2 =1 gb∫ du 1− u 2 =1 gb∫ 1 2( 1 1+u + 1 1−u )du=12 gb[ln(1+u)−ln(1−u)]=(1 gb )[12ln1+u1−u]$ ........(8)

Use the identity,

$arctanh u=12ln1+u1−u$

Hence write equation (8) as,

$∫ d v y g−b v y 2 =(1 gb )[12ln1+u1−u]=(1 gb )(arctanhu)$

Therefore, equation (4) can be written as ,

$(1 gb)(arctanhu)=t+C$ .....(9)

Substitute the value of $u$ in equation (9).

$(1 gb)(arctanh( b g vy))=t+C$ ........(10)

Apply the initial conditions of the sky diver $(vy)t=0=0$ in equation (10).

$(1 gb )(arctanh( b g ×0))=0+CC=0$

Therefore, equation (10) can be written as,

$(1 gb)(arctanh( b g vy))=t$ ........(11)

Therefore,

$arctanh(bgvy)=gbt$

$tanh(gbt)=bgvy$ ........(12)

From equations (2) $vt=gb$ and (3) $T=vtg$ ,

$T=vtg=( g b )(1g)=1 gb$

Therefore, equation (12) can be written as,

$tanh( gbt)=bgvytanh(t 1 gb )=vy g b tanh(tT)=vyvt$

Hence,

$vy=vttanh(tT)$

Conclusion:

Thus, it is proved that the speed of a sky diver, whose acceleration is given by the equation $ay=g−bvy2$ is given by $vy(t)=vttanh(tT)$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

(b)

To determine

To Find:The fraction of the terminal speed is the speed of the sky diver at time $t=T$ if $T$ is the time scale parameter given by $T=vtg$ .

(b)

Expert Solution

Answer to Problem 119P

At time $t=T$ the speed of the sky diver is $0.762$ times the terminal speed.

Explanation of Solution

Given:

The equation for the speed of the sky diver

$vy=vttanh(tT)$

Calculation:

Substitute $t=T$ in the given equation:

$vy=vttanh(tT)=vttanh(1)=vt(0.762)$

Conclusion:

Thus, at time $t=T$ the speed of the sky diver is $0.762$ times the terminal speed.

Answer 13

(b)

To determine

To Find:The fraction of the terminal speed is the speed of the sky diver at time $t=T$ if $T$ is the time scale parameter given by $T=vtg$ .

Answer 14

(b)

Expert Solution

Answer to Problem 119P

At time $t=T$ the speed of the sky diver is $0.762$ times the terminal speed.

Answer 15

(b)

Expert Solution

Answer 16

(b)

Expert Solution

Answer 17

Expert Solution

Answer 18

Explanation of Solution

Given:

The equation for the speed of the sky diver

$vy=vttanh(tT)$

Calculation:

Substitute $t=T$ in the given equation:

$vy=vttanh(tT)=vttanh(1)=vt(0.762)$

Conclusion:

Thus, at time $t=T$ the speed of the sky diver is $0.762$ times the terminal speed.

Answer 19

(c)

To determine

To graph: $vy$ as a function of time assuming the value of terminal velocity as $56 m/s$ .

(c)

Expert Solution

Explanation of Solution

Given:

The equation for the speed of the sky diver

$vy=vttanh(tT)$

Calculation:

In a spread sheet, calculate the values of $vy$ by substituting $56 m/s$ for $vt$ and plot a graph of $vy$ as a function of time as shown below:

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 2, Problem 119P

Conclusion:

Thus, it can be seen from the graph that the sky diver attains the terminal velocity at around 20 s.

Answer 20

(c)

To determine

To graph: $vy$ as a function of time assuming the value of terminal velocity as $56 m/s$ .

Answer 21

(c)

Expert Solution

Explanation of Solution

Given:

The equation for the speed of the sky diver

$vy=vttanh(tT)$

Calculation:

In a spread sheet, calculate the values of $vy$ by substituting $56 m/s$ for $vt$ and plot a graph of $vy$ as a function of time as shown below:

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 2, Problem 119P

Conclusion:

Thus, it can be seen from the graph that the sky diver attains the terminal velocity at around 20 s.

Answer 22

(c)

Expert Solution

Answer 23

(c)

Expert Solution

Answer 24

Expert Solution

Answer 25

Ch. 2 - Prob. 1P Ch. 2 - Prob. 2P Ch. 2 - Prob. 3P Ch. 2 - Prob. 4P Ch. 2 - Prob. 5P Ch. 2 - Prob. 6P Ch. 2 - Prob. 7P Ch. 2 - Prob. 8P Ch. 2 - Prob. 9P Ch. 2 - Prob. 10P

The speed of a sky diver, whose acceleration is given by the equation a y = g − b v y 2 is given by v y ( t ) = v t tanh ( t T ) ,when expressed as a function of time, where v t is the terminal speed which is given by v t = g b and T is the time scale parameter given by T = v t g .

Concept explainers

Videos

To show:The speed of a sky diver, whose acceleration is given by the equation $ay=g−bvy2$ is given by $vy(t)=vttanh(tT)$ ,when expressed as a function of time, where $vt$ is the terminal speed which is given by $vt=gb$ and $T$ is the time scale parameter given by $T=vtg$ .

Explanation of Solution

To Find:The fraction of the terminal speed is the speed of the sky diver at time $t=T$ if $T$ is the time scale parameter given by $T=vtg$ .

Answer to Problem 119P

Explanation of Solution

To graph: $vy$ as a function of time assuming the value of terminal velocity as $56 m/s$ .

Explanation of Solution

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Chapter 2 Solutions

The speed of a sky diver, whose acceleration is given by the equation a y = g − b v y 2 is given by v y ( t ) = v t tanh ( t T ) ,when expressed as a function of time, where v t is the terminal speed which is given by v t = g b and T is the time scale parameter given by T = v t g .

Concept explainers

Videos

Explanation of Solution

Answer to Problem 119P

Explanation of Solution

To graph: vy<m:math><m:mrow><m:msub><m:mi>v</m:mi><m:mi>y</m:mi></m:msub></m:mrow></m:math> as a function of time assuming the value of terminal velocity as 56 m/s<m:math><m:mrow><m:mn>56</m:mn><m:mtext> m/s</m:mtext></m:mrow></m:math> .

Explanation of Solution

Want to see more full solutions like this?

Chapter 2 Solutions

To graph: $vy$ as a function of time assuming the value of terminal velocity as $56 m/s$ .