Chapter 6.4, Problem 57E

(a)

To determine

To show: The given function is a solution of initial value problem.

Answer to Problem 57E

The given function is a solution of initial value problem has been showed.

Explanation of Solution

Given:

The mass of a body is m and the velocity is v.

Calculation:

The differential equation $m\frac{dV}{dt}=mg-k{V}^2$ has the solution $\sqrt{\frac{mg}{k}}\tanh \left(\sqrt{\frac{gk}{m}}t\right)$ can be showed as;

Solve the left hand side of the equation first:

  $\begin{array}{l}m\frac{dV}{dt}=m.\sqrt{\frac{mg}{k}}\sec h^2\left(\sqrt{\frac{gk}{m}}t\right)\sqrt{\frac{gk}{m}}\\ \text{ = m}\sqrt{g^2}\sec h^2\left(\sqrt{\frac{gk}{m}}t\right)\\ \text{ = }mg\sec h^2\left(\sqrt{\frac{gk}{m}}t\right)\end{array}$

Now, solve the right hand side of the equation:

  $\begin{array}{l}mg-k{V}^2\\ mg-k\frac{mg}{k}{\tanh }^2\left(\sqrt{\frac{gk}{m}}t\right)\\ mg-mg\left(1-\sec h^2(\sqrt{\frac{gk}{m}}t\right)\\ mg-mg+mg\sec h^2\left(\sqrt{\frac{gk}{m}}t\right)\\ mg\sec h^2\left(\sqrt{\frac{gk}{m}}t\right)\end{array}$

Since, both lest hand side and the right hand side of the equation becomes equal. Hence, it can be said that The differential equation $m\frac{dV}{dt}=mg-k{V}^2$ has the solution $\sqrt{\frac{mg}{k}}\tanh \left(\sqrt{\frac{gk}{m}}t\right)$

Conclusion:

The given function is a solution of initial value problem has been showed.

(b)

To determine

The body limiting velocity.

Answer to Problem 57E

The body limiting velocity is $\sqrt{\frac{mg}{k}}$ .

Explanation of Solution

Given:

The mass of a body is m and the velocity is v.

Calculation:

  $\begin{array}{l}{\lim }_{t\to \infty }V=\sqrt{\frac{mg}{k}}{\lim }_{t\to \infty }\tanh \left(\sqrt{\frac{gk}{m}}t\right)\\ \text{ = }\sqrt{\frac{mg}{k}}\end{array}$

Conclusion:

The body limiting velocity is $\sqrt{\frac{mg}{k}}$ .

(c)

To determine

Diver’s limiting velocity in feet per second and in miles per hour.

Answer to Problem 57E

Diver’s limiting velocity in feet per second is $179\frac{ft}{s}$ and in miles per hour is $\text{122}\text{.04 }\frac{miles}{hour}$ .

Explanation of Solution

Given:

Diver’s mass is $160\text{ lbs}$ .

Calculation:

  $\begin{array}{l}{V}_{\lim }=\sqrt{\frac{160}{0.005}}\\ \text{ = }80\sqrt{5}\\ =179\frac{ft}{s}\\ \text{ = 122}\text{.04 }\frac{miles}{hour}\end{array}$

Conclusion:

Diver’s limiting velocity in feet per second is $179\frac{ft}{s}$ and in miles per hour is $\text{122}\text{.04 }\frac{miles}{hour}$ .