Chapter 6, Problem 6.65CP

A 9.00-kg object starting from rest falls through a viscous medium and experiences a resistive force given by Equation 6.2. The object reaches one half its terminal speed in 5.54 s. (a) Determine the terminal speed. (b) At what time is the speed of the object three-fourths the terminal speed? (c) How far has the object traveled in the first 5.54 s of motion?

To determine

The terminal speed of the object.

Answer to Problem 6.65CP

The terminal speed of the object is $78.3\text{m}/\text{s}$.

Explanation of Solution

The mass of the object is $9.00\text{kg}$, the time to reach its terminal speed is $5.54\text{sec}$.

Write the expression for the terminal speed of the object

    $v_{\text{T}}=\frac{mg}{b}$                                                                                                     (I)

Here, $m$ is the mass of the object, $b$ is a constant., $g$ is the acceleration due to gravity and $v_{\text{T}}$ is the terminal speed.

Write the expression for the speed of the object at an instant of time

    $v=v_{\text{T}}\left(1-e^{\frac{-bt}{m}}\right)$                                                                                        (II)

Here, $t$ is the time take and $v$ is the instantaneous speed.

From the given condition, it is clear that the speed of the object is one half of its terminal speed at $t=5.54\text{sec}$.

Substitute $\frac{1}{2}{v}_{\text{T}}$ for $v$ in equation (II).

    $\begin{array}{c}\frac{1}{2}{v}_{\text{T}}=v_{\text{T}}\left(1-e^{\frac{-bt}{m}}\right)\\ \frac{1}{2}=1-e^{\frac{-bt}{m}}\\ e^{\frac{-bt}{m}}=\frac{1}{2}\end{array}$

Further solve the above expression.

    $\begin{array}{c}-\frac{bt}{m}=\ln \left(\frac{1}{2}\right)\\ \frac{m}{b}=\frac{t}{\ln \left(2\right)}\end{array}$

Substitute $\frac{t}{\ln \left(2\right)}$ for $\frac{m}{b}$ in equation (I).

    $v_{\text{T}}=\left(\frac{t}{\ln \left(2\right)}\right)g$

Conclusion:

Substitute $5.54\text{sec}$ for $t$ and $9.8\text{m}/{\text{s}}^2$ for $g$ in the above equation.

    $\begin{array}{c}{v}_{\text{T}}=\left(\frac{5.54\text{sec}}{\ln \left(2\right)}\right)\left(9.8\text{m}/{\text{s}}^2\right)\\ =78.3\text{m}/\text{s}\end{array}$

Therefore, the terminal speed of the object is $78.3\text{m}/\text{s}$.

To determine

The time at which the speed of the object is three-fourth of the terminal speed.

Answer to Problem 6.65CP

The time at which the speed of the object is three-fourth of the terminal speed is $11.1\text{sec}$.

Explanation of Solution

From the given condition, it is clear that the speed of the object is three-fourth of its terminal speed.

Substitute $\frac{3}{4}{v}_{\text{T}}$ for $v$ in equation (II).

    $\begin{array}{c}\frac{3}{4}{v}_{\text{T}}=v_{\text{T}}\left(1-e^{\frac{-bt}{m}}\right)\\ \frac{3}{4}=1-e^{\frac{-bt}{m}}\\ e^{\frac{-bt}{m}}=\frac{1}{4}\\ -\frac{bt}{m}=\ln \left(\frac{1}{4}\right)\end{array}$

Rearrange the above expression for $t$.

    $t=-\left(\frac{m}{b}\right)\ln \left(0.25\right)$

Conclusion:

Substitute $\frac{5.54\text{sec}}{\ln \left(2\right)}$ for $\frac{m}{b}$ in above equation.

    $\begin{array}{c}t=-\left(\frac{5.54\text{sec}}{\ln \left(2\right)}\right)\ln \left(0.25\right)\\ =11.07\text{sec}\\ \approx 11.1\text{sec}\end{array}$

Therefore, the time at which the speed of the object is three-fourth of the terminal speed is $11.1\text{sec}$.

To determine

The distance travelled by the object in first $5.54\text{sec}$ of motion.

Answer to Problem 6.65CP

The distance travelled by the object in first $5.54\text{sec}$ of motion is $121\text{m}$.

Explanation of Solution

Write the expression for the speed of an object

    $v=\frac{dr}{dt}$

Here, $r$ is the distance travelled by the object.

Rearrange the above expression for $r$.

    $r={\displaystyle \underset{0}{\overset{t}{\int }}vdt}$                                                                               (III)

Recall equation (II)

    $v=v_{\text{T}}\left(1-e^{\frac{-bt}{m}}\right)$

Substitute $v_{\text{T}}\left(1-e^{\frac{-bt}{m}}\right)$ for $v$ in equation (III).

    $\begin{array}{c}r={\displaystyle \underset{0}{\overset{t}{\int }}{v}_{\text{T}}\left(1-e^{\frac{-bt}{m}}\right)dt}\\ =v_{\text{T}}\left[t+\frac{m}{b}\left(e^{\frac{-bt}{m}}-1\right)\right]\end{array}$

Conclusion:

Substitute $78.34\text{m}/\text{s}$ for $v_{\text{T}}$, $\frac{5.54\text{sec}}{\ln \left(2\right)}$ for $\frac{m}{b}$, and $5.54\sec$ for $t$ in above expression.

    $\begin{array}{c}r=\left(78.34\text{m}/\text{s}\right)\left[5.54\text{sec}+\frac{5.54\text{sec}}{\ln \left(2\right)}\left(e^{-\frac{\ln \left(2\right)\left(5.54\text{sec}\right)}{5.54\text{sec}}}-1\right)\right]\\ =120.9\text{m}\\ \approx 121\text{m}\end{array}$

Therefore, the distance travelled by the object in first $5.54\text{sec}$ of motion is $121\text{m}$.