Chapter 6, Problem 45CP

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 45CP

The terminal speed of the object is $78.34\text{m}/\text{s}$.

Explanation of Solution

Given information:

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

The terminal speed of the object is given as,

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

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

The speed of the object at an instant of time is given as,

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

• $t$ is the time take.
• $v$ 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 $t=5.54\text{sec}$.

Therefore, the velocity of the object is,

$v=\frac{1}{2}{v}_{\text{T}}$

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).

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

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

$\begin{array}{c}{v}_{\text{T}}=\left(\frac{5.54\text{sec}}{\ln \left(2\right)}\right)\left(9.8\text{m}/{\text{s}}^2\right)\\ =78.34\text{m}/\text{s}\end{array}$

Conclusion:

Therefore, the terminal speed of the object is $78.34\text{m}/\text{s}$.

To determine

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

Answer to Problem 45CP

The time at which the speed of the object is three-fourth of the terminal speed is $11.07\text{sec}$.

Explanation of Solution

Given information:

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

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

Therefore, the velocity of the object is,

$v=\frac{3}{4}{v}_{\text{T}}$

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 $t$.

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

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

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

Conclusion:

Therefore, the time at which the speed of the object is three-fourth of the terminal speed is $11.07\text{sec}$.

To determine

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

Answer to Problem 45CP

The distance travelled by the object in first $5.54\text{sec}$ of motion is $120.9\text{m}$.

Explanation of Solution

Given information:

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

The speed of an object is given as,

$v=\frac{dr}{dt}$

• $r$ is the distance travelled by the object.

Rearrange the above expression for $r$.

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

The speed of the object is given as,

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

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

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

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.

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

Conclusion:

Therefore, the distance travelled by the object in first $5.54\text{sec}$ of motion is $120.9\text{m}$.