Question

65 A 9.00kgobject 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 travelled in the first 5.54sof motion?

Short answer

1. The terminal speed is 78 m/s.
2. The time required to become the speed of the object three-fourth of the

terminal speed is 11.0799.

1. The distance travelled by the object is 121m.

Step-by-step solution

Defining resistive force

Calculating the terminal speed

(a)

Now it is given that the speed of the object is one-half of the terminal speed.

Substitute $\frac{mg}{2b}$for v in equation (1)

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

Take on both sides.

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

Substitute 5.54s for t in the above equation.

$\begin{array}{c}\frac{b\left(5.54\right)}{m}=\ln 2\\ \frac{m}{b}=\frac{5.54}{\ln 2}\\ =7.9925 \mathrm{s}\end{array}$

The terminal speed is given by

$v_T=\frac{mg}{b}$

Substitute 7.9925s for $\frac{m}{b}$and for 9.8$m/s^2$g.

Therefore, the terminal speed is 78.3 m/s.

Calculating the required time

(b)

Now it is given that the speed of the object is three-fourth of the terminal speed.

Calculating distance travelled by the object

(c)

The velocity of the object can be expressed as

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

Then the distance travelled by the object in time t is

$r=\underset{0}{\overset{t}{\int }}vdt$