percibaa8

2021-12-29

Application of First-Order Differential Equations 1. Any time a body leaves its resting position, it encounters resistance proportional to the cube of the speed. Find the time needed to reach a speed of 2 m/s if the limiting speed is 3 m/s.

usumbiix

Step 1
It is assumed that resistance is proportional to the speed's cube at all times.
$F=C{v}^{3}$
At the limiting speed ${v}_{0}^{3}$ is 3m/s, the force F balances the weight of the body.
$C{v}_{0}^{3}=mg$
$C=\frac{mg}{{v}_{0}^{3}}$
The forces in equilibrium is,
$mg-F=ma$

Step 2
Reduce the complexity of the equation to determine the time needed to reach a 2 m/s speed.

$\frac{9.81}{27}t=0.0807$
$t=\frac{0.0807×27}{9.81}$

trisanualb6

Step 1
(Note: We are entitled to solve only one question at a time.)
Given:
A body falls from the rest against a resistance propotional to the cube of the speed at any instant.
To Find:
If the limiting speed is 3m/s, find the time required to attain a speed of 2m/s.
Step 2
Explanation:
Let F be the resistance force acting on the body and v be the speed of the body and $\therefore$ resistance acting on the body is proportional to cube of speed.
$⇒F=C{v}^{3}$ (i)
where C is proporanaility constant direction of force is upward and ${v}_{0}=3\frac{m}{s}$ is the limiting speed at the limiting speed the force F, balances the weight of the body
$⇒C{v}_{0}^{3}=mg$
$⇒C=\frac{mg}{{v}_{0}^{3}}$ (ii)
Now, at any instant of time t, the speed is v $\therefore mg-F=ma$, where a : acceleration
$⇒mg-C{v}^{3}=m\frac{dv}{dt}$
$⇒g\left(\frac{{v}_{0}^{3}-{v}^{3}}{{v}_{0}^{3}}\right)=\frac{dv}{dt}$
$⇒\frac{9.8}{{v}_{0}^{3}}dt=\frac{dv}{{v}_{0}^{3}-{v}^{3}}$
put ${v}_{0}\right\}=3$ in above, we get
$⇒\frac{9.8}{27}dt=\frac{dv}{{v}_{0}^{3}-{v}^{3}}$
integrating both side we get
${\int }_{0}^{t}\frac{9.8}{27}dt={\int }_{0}^{2}\frac{dv}{{\left(3\right)}^{3}-{v}^{3}}$
$⇒0.363t=0.0807$
$⇒t=0.22\mathrm{sec}$
Step 3
The time required to attain the speed of 2m/s is 0.22 sec.

karton

$\begin{array}{}F=C{v}^{3}\\ C{v}_{0}^{3}=mg\\ C=\frac{mg}{{v}_{0}^{3}}\\ mg-F=ma\\ mg-C{v}^{3}=m\frac{dv}{dt}\\ mg-\frac{mg}{{v}_{0}^{3}}{v}_{0}^{3}=m\frac{dv}{dt}\\ g-\frac{g}{{v}_{0}^{3}}=\frac{dv}{dt}\\ \frac{gdt}{{v}_{0}^{3}}=\frac{dv}{\left({v}_{0}^{3}-{v}^{3}\right)}\\ {\int }_{0}^{4}\frac{gdt}{{3}^{3}}={\int }_{0}^{2}\frac{dv}{\left({3}^{3}-{v}^{3}\right)}\\ \frac{9.81}{27}t=0.0807\\ t=\frac{0.0807×27}{9.81}\\ =0.22sec\end{array}$