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

Beginner2021-12-30Added 33 answers

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,

$mg-F=ma$

$mg-C{v}^{3}=m\frac{dv}{dt}$

$mg-\frac{mg}{{v}_{0}^{3}}{v}_{0}^{3}=m\frac{dv}{dt}$

Step 2

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

$g-\frac{g}{{v}_{0}^{3}}=\frac{dv}{dt}$

$\frac{gdt}{{v}_{0}^{3}}=\frac{dv}{({v}_{0}^{3}-{v}^{3})}$

$\int}_{0}^{4}\frac{gdt}{{3}^{3}}={\int}_{0}^{2}\frac{dv}{({3}^{3}-{v}^{3})$

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

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

$=0.22\mathrm{sec}$

trisanualb6

Beginner2021-12-31Added 32 answers

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.

$\Rightarrow F=C{v}^{3}$ (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

$\Rightarrow C{v}_{0}^{3}=mg$

$\Rightarrow C=\frac{mg}{{v}_{0}^{3}}$ (ii)

Now, at any instant of time t, the speed is v$\therefore mg-F=ma$ , where a : acceleration

$\Rightarrow mg-C{v}^{3}=m\frac{dv}{dt}$

$\Rightarrow g\left(\frac{{v}_{0}^{3}-{v}^{3}}{{v}_{0}^{3}}\right)=\frac{dv}{dt}$

$\Rightarrow \frac{9.8}{{v}_{0}^{3}}dt=\frac{dv}{{v}_{0}^{3}-{v}^{3}}$

put${v}_{0}\}=3$ in above, we get

$\Rightarrow \frac{9.8}{27}dt=\frac{dv}{{v}_{0}^{3}-{v}^{3}}$

integrating both side we get

$\int}_{0}^{t}\frac{9.8}{27}dt={\int}_{0}^{2}\frac{dv}{{\left(3\right)}^{3}-{v}^{3}$

$\Rightarrow 0.363t=0.0807$

$\Rightarrow t=0.22\mathrm{sec}$

Step 3

Answer:

The time required to attain the speed of 2m/s is 0.22 sec.

(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

where C is proporanaility constant direction of force is upward and

Now, at any instant of time t, the speed is v

put

integrating both side we get

Step 3

Answer:

The time required to attain the speed of 2m/s is 0.22 sec.

karton

Expert2022-01-09Added 613 answers

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