# How is the fuction v(t)? And how to set the graph?

## We have these information: $m \frac{\mathrm{dv}}{\mathrm{dt}} = g - R \left(v\right)$ $v \left(0\right) = 0$ We only know that $R \left(v\right)$ is a increasing, continuous function, and $R \left(0\right) = 0$. _ What are the proprieties of $R \left(v\right)$? What are the proprieties of $v \left(t\right)$? How is the graph of $v \left(t\right)$? Thank you.

Mar 17, 2016

See below.

#### Explanation:

This looks similar to falling of an object from height, maybe a paratrooper or a skydiver.

Since left hand side of the given expression has dimensions of force, and $g$ is normally used to denote acceleration due to gravity, I will take the liberty of rewriting the equation as

$m \frac{\mathrm{dv}}{\mathrm{dt}} = m g - R \left(v\right)$
with $v \left(0\right) = 0$
We only know that $R \left(v\right)$ is a increasing, continuous function, and $R \left(0\right) = 0$.
These two initial conditions and expression fits in the description well.

The weight of the object $m g$ acts downwards whereas, drag provided to the falling object by the atmosphere acts in the opposing direction. Hence, preceded by $- v e$ sign

1. At very low speeds for small objects, air resistance is approximately proportional to velocity and can be expressed in the form

$R \left(v\right) = - b \cdot v$, where $b$ is a constant of proportionality.

For higher velocities and larger objects the frictional drag is approximately proportional to the square of the velocity and can be expressed as
$R \left(v\right) = \frac{1}{2} \rho {v}^{2} {C}_{D} A$

where $\rho$ is the density of the air, $A$ is the area of cross section, and ${C}_{D}$ is the drag coefficient and is a dimensionless number.
2. When the body starts falling down its velocity keeps on increasing due to effect of gravity. With the increase in velocity the drag force increases as stated above, and it slows the object down.

There approaches a velocity when downwards force is equal to the drag force. This velocity ${v}_{T}$ is called terminal velocity. Thereafter, due to net force being zero, the object continues its descent with this velocity.
The expression reduces to
$m g = \frac{1}{2} \rho {v}_{T}^{2} {C}_{D} A$
or Terminal velocity ${v}_{T} = \sqrt{\frac{2 m g}{\rho {C}_{D} A}}$
3.

Any object falling through viscous medium accelerates quickly towards its terminal velocity, approaching gradually as the velocity gets nearer to the it.