# Dynamics 1

• September 2nd 2009, 01:08 AM
Modernized
Dynamics 1

A particle has an initial velocity U and after travelling a distance d in time T along a straight path its velocity is V. The retardation of the particle is proportional to its velocty at any time. Show that:
U = V e ^ ( T(U-V)/d )
• September 2nd 2009, 04:16 AM
CaptainBlack
Quote:

Originally Posted by Modernized

A particle has an initial velocity U and after travelling a distance d in time T along a straight path its velocity is V. The retardation of the particle is proportional to its velocty at any time. Show that:
U = V e ^ ( T(U-V)/d )

The accelleration is:

$
\dot V=-kV
$

Solve this for $V$ using the condition that at $T=0$; $V=U$ . Now integrate again to get the distance as a function of $T$ and use this to eliminate $k$ from the velocity equation.

CB
• September 2nd 2009, 04:21 AM
mr fantastic
Quote:

Originally Posted by Modernized

A particle has an initial velocity U and after travelling a distance d in time T along a straight path its velocity is V. The retardation of the particle is proportional to its velocty at any time. Show that:
U = V e ^ ( T(U-V)/d )

$a = -kv \Rightarrow \frac{dv}{dt} = -kv$.

Solve $\frac{dt}{dv} = - \frac{1}{k} \cdot \frac{1}{v}$ for v, subject to v = U when t = 0. Substitute v = V when t = T to get the value of k.

Solve $\frac{dx}{dt} = v$ for x, subject to x = 0 when t = 0. Substitute x = d when t = T to get $U = V e^{\frac{T(U-V)}{d}}$.
• September 2nd 2009, 04:30 AM
Hello Modernized
Quote:

Originally Posted by Modernized

A particle has an initial velocity U and after travelling a distance d in time T along a straight path its velocity is V. The retardation of the particle is proportional to its velocty at any time. Show that:
U = V e ^ ( T(U-V)/d )

In this problem we use two different ways to express acceleration; first in terms of time; secondly in terms of distance. They are:

Acceleration = $\frac{dv}{dt}\,\Big(=\frac{dx}{dt}\times\frac{dv}{ dx}\Big)\,=v\frac{dv}{dx}$.

The first part of this problem connects velocity, time, and acceleration, so we express the phrase:
Quote:

The retardation of the particle is proportional to its velocty at any time
in the form

$\frac{dv}{dt}=-kv$ where $k$ is a positive constant

$\Rightarrow \int_U^V\frac{dv}{v}=-k\int_0^T dt$, noting that the boundary conditions have been incorporated into the limits of the integrals.

$\Rightarrow \ln(V)-\ln(U) = -kT$

$\Rightarrow k = \frac{1}{T}\ln\Big(\frac{U}{V}\Big)$ ... (1)

Next we use the alternative expression for acceleration, in order to relate velocity, acceleration and distance:

$v\frac{dv}{dx}= -kv$, where $k$ is a positive constant.

$\Rightarrow \int_U^Vdv=-k\int_0^d dx$, noting again the boundary conditions as limits.

From here we can find another expression for $k$, in terms of V, U and d (it's very easy!). And hence, from equation (1), we can get the result.

Can you complete it now?

• September 2nd 2009, 04:33 AM
CaptainBlack
Quote:

Originally Posted by mr fantastic
$a = -kv \Rightarrow \frac{dv}{dt} = -kv$.

Solve $\frac{dt}{dv} = - \frac{1}{k} \cdot \frac{1}{v}$ for v, subject to v = U when t = 0. Substitute v = V when t = T to get the value of k.

This last step gives the impression of making the simple complicated (well more complicated that it need be), or is that impression just because I know the solution to $\dot{v}=-kv$ backwards and forwards anagrammatized.

CB
• September 2nd 2009, 04:38 AM
mr fantastic
Quote:

Originally Posted by CaptainBlack
This last step gives the impression of making the simple complicated (well more complicated that it need be), or is that impression just because I know the solution to $\dot{v}=-kv$ backwards and forwards anagrammatized.

CB

Well, I like to make the simple complicated and the complicated even more complicated.

(Actually, I think the answer is the latter rather than the former. But is that giving you a complicated answer to a simple question ....? In which case ..... (Thinking))