# Math Help - Dynamics 1

1. ## 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 )

2. 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

3. 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 $U = V e^{\frac{T(U-V)}{d}}$.

4. Hello Modernized
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:
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?

5. 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

6. 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 ..... )