Velocity-Based Differentiation for B(s) in a Simple Homework Problem

[Hemmer]

Homework Statement

I have nearly finished a problem. I am at the stage where I have:

B(s)\propto\frac{1}{s^{3}}

v=\frac{ds}{dt}

I want to find the rate of change of B(s), but expressed in terms of a velocity, rather than a displacement.

Homework Equations

So for displacement:

B'(s) \propto\frac{-3}{s^{2}}

The Attempt at a Solution

How can I express this in terms of v instead? My guess would be that:

B'(s) \propto\frac{1}{v^{3}}

but I'm not sure that's correct. This does seem like quite a simple problem but its really got me stumped.

Thanks in advance, Ewan

 

[azatkgz]

Is the velocity constant?
Do you mean by rate of change

\frac{dB(s)}{dt}?
And I think it would be better if you post whole question.

 

[Hemmer]

Is the velocity constant?
Do you mean by rate of change

\frac{dB(s)}{dt}?
And I think it would be better if you post whole question.

Yes the velocity is constant. The rest of the question isn't relevant to my problem, but I can post if you want. And yes by rate of change I mean:

\frac{dB(s)}{dt}

 

[azatkgz]

Can you post?

 

[Hemmer]

Never mind I managed to solve it (i think!).

B \propto \frac{-3}{s^{3}} \frac{dv}{dt}

using substitution of s for v.