# Calculating the time taken for variable declaration

1. Apr 14, 2012

### CuriousQuazim

Hey guys ^^, basically there's this problem I've come across and I am somewhat curious of your opinions on it.

There's a bike that was previously on an incline that begins to run on a level surface. Assuming that the only resistive forces acting on the bike are due to aerodynamic drag I want to find the time taken for the bike to decelerate from speed A to a lower speed B.

Using the Standard equation for Drag Fd=1/2*Cd*ρ*U^2*A

I can clearly see that the force acting upon this bike is variable, and thus the acceleration is non linear, I've assumed the Drag coefficient is constant for the speeds used (there's not that significant a difference) and given we know the initial and final speed, along with the initial force... how would one go about finding the time taken to decelerate to a given speed? I'm thinking the integral approach but I'm not quite sure how to go about doing it...

Thanks for your help in advance guys ^^, the reason I haven't given values is because I would only like an approach rather than a worked solution, prefer to use my own noodles if I have any say in things haha.

2. Apr 14, 2012

### tiny-tim

Hi CuriousQuazim!

(try using the X2 button just above the Reply box )
Is U the speed?

Then use acceleration = dU/dt.

3. Apr 14, 2012

### AlephZero

Start from the equation of motion, force = mass x acceleration
You have the force from the engine (let's assume it is constant to keep it simple) and the drag force. Use the fact that acceleration is the rate of change of velocity. So
$$M \frac{dv}{dt} = F - Av^2$$ where A is the constant from your drag equation.
You can solve that for $v$ as a function of $t$, and then integrate again to find the distance travelled.

4. Apr 14, 2012

### CuriousQuazim

Hi Tinytim! Thanks for the heads up on the user interface ;)

Haha I'm ashamed to say that really was the obvious answer ^^, thanks I've solved it now. Sometimes the one thing you assume it can't be really is what it is...

Hey Alpha Zero
Thanks you very much for the suggestion, I never did think about if from that stand point. However let me just clarify that this is a bicycle that was running purely off gravitational force when it was on an inclined surface; it has now reached a level surface and as such has no driving force and is only experiencing the drag force.

Thank you again guys for your time and input ^^,