# Solving problems in Kinematics with simple calculus

1. May 12, 2013

### snath_98

Hi,

I have a generic question about solving kinematic problems using simple calculus. For eg.
A variable force F = bv (v is instantaneous speed at some time t) acts on a vehicle moving with speed v. Suppose the initial speed at t = 0 of the vehicle is v0. Calculate the distance through which it moves before stopping.
By work energy theorem, (1/2) m (V0)2 = ∫ F. ds
Now how do we integrate the right side and move forward? That's my main question. Any help is appreciated

Thanks
Sanjay

2. May 12, 2013

### snath_98

b above is a constant.

3. May 12, 2013

### Philip Wood

I wouldn't start from here. I think (though I may be wrong) that the theorem in this form is not much use when the force is velocity-dependent.

Instead, assuming that the vehicle is confined to move in a straight line, you can use Newton's second law in the form F= ma. Substitute bv for F (remembering that b is negative - I'd write it as -β] and substitute dv/dt for a. This gives you a differential equation, which is easily solved to give v as a function of t. Integrate this wrt time, between zero and infinity to give you the distance gone.

This is quite interesting, as it shows that the vehicle goes a finite distance, but takes an infinite time to do it! (Well, to do every last nanometre). This is not what a real vehicle would do, because F = -βv does not model the usual rolling resistance, bearing friction etc. very well. These will not tend to zero as the vehicle's speed goes to zero.

Last edited: May 12, 2013
4. May 12, 2013

### voko

A note on terminology. Any problem involving forces is dynamical, not kinematic.