# Forming a DE from F=ma

#### craig

A particle of mass $$\displaystyle m$$ is projected with speed $$\displaystyle u$$ along a straight horizontal track. The first section of the track has length $$\displaystyle d$$. On this section of the track the motion is resisted by a
constant force of magnitude $$\displaystyle mk$$, where $$\displaystyle k$$ is a positive constant. The particle does not come to rest on this first section of the track.

Show that the speed $$\displaystyle V$$ of the particle at the end of the first section of the track is given by:

$$\displaystyle V = \sqrt{u^2 - 2kd}$$

Using the equation $$\displaystyle ma = F$$, I got the following equation.

If we let the speed of the particle = $$\displaystyle v$$

$$\displaystyle m\frac{dv}{dt} = v-mk$$, however in the solutions they just have $$\displaystyle m\frac{dv}{dt} = -mk$$, what's happened to the $$\displaystyle v$$, I would have thought that the forces acting on the paticle would be $$\displaystyle v$$ and $$\displaystyle -mk$$?

Thanks for the help

#### roninpro

If you think about your units for a second, $$\displaystyle v$$ cannot possibly be a force.

• craig

#### craig

Ohh yeh, actually that's rather obvious isn't it :S Think I must be used to dealing with a constant force produced by the engine.

Thanks for spotting that obvious mistake!