Forming a DE from F=ma

Apr 2008
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
Nov 2009
If you think about your units for a second, \(\displaystyle v\) cannot possibly be a force.
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Reactions: craig
Apr 2008
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!