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