Question

A 900 kg roller coaster car is travelling westward on the track and begins to slow down from 18m/s to 2m/s.

Answer 1

Use Newton's second law to determine the acceleration applied by the stopping force:

$F=ma\implies7024\,\mathrm N=(900\,\mathrm{kg})a\implies a\approx7.8\,\frac{\rm m}{\mathrm s^2}$

Then recall that

${v_f}^2-{v_i}^2=2a\Delta x$

Take the direction in which the car is traveling to be positive, so that its acceleration points in the opposite direction and $a$ has negative sign. Then

$\left(2\,\frac{\rm m}{\rm s}\right)^2-\left(18\,\frac{\rm m}{\rm s}\right)^2=2\left(-7.8\,\frac{\rm m}{\mathrm s^2}\right)\Delta x\implies\Delta x\approx21\,\mathrm m$