## Essential University Physics: Volume 1 (3rd Edition)

We know that the tension in the rope is given as $T=Kx$ We plug in the known values to obtain: $T=(1300)(0.55)=715N$ Now we can find the acceleration of the cars as $a=\frac{T}{m}$ $a=\frac{715}{1900}=0.3763\frac{m}{s^2}$ The distance covered by cars is $d=(\frac{1}{2})at^2$ We plug in the known values to obtain: $d=(\frac{1}{2})(0.3763)(60)^2=680m$