A fairground ride ends with the car moving up a ramp at a slope of 30 degrees. Given that the deceleration of the car is 4.905 m/s, and that the car enters the bottom of the ramp at 18 m/s, calculate the minimum length of the ramp for the car to stop before it reaches the end. It seems if I take different methods I get two different answers, one a factor of three out from the other. If I try to first work out the time taken, I do as follows.. V=u+at, so 0=18+(-4.905t), so t=3.669724771 S=ut+1/2at^2, so s=18x3.66..+1/2*4.905*3.66..^2 So s = 99.08256881 BUT if I use.. v^2 = u^2 + 2as I get as follows.. 0^2 = 18^2 + 2*4.905*s So s = 33.02752294 Obviously one of these methods must have a flaw somewhere, but I can't for the life of me figure out where! Any help would be greatly appreciated! Thanks!