Calculating the Minimum Ramp Length for a Fairground Ride

[aguycalledwil]

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!

 

[ehild]

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

a is negative. Replace + by minus.

ehild

 

[aguycalledwil]

Ah of course! Thanks so much, that really helps.