- #1
aguycalledwil
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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!
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!
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