SOLUTION: This is the problem that I have trouble solving. I have an idea of how to solve it but I am not sure where to beign to find the x cooridinates. You are riding on a roller coast

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons -> SOLUTION: This is the problem that I have trouble solving. I have an idea of how to solve it but I am not sure where to beign to find the x cooridinates. You are riding on a roller coast      Log On


   



Question 32212: This is the problem that I have trouble solving. I have an idea of how to solve it but I am not sure where to beign to find the x cooridinates.
You are riding on a roller coaster whose path can be modeled by
y=.005x2(squared)-2x+200
where x and y are measured in feet. When t=0 sec, you are at the point (0,200) going downhill.
a. when t=4 sec, the y coordinate of your locations is 32. What is the x coordinate?
b. when t=4.23sec, the y coordinate of your location is 18. What is the y coordinate?
c.Use pythagorean theorem to approximate the lenght of the track between the two points.
d. what was the average speed betweem the two points?
Thanks.

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
y=0.005x^2-2x+200
a)32=0.005x^2-2x+200
0=0.005x^2-2x+168
Graphing this I find x=120, giving the point (120,32)
b)18=0.005x^2 etc.
0 =0.005x^2-2x+182
Graphing this I find x=140, giving the point (140,18)
c) d=sqrt[(32-18)^2 + (120-140)^2]= 24.4 ft
d) The average is the distance divided by the time change:
ave.= 24.4 ft/ (0.23) sec. = 106.09 ft./sec.
Cheers,
stan H.