1. The problem statement, all variables and given/known data At the Olympic bob-sleigh competition, a new (I admit, fairly boring...) track has been built, as drawn in the figure. It is composed of straight lines with lengths as indicated in the figure and 90.0 degree turns, each with the same radius of curvature r1 = r2 = r3 = r4 = 40,0 m. The track slopes evenly downwards all the way down (including through the corners), a total altitude drop of 100 m over a distance of 1.50 km, so between the starting point (A) and the end point (E). You may assume that the ice is frictionless. There is gravity, g = 9:80 m/s2. a) Assuming that the sleigh starts from rest at (A), what is its speed when it reaches (B)? How long does the whole run take (from A to E)? (To keep it simple, we will ignore that bobsleighs are usually pushed to begin with by the crew. Here, it is just sliding under the effect of gravity). b) Consider the point when the sleigh enters the first corner (B). What is the required centripetal acceleration, that the track has to provide immediately after (B)? What is the tangential acceleration during the turn? What is the required centripetal acceleration immediately before reaching (C)? (c) At what angle with the horizontal does the sleigh need to slide up the side of the track in order to have the required centripetal force at (B) and (C)? d) What is the total acceleration of the sleigh as it comes out of the fourth corner (immediately before (D))? In units of g? 3. The attempt at a solution a) My solutions for a: In every meter the altitude drops down 0.06666667 m in 200 m it drops down = 13.3 m so θ= 4 v=sqrt(2gdsinθ)=16,5m/s from A to B Time taken from A to E: t=sqrt(2d/gsinθ)= 66,2 seconds b) centripetal acceleration immediately after B: ac=v^2/r= 6,8 m/s^2 tangential acceleration: a=sqrt(ar^2+at^2) 9.80m/s^2=sqrt(6.8^2+at^2) at=sqrt(6.8^2-9.80^2)=12m/s^2 centripetal acceleration before reaching C: ar=sqrt(-9.80^2+12^2)= 15,4 m/s^2 c) angle required: -mgcosθ=macsinθ tanθ=-g/ac=-55 degrees d) total acceleration a=sqrt(ar^2+at^2)=sqrt(15,4^2+12^2)=19,5 m/s^2 All answer are appreciated!