The Indianapolis speedway consists of a 2.5 mile track having four turns, each 0.25 mile long and banked at 9 12'

Your question is missing below is the question

Question : What is the no-friction needed speed (in m/s ) for these turns?

20.1 m/s

Explanation:

2.5 mile track

number of turns = 4

length of each turn = 0.25 mile

banked at 9 12'

Determine the no-friction needed speed

First step : calculate the value of R

2πR / 4 =  πR / 2

note : πR / 2 = 0.25 mile

∴ R = ( 0.25 * 2 ) / π

      = 0.159 mile ≈ 256 m

Finally no-friction needed speed

tan θ = v^2 / gR

∴ v^2 = gR * tan θ

 v = √9.81 * 256 * tan(9.2°)  = 20.1 m/s

prolly a beacuse the momentuem the car has increases while going down hill

i believe the answer is a hope it and is correct