*Units must be converted to meters A train enters a curved horizontal section of

Question

*Units must be converted to meters A train enters a curved horizontal section of track at a speed of 100 km/h and slows down with constant deceleration to 50 km/h in 12 seconds. An accelerometer mounted inside the train records a horizontal acceleration of 2 m/s^2 when the train is 6 seconds into the curve. Calculate the radius of curvature p of the track for this instant. I know the answer is 266 m but I can't figure out how to work through it. Please be descriptive and pictures/diagrams would be very helpful! Thanks in advance

Explanation / Answer

v2 = V1 - a*t

V1 = 100km/hr = 27.77 m/s

V2 = 27.77 - (27.77 - 27.77*0.5)*0.5 = 20.83 m/s

2 = V2^2/r

2 = 20.83^2/r

r = 217.05 m

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