86 km/h = 23.9m/s
44 km/h = 12.2 m/s
tangential acceleration = -(23.9-12.2)/13
=-0.9 m/s^2
radial acceleration = v^2/R=12.2^2/140
= 1.06 m/s^2
|a| = sqrt(.9^1+1.06^2)
= 1.94 m/s^2
A train slows down as it rounds a sharp horizontal turn, going from 86.0 km/h to 44.0 km/h in the 13.0 s that it takes to round the bend. The radius of the curve is 140 m. Compute the acceleration at the moment the train speed reaches 44.0 km/h. Assume the train continues to slow down at this time at the same rate.
When i do this question i get accleration as 1.32 m/s^2. It tells me I am within 10% of the right answer but I m not sure how I'm wrong. (btw it is online homework)
Thanks.
1 answer
1 answer