Tutorial Exercise A train slows down as it rounds a sharp horizontal turn, going

Question

Tutorial Exercise A train slows down as it rounds a sharp horizontal turn, going from 97.0 km/h to 53.0 km/h in the 12.5 s it takes to round the bend. The radius of the curve is 155 m. Compute the acceleration at the moment the train speed reaches 53.0 km/h. Assume the train continues to slow down at this time at the same rate Part 1 of 5 Conceptualize If the train is taking this turn at a safe speed, then its acceleration should be significantly less than g perhaps a few m/s Otherwise, it might jump the tracks! As shown in the figure, the acceleration components are directed toward the center of the curve and in the opposite direction of the train's movement, since the train is slowing down. Part 2 of 5 Categorize Because the train is changing both its speed and direction, the acceleration vector will be the v ector sum of the tangential and radial acceleration components. The tangential acceleration can be found from the changing speed and elapsed time, while the radial acceleration can be found from the radius of curvature of the path and the speed of the train.

Explanation / Answer

1km = 1000 m ; 1 hour = 60 min x 60sec = 3600

1km/hr = 1000/3600 m/s = 0.278 m/s

vi = 97 km/h (1000/1km) x (1h/3600 s) = 26.94 m/s

Hence, vi = 26.94 m/s

similarly, vf = 53 km/h (1000/1 km) (1hr/3600 s) = 14.72 m/s

hence, vf = 14.72 m/s

a = v^2/R = 14.72^2/155 = 1.4 m/s^2

a = 1.4 m/s^2 (centripital acc)

tangential acc: a = ( - 26.91 + 14.72)/12.5 = - 0.98 m/s^2

a(net) = sqrt (1.4^2 + -0.98^2) = 1.71 m/s^2

Hence, a = 1.71 m/s^2

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.