Question

what must be the acceleration of a train in order for it to stop from 12 m/s in a distance of 541 m? (Assume the train accelerates uniformly.)

Answer 1

-.133 m/s^2

Step-by-step explanation:

Acceleration will be negative because it is SLOWING to a stop.

vf = 0

vo = 12

vf = vo + at at = vf-vo = -12 < == sub in below

d = vo t + 1/2 a t^2

541 = 12 t+ 1/2 at t

541 = 12t + 1/2 (-12) t

541 = 6t ====> t = 90 1/6 s

vf = vo + at

0 = 12 + a ( 90 1/6)

a = - .133 m/s^2