What must be the acceleration of a train in order for it to stop from 12m/s in a distance of 541m

x = 541 m

Dt = ?

vi = 12 m/s

vf = 0 (stop)

a = ?

Use vf^2 = vi^2 + 2a(Dx): 0^2 = (12 m/s)^2 + 2a(541 m), a = -144 m^2/s^2, a = -144 m^2/s^2/(1082 m) = -.1331 m/s/s or -.13 m/s/s

aceeleration =0.133m/s^2

Explanation:

they should be similar or the same

pe = weightxheight = 2000x1000=2,000,000joules