A train which is moving with uniform acceleration is observed to take 20 and 30 seconds to travel succesive 0.4km.how much far then will it travel before coming to rest, if the acceleration remains constant

Answer 1

V1 = 400m/20s = 20 m/s.
V2 = 400m/30s = 13.33 m/s.

a = (V2-V1)/(T2-T1) = (13.33-20)/(30-20) = -0.667 m/s^2.

V2 = V1 + 2a*d.
13.33 = 20 - 2*0.667*d. d = ?.

Answer 2

But answer is 162m

Answer 3

V1 = 400m/20s = 20 m/s.
V2 = 400m/30s = 13.33 m/s.

a = (V2-V1)/(T2-T1) = (13.33-20)/(30-20) = -0.667 m/s^2.

V2 = V1 + 2a*d.
13.33 = 20 - 2*0.667*d. d = ?.