The Speed of train is reduced from 60km/h at the same time as it travels a distance of 450 m, If the reduction in speed is uniform , find how much further it will travel (approx) before coming to rest?

Question

Answer is 30m
but how?

See my solution ,

i solved it in two ways .
1:
Vi=16 m / s , s= 450 , vf = 0
450/16 = 28.125 (approx) = 30m

2:
2aS=vf^2-Vi^2

a=0.28 = 28 is same so approx= 30m

some teachers are saying this is wrong quesiton but let me know if its right

Answer 1

I believe there is an error in the problem. The problem should read: The
speed of a train is reduced from 60km/h
to 15km/h(for example) at the same time
as it travels a distance of 450 m. If
the reduction in speed is uniform, find how much further it will travel before coming to rest.

Note: The final velocity(Vf) is not zero
at 450 m, because the train has not yet come to rest.

Answer 2

Vo = 60km/h = 60,000m/3600s = 16.67 m/s.

V = 15km/h = 15,000m/3600s = 4.17 m/s.

a = (V^2-Vo^2)/2d = (4.17^2-(16.67^2))/900 = -0.289 m/s^2.

d = (Vf^2-Vo^2)/2a
d = (0-(16.67^2)/-0.579 = 480 m. To stop.

480-450 = 30 m. Further before coming to rest.

Answer 3

u=60×5÷18=50÷3m/s
v=15×5÷18=25÷6m/s

Using v^2=u^2+2as,we get

(50÷3)^2=(25÷6)^+2×a×450

a=-125÷36×12m/s^2

s1=v^2÷2a=25÷6×25×36×12÷6×2×125=30m

s1 is the further distance travelled before coming to rest.

Answer 4

But how did u get 2nd velocity??🤔