Let the train remains under acceleration for time t₁ and under deceleration for time t₂ .
The final velocity under acceleration becomes the initial velocity of deceleration.
.1 t₁ = .5 t₂
t₁ = 5 t₂
Now , total distance travelled is 1000 m
1/2 a₁t₁² + 1/2 a₂ t₂² = 1000
a₁t₁² + a₂ t₂² = 2000
.1 x 25 t₂² + .5 t₂² = 2000
3 t₂² = 2000
t₂² = 666.67
t₂ = 25.82 s
t₁ = 25.82 x 5 = 129.1 s
Total time = 154.92 s