Answer 1

Initial velocity, u = 0
Acceleration, a = 2 m/s²
Let the final velocity be v before the brakes are applied.
Now,           
t = 30 s
v = u + at
v = 0 + 2 × 30
 ⇒ v = 60 m/s

Half the maximum speed \[= \frac{60}{2} = 30 \text{ m } /s\]

When the train is accelerating with an acceleration of 2 m/s²:
Distance,

\[s = \frac{v^2 - u^2}{2a'}\]

\[= \frac{{30}^2 - 0^2}{2 \times 2}\]

⇒ s = 225 m

When the train is decelerating with an acceleration of - 1 m/s²:

Distance,

\[s = \frac{v^2 - u^2}{2a'}\]

\[= \frac{{30}^2 - {60}^2}{2\left( - 1 \right)}\]

⇒ s = 1350 m
Position from the starting point = 900 + 1350 = 2250
                                                   = 2.25 km