Short Note

A train starts from rest and moves with a constant acceleration of 2.0 m/s^{2} for half a minute. The brakes are then applied and the train comes to rest in one minute. Find the position(s) of the train at half the maximum speed.

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#### Solution

Initial velocity, *u* = 0

Acceleration, *a* = 2 m/s^{2}

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/s^{2}:

Distance,

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

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

⇒

*s =*225 mWhen the train is decelerating with an acceleration of - 1 m/s^{2}:

Distance,

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

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

⇒

Position from the starting point = 900 + 1350 = 2250

= 2.25 km

*s =*1350 mPosition from the starting point = 900 + 1350 = 2250

= 2.25 km

Is there an error in this question or solution?

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