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What distance is required for a train to stop if its initial velocity is 23 m/s and its
deceleration is 0.25 m/s²? (Assume the train decelerates at a constant rate.)

Respuesta :

AL2006

-- The train starts at 23 m/s and slows down by 0.25 m/s every second.

So it'll take (23/0.25) = 92 seconds to stop.

-- Its average speed during that time will be (1/2)(23+0) = 11.5 m/s

-- Moving at an average speed of 11.5 m/s for 92 sec, the train will cover

(11.5 m/s) x (92 sec)  =  1,058 meters .

onyebuchinnaji

The distance required for the train to travel before stopping is 1,058 m.

The given parameters;

initial velocity of the train, u = 23 m/s

deceleration of the train, a = 0.25 m/s

The distance traveled by the train is calculated as follows;

[tex]v^2 = u^2 -2as[/tex]

where;

v is the final velocity of the train

The final velocity of the train is zero when the train stops

[tex]v^2 = u^2 -2as\\\\0 = 23^2 - 2(0.25)s\\\\0.5s = 529\\\\s = \frac{529}{0.5} \\\\s = 1,058 \ m[/tex]

Thus, the distance required for the train to travel before stopping is 1,058 m.

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