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.)

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;

where;

v is the final velocity of the train

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

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

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-- 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 .

b.

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get the total number of pieces of fruit and divide it by 4 to get the average. the result is the answer to the problem. 14 total number of fruits and there are 4 members. 14 divided by 4 is equals to 3.5.