Need help w/ Acceleration of a Freight Train problem

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The freight train initially travels at a speed of 29.8 m/s and decelerates to 26.2 m/s over a period of 9.2 seconds, resulting in an acceleration of -0.391 m/s². To calculate the additional time required to bring the train to a complete stop, the formula \( t = \frac{v}{a} \) is applied, yielding approximately 67.0 seconds. The total distance traveled during the entire slowing phase is calculated using the formula \( d = v_i t + \frac{1}{2} a t^2 \), resulting in a distance of approximately 1,200 meters.

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Question: A freight train has a speed of 29.8 m/s at a given instant and 9.2 seconds later its speed is 26.2 m/s.
train's acceleration is -3.91×10-1 m/s^2

What additional time would be necessary to bring the train to a complete stop, if it continues to slow at the rate given?

What total distance is traveled in stopping during the entire slowing phase?
 
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