Need help w/ Acceleration of a Freight Train problem

[junesmrithi]

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?

 

[Hootenanny]

Please show some work

 

[m2003]

I would recommend using the equation v = u + at to solve for the acceleration of the train. In this case, the initial velocity (u) is 29.8 m/s, the final velocity (v) is 26.2 m/s, and the time (t) is 9.2 seconds. Plugging these values into the equation, we get a value of -3.91×10-1 m/s^2 for the train's acceleration.

To calculate the additional time needed to bring the train to a complete stop, we can use the equation v = u + at again. This time, the final velocity (v) is 0 m/s and the initial velocity (u) is 26.2 m/s. We can solve for t and get a value of 26.2 seconds. Therefore, it would take an additional 26.2 seconds for the train to come to a complete stop if it continues to slow down at the same rate.

To find the total distance traveled during the entire slowing phase, we can use the equation s = ut + 1/2at^2. Again, the initial velocity (u) is 29.8 m/s, the final velocity (v) is 0 m/s, and the time (t) is 26.2 seconds. Plugging these values into the equation, we get a total distance of approximately 345.8 meters traveled during the entire slowing phase.