Asubway train moving at 75 m/s begins to accelerate at a rate of -2.5 m/s squared for 5.0 seconds. it then travels at a constant speed for 100.0 seconds. it finally slows down to a stop at -5 m/s squared. find the total distance covered and any other values needed to determine the total distance.

For the first 5.0 seconds of its motion, the train covers a distance of

After the first 5.0 seconds, the train's velocity is

For the next 100.0 seconds, the trains covers a distance of

and thus a total distance of 6594 meters.

After the first 105.0 seconds, the train's velocity at time (where corresponds to the 105.0 second mark) is given by

and its position at time (again, corresponds to the 105.0 second mark) by

The time it takes for the train to stop is such that , for which we have

Then the distance covered in the final slow-down phase of the train is

giving a total distance covered of about 6990 meters.