Kinematics question

Question

A train normally travels at a uniform speed of 72 km/h on a long stretch of straight, level track. On a particular day, the train must make a 2.0 min stop at a station along this track. If the train decelerates at a uniform rate of 1.1 m/s^2 and, after the stop, accelerates at a rate of 0.60 m/s^2 , how much time is lost because of stopping at the station?

Answer 1

Normal speed 72 km/hr

Since our deceleration and acceleration are in m/s²

We need to get our speed units into m/s:

 

72 km/hr = 72000 m/hr

(72000 m/hr)/(3600 sec/hr) = 20 m/s

 

Normal speed in m/sec is 20 m/s

 

Deceleration at 1.1 m/s²

(20 m/sec)/(1.1 m/sec²) = 18.18 sec

 

2 minutes at the stop

 

Acceleration:  (.60 m/s²)(x sec) = 20

                     x = 20/.6

                     x = 33.33 sec

                    

total time Lost = deceleration + stop time + acceleration

                     = 0 mins 18.18 sec

                     + 2  mins

                     + 0  mins 33.33 sec

                   ---------------------------

                    2 mins 51.51 seconds