The first thing to do is to calculate the acceleration of the train. We do that by using the acceleration formula, which states
#a=(u-v)/t#, where #a# is the acceleration, #v# is the initial velocity in m/s, #u# is the final velocity in m/s and #t# is the time taken in seconds. Looking at our question, we see that #t=7.94#, #u=12.3864 and v=0# (since the train starts from rest). Plug these into the formula, and we get
#a=(12.3864-0)/7.94#
#a=12.3864/7.94#
#a=39/25=1.56ms^-2#
The last thing to do is to calculate the distance travelled. To do that we use this formula (which works with displacement as long as the displacement = distance, as it does in this question):
#s=ut+ (at^2)/2#, where #s# is the distance or displacement in metres, #u# is the initial velocity in m/s, #t# is the time in seconds and #a# is the acceleration in m/#s^2#. Looking at our equation, we see that #t=5.57388, u=0 and a=1.54#. Plug these into our equation to get
#s=0*5.57388 +(1.56*5.57388^2)/2#
#s=(1.56*5.57388^2)/2#
#s=48.46629568/2#
#s~~24.233m#, the total displacement of the train.
http://www.dummies.com/education/science/physics/how-to-calculate-time-and-distance-from-acceleration-and-velocity/
I hope I helped!
