Answer:
81.22 s
Step-by-step explanation:
Period during which train accelerates t
=

=11.91 s
distance travelled during acceleration
= 1/2 a t² =.5 x 1.418 x 11.91² =100.57 m
Train decelerates during the period
=

= 8.445
Distance travelled during the period of deceleration
= v² /2 a = 16.89² / 2 x 2 =71.32 m.
Distance during which train travels uniformly
= 1200 - ( 100.57 + 71.32 ) = 1028.11 m
time taken in this part of journey
=

= 60.87 s
Total journey period = 60.87 + 8.445 + 11.91 = 81.22 s