Question

ans.............. .

Answer 1

maximum velocity is attained when the train stops acceleration. v=0+at
60=at.let total time taken=10x
then 60=ax.......(1)
distance traveled in first x time, s1=0+1/2ax^2.
as the train stops acceleration and moves uniformly for time 8x ,distance travelled=speed reached into time taken=60*8x=480x.
distance traveled during retardation s3=vx-1/2ax^2=60*x-1/2ax^2.
now from (1) a=60/x
so s1+s2+s3=1/2(60/x)x^2+480x+60x-1/2(60/x)x^2=30x+480x+30x=540x
average speed=total distance/total time
=540x/10x=54kmph