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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
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
Anonymous:
any problem regarding it.
let me check it
here total time u take 10x how
as the sum of 1+8+1=10, any of the ratio will be a multiple of a number let x
ok
so total =10x
thank you
: )
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