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station A and station B apart from rest st station A and acceleration at A constants rate of a=1.2 m.s.s until the mid point to...Asked by ibra
                station A and station B apart from rest st station A and acceleration at A constants rate of a=1.2 m.s.s 
until the mid point to station B it slows down at the same rate a and stop at b final?
a) the time of train between station ?
b)The maximum speed of the train?
            
        until the mid point to station B it slows down at the same rate a and stop at b final?
a) the time of train between station ?
b)The maximum speed of the train?
Answers
                    Answered by
            Steve
            
    geez - same problem, no fixes made.
    
                    Answered by
            bobpursley
            
    I think you need the distance between A and B. Lets call the distance x.
V(x/2)^2=2 a (x/2)
or v at half way is sqrt(2*1.2)(x^2/4)
so v average for the first half trip is half that, or
vavg=sqrt(2*1.2*x^2/16)
but this is the same average speed on the second half...
so time=x/vavg=x/sqrt(2.4x^2/16)
= x^2/4 sqrt (2.4)
max speed? we did that first.
    
V(x/2)^2=2 a (x/2)
or v at half way is sqrt(2*1.2)(x^2/4)
so v average for the first half trip is half that, or
vavg=sqrt(2*1.2*x^2/16)
but this is the same average speed on the second half...
so time=x/vavg=x/sqrt(2.4x^2/16)
= x^2/4 sqrt (2.4)
max speed? we did that first.
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